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kt4Qsk8Vq7TCfuAU | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | The two lines $$x=ay+b,z=cy+d$$ and $$x = a'y + b',z = c'y + d'$$ will be perpendicular, if and only if : | [{"identifier": "A", "content": "$$aa' + cc' + 1 = 0$$ "}, {"identifier": "B", "content": "$$aa' + bb'cc' + 1 = 0$$ "}, {"identifier": "C", "content": "$$aa' + bb'cc' = 0$$ "}, {"identifier": "D", "content": "$$\\left( {a + a'} \\right)\\left( {b + b'} \\right) + \\left( {c + c'} \\right) = 0$$ "}] | ["A"] | null | $${{x - b} \over a} = {y \over 1} = {{z - d} \over c};$$
<br/><br/>$${{x - b'} \over {a'}}$$
$$ = {y \over 1} = {{z - d'} \over c'}$$
<br><br>For perpenedicularity of lines $$aa' + 1 + cc' = 0$$ | mcq | aieee-2003 |
UTyglZ3FzG2VUgaZ | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | A line makes the same angle $$\theta $$, with each of the $$x$$ and $$z$$ axis.
<br/><br/>If the angle $$\beta \,$$, which it makes with y-axis, is such that $$\,{\sin ^2}\beta = 3{\sin ^2}\theta ,$$ then $${\cos ^2}\theta $$ equals : | [{"identifier": "A", "content": "$${2 \\over 5}$$ "}, {"identifier": "B", "content": "$${1 \\over 5}$$"}, {"identifier": "C", "content": "$${3 \\over 5}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["C"] | null | <b>Concept :</b> If a line makes the angle $$\alpha ,\beta ,\gamma $$ with x, y, z axis respectively then
$$${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$$$
<br><br>In this question given that the line makes angle θ with x and z-axis and β with y−axis.
<br><br>$$\therefore\: cos^2\theta+cos^2\beta+cos^2\theta=1$$
<br><br>$$\Rightarrow\:2cos^2\theta=1-cos^2\beta$$
<br><br>$$ \Rightarrow 2{\cos ^2}\theta = {\sin ^2}\beta $$
<br><br>But given that $$sin^2\beta=3sin^2\theta$$
<br><br>$$\therefore$$ $$2{\cos ^2}\theta = 3{\sin ^2}\theta $$
<br><br>$$ \Rightarrow 2{\cos ^2}\theta = 3\left( {1 - {{\cos }^2}\theta } \right)$$
<br><br>$$ \Rightarrow 2{\cos ^2}\theta = 3 - 3{\cos ^2}\theta $$
<br><br>$$ \Rightarrow 5{\cos ^2}\theta = 3$$
<br><br>$$ \Rightarrow {\cos ^2}\theta = {3 \over 5}$$ | mcq | aieee-2004 |
YwetrlhmGeDvUihA | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | If a line makes an angle of $$\pi /4$$ with the positive directions of each of $$x$$-axis and $$y$$-axis, then the angle that the line makes with the positive direction of the $$z$$-axis is : | [{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$ "}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$${\\pi \\over 3}$$"}] | ["B"] | null | Let the angle of line makes with the positive direction of $$z$$-axis is $$\alpha $$ direction cosines of line with the $$+ve$$ directions of $$x$$-axis, $$y$$-axis, and $$z$$-axis is $$l,$$ $$m,$$ $$n$$ respectively.
<br><br>$$\therefore$$ $$l = \cos {\pi \over 4},m = \cos {\pi \over 4},\,\,n = cos\,\alpha $$
<br><br>as we know that, $${l^2} + {m^2} + {n^2} = 1$$
<br><br>$$\therefore$$ $${\cos ^2}{\pi \over 4} + {\cos ^2}{\pi \over 4} + {\cos ^2}\alpha = 1$$
<br><br>$$ \Rightarrow {1 \over 2} + {1 \over 2} + {\cos ^2}\alpha = 1$$
<br><br>$$ \Rightarrow {\cos ^2}\alpha = 0 \Rightarrow \alpha = {\pi \over 2}$$
<br><br>Hence, angle with positive direction of the $$z$$-axis is $${\pi \over 2}$$ | mcq | aieee-2007 |
6XCk3G13gwVOzXa9 | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | Let $$L$$ be the line of intersection of the planes $$2x+3y+z=1$$ and $$x+3y+2z=2.$$ If $$L$$ makes an angle $$\alpha $$ with the positive $$x$$-axis, then cos $$\alpha $$ equals | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$ "}] | ["C"] | null | Let the direction cosines of line $$L$$ be $$l,m,n,$$
<br><br>then $$2l+3m+n=0$$ $$\,\,\,\,\,\,\,....\left( i \right)$$
<br><br>and $$l + 3m + 2n = 0\,\,\,\,\,\,\,\,\,\,....\left( {ii} \right)$$
<br><br>on solving equation $$(i)$$ and $$(ii),$$ we get
<br><br>$${l \over {6 - 3}} = {m \over {1 - 4}} = {n \over {6 - 3}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\, \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3}$$
<br><br>Now $$ \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3} = {{\sqrt {{l^2} + {m^2} + {n^2}} } \over {\sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {3^2}} }}$$
<br><br>As $${l^2} + {m^2} + {n^2} = 1$$
<br><br>$$\therefore$$ $${l \over 3} = {m \over { - 3}} = {n \over 3} = {1 \over {\sqrt {27} }}$$
<br><br>$$ \Rightarrow l = {3 \over {\sqrt {27} }} = {1 \over {\sqrt 3 }},\,\,m = - {1 \over {\sqrt 3 }},n = {1 \over {\sqrt 3 }}$$
<br><br>Line $$L,$$ makes an angle $$\alpha $$ with $$+ve$$ $$x$$-axis
<br><br>$$\therefore$$ $$l = \cos \,\alpha \,\,\,\, \Rightarrow \,\,\,\cos \alpha \,\, = {1 \over {\sqrt 3 }}$$ | mcq | aieee-2007 |
liW5CBWFat8KJOch | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | The projections of a vector on the three coordinate axis are $$6,-3,2$$ respectively. The direction cosines of the vector are : | [{"identifier": "A", "content": "$${6 \\over 5},{{ - 3} \\over 5},{2 \\over 5}$$ "}, {"identifier": "B", "content": "$${6 \\over 7 },{{ - 3} \\over 7},{2 \\over 7}$$"}, {"identifier": "C", "content": "$${- 6 \\over 7 },{{ - 3} \\over 7},{2 \\over 7}$$ "}, {"identifier": "D", "content": "$$6, -3, 2$$ "}] | ["B"] | null | Let $$P\left( {{x_1},{y_1},{z_1}} \right)$$ and $$Q\left( {{x_2},{y_2},{z_2}} \right)$$ be the initial and final points of the vector whose projections on the three coordinates axes are $${6, - 3,2}$$ then
<br><br>$${x_2} - {x_1}, = 6;\,\,{y_2} - {y_1} = - 3;\,\,{z_2} - {z_1} = 2$$
<br><br>So that directions ratios of $$\overrightarrow {PQ} $$ are $${6, - 3,2}$$
<br><br>$$\therefore$$ Direction cosines of $$\overrightarrow {PQ} $$ are
<br><br>$${6 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},{{ - 3} \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},$$
<br><br>$$\,\,\,\,\,\,\,\,$$ $${2 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }} = {6 \over 7},{{ - 3} \over 7},{2 \over 7}$$ | mcq | aieee-2009 |
aHh6mgnU6ToKRQPj | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | A line $$AB$$ in three-dimensional space makes angles $${45^ \circ }$$ and $${120^ \circ }$$ with the positive $$x$$-axis and the positive $$y$$-axis respectively. If $$AB$$ makes an acute angle $$\theta $$ with the positive $$z$$-axis, then $$\theta $$ equals : | [{"identifier": "A", "content": "$${45^ \\circ }$$"}, {"identifier": "B", "content": "$${60^ \\circ }$$ "}, {"identifier": "C", "content": "$${75^ \\circ }$$"}, {"identifier": "D", "content": "$${30^ \\circ }$$"}] | ["B"] | null | Direction cosines of the line :
<br><br>$$\ell = \cos {45^ \circ } = {1 \over {\sqrt 2 }},m = \cos {120^ \circ } = {{ - 1} \over 2},\pi = \cos \theta $$
<br><br>where $$\theta $$ is the angle, which line makes with positive $$z$$-axis.
<br><br>Now $${\ell ^2} + {m^2} + {n^2} = 1$$
<br><br>$$ \Rightarrow {1 \over 2} + {1 \over 4} + {\cos ^2}\theta = 1,\,\,{\cos ^2}\theta = {1 \over 4}$$
<br><br>$$ \Rightarrow \cos \theta = {1 \over 2}\,\,\,\,\left( \theta \right.$$ being acute)
<br><br>$$ \Rightarrow 0 = {\pi \over 3}$$ | mcq | aieee-2010 |
xarDenvP2mwduBif | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | The angle between the lines whose direction cosines satisfy the equations $$l+m+n=0$$ and $${l^2} = {m^2} + {n^2}$$ is : | [{"identifier": "A", "content": "$${\\pi \\over 6}$$"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$ "}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["C"] | null | Given
<br><br>$$l + m + n = 0$$ and $${l^2} = {m^2} + {n^2}$$
<br><br>Now, $${\left( { - m - n} \right)^2} = {m^2} + {n^2}$$
<br><br>$$ \Rightarrow mn = 0 \Rightarrow m = 0\,\,$$ or $$\,\,n = 0$$
<br><br>If $$m=0$$ then $$l=-n$$
<br><br>We know
<br><br>$${l^2} + {m^2} + {n^2} = 1 \Rightarrow n = \pm {1 \over {\sqrt 2 }}$$
<br><br>i.e.$$\left( {{l_1},{m_1},{n_1}} \right) = \left( { - {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }}} \right)$$
<br><br>If $$n=0$$ then $$l=-m$$
<br><br>$${l^2} + {m^2} + {n^2} = 1\,\,\, \Rightarrow 2{m^2} = 1$$
<br><br>$$ \Rightarrow m = \pm {1 \over {\sqrt 2 }}$$
<br><br>Let $$m = {1 \over {\sqrt 2 }} \Rightarrow l = - {1 \over {\sqrt 2 }}$$
<br><br>and $$n=0$$
<br><br>$$\left( {{l_2},{m_2},{n_2}} \right) = \left( { - {1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }},0} \right)$$
<br><br>$$\therefore$$ $$\cos \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 3}$$ | mcq | jee-main-2014-offline |
4yK9enk8xtYf7Hy4fE6wX | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | ABC is a triangle in a plane with vertices
<br/><br> A(2, 3, 5), B(−1, 3, 2) and C($$\lambda $$, 5, $$\mu $$).
<br/><br/>If the median through A is equally inclined to the coordinate axes, then the value of ($$\lambda $$<sup>3</sup> + $$\mu $$<sup>3</sup> + 5) is : </br> | [{"identifier": "A", "content": "1130"}, {"identifier": "B", "content": "1348"}, {"identifier": "C", "content": "676"}, {"identifier": "D", "content": "1077"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267089/exam_images/rmtgi7eihcbphpv9sdfi.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Mathematics - 3D Geometry Question 270 English Explanation">
<br><br>DR's of AD are
<br><br>$${{\lambda - 1} \over 2} - 2,{{5 + 3} \over 2} - 3,{{\mu + 2} \over 2} - 5$$
<br><br>i.e. $${{\lambda - 5} \over 2},\,\,1,\,\,{{\mu - 8} \over 2}$$
<br><br>As medium is making equal angles with coordinate axes,
<br><br>$$ \therefore $$ $${{\lambda - 5} \over 2} = 1 = {{\mu - 8} \over 2}$$
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 7, $$\mu $$ = 10
<br><br>$$ \therefore $$ $$\lambda $$<sup>3</sup> + $$\mu $$<sup>3</sup> + 5 = 7<sup>3</sup> + 10<sup>3</sup> + 5 = 1348 | mcq | jee-main-2016-online-10th-april-morning-slot |
jUxJqw5CxzEzFWDDqXW8u | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | An angle between the lines whose direction cosines are gien by the equations,
<br/>$$l$$ + 3m + 5n = 0 and 5$$l$$m $$-$$ 2mn + 6n$$l$$ = 0, is : | [{"identifier": "A", "content": "$${\\cos ^{ - 1}}\\left( {{1 \\over 3}} \\right)$$"}, {"identifier": "B", "content": "$${\\cos ^{ - 1}}\\left( {{1 \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{1 \\over 6}} \\right)$$"}, {"identifier": "D", "content": "$${\\cos ^{ - 1}}\\left( {{1 \\over 8}} \\right)$$"}] | ["C"] | null | Given
<br><br>l + 3m + 5n = 0
<br><br>and 5$$l$$m $$-$$ 2mn + 6n$$l$$ = 0
<br><br>From eq. (1) we have
<br><br>$$l$$ = $$-$$ 3m $$-$$ 5n
<br><br>Put the value of $$l$$ in eq. (2), we get ;
<br><br>5 ($$-$$3m $$-$$5n) m $$-$$ 2mn + 6n ($$-$$ 3m $$-$$ 5n) = 0
<br><br>$$ \Rightarrow $$ 15m<sup>2</sup> + 45mn + 30n<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ m<sup>2</sup> + 3mn + 2n<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ m<sup>2</sup> + 2mn + mn + 2n<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ (m + n) (m + 2n) = 0
<br><br>$$ \therefore $$ m = $$-$$ n or m = $$-$$ 2n
<br><br>For m = $$-n, $$ $$l$$ = $$-$$ 2n
<br><br>And for m = $$-$$ 2n, $$l$$ = n
<br><br>$$ \therefore $$ ($$l$$, m, n) = ($$-$$2n, $$-$$n, n) Or ($$l$$, m, n) = (n, $$-$$ 2n, n)
<br><br>$$ \Rightarrow $$ ($$l$$, m, n) = ($$-$$2, $$-$$1, 1) Or ($$l$$, m, n) = (1, $$-$$ 2, 1)
<br><br>Therefore, angle between the lines is given as :
<br><br>cos ($$\theta $$) = $${{\left( { - 2} \right)\left( 1 \right) + \left( { - 1} \right).\left( { - 2} \right) + \left( 1 \right)\left( 1 \right)} \over {\sqrt 6 .\sqrt 6 }}$$
<br><br>$$ \Rightarrow $$ cos ($$\theta $$) = $${1 \over 6}$$ $$ \Rightarrow $$ $$\theta $$ =cos<sup>$$-$$1</sup> $$\left( {{1 \over 6}} \right)$$ | mcq | jee-main-2018-online-15th-april-evening-slot |
3FhkSd7tirtXKNl5jLwyP | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | If a point R(4, y, z) lies on the line segment joining
the points P(2, –3, 4) and Q(8, 0, 10), then the
distance of R from the origin is : | [{"identifier": "A", "content": "$$2 \\sqrt {14}$$"}, {"identifier": "B", "content": "$$ \\sqrt {53}$$"}, {"identifier": "C", "content": "$$2 \\sqrt {21}$$"}, {"identifier": "D", "content": "6"}] | ["A"] | null | Equation of PQ is
<br><br> $${{x - 2} \over {8 - 2}} = {{y + 3} \over {0 - \left( { - 3} \right)}} = {{z - 4} \over {10 - 4}}$$
<br><br>$$ \Rightarrow $$ $${{x - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
<br><br>Point R (4, y, z) lies on this
<br><br>$$ \therefore $$ $${{4 - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
<br><br>$$ \Rightarrow $$ $${1 \over 3} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
<br><br>y = -2 and y = 6
<br><br>$$ \therefore $$ R = (4, -2, 6)
<br><br>Distance of R(4, -2, 6) from the origin O(0, 0, 0) is
<br><br>RO = $$\sqrt {{4^2} + {{\left( { - 2} \right)}^2} + {6^2}} $$
<br><br>= $$\sqrt {56} = 2\sqrt {14} $$ | mcq | jee-main-2019-online-8th-april-evening-slot |
5BKZvRMs9hDCERctl67k9k2k5iu7qri | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | The projection of the line segment joining the
points (1, –1, 3) and (2, –4, 11) on the line
joining the points (–1, 2, 3) and (3, –2, 10)
is ____________. | [] | null | 8 | Let A (1, – 1, 3), B(2, – 4, 11), C (–1, 2, 3) & D (3, –2, 10)
<br><br>$$ \therefore $$ $$\overrightarrow {AB} = \widehat i - 3\widehat j + 8\widehat k$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow {CD} = 4\widehat i - 4\widehat j + 7\widehat k$$
<br><br>Projection of $$\overrightarrow {AB} $$ on $$\overrightarrow {CD} $$ = $${{\overrightarrow {AB} .\overrightarrow {CD} } \over {\left| {\overrightarrow {CD} } \right|}}$$
<br><br> = $${{4 + 12 + 56} \over {\sqrt {16 + 16 + 49} }}$$
<br><br> = $${{72} \over 9}$$
<br><br> = 8 | integer | jee-main-2020-online-9th-january-morning-slot |
rAho8VIbIc91mUekGJ1kls51pef | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | Let $$\alpha$$ be the angle between the lines whose direction cosines satisfy the equations l + m $$-$$ n = 0 and l<sup>2</sup> + m<sup>2</sup> $$-$$ n<sup>2</sup> = 0. Then the value of sin<sup>4</sup>$$\alpha$$ + cos<sup>4</sup>$$\alpha$$ is : | [{"identifier": "A", "content": "$${{3 \\over 8}}$$"}, {"identifier": "B", "content": "$${{3 \\over 4}}$$"}, {"identifier": "C", "content": "$${{1 \\over 2}}$$"}, {"identifier": "D", "content": "$${{5 \\over 8}}$$"}] | ["D"] | null | $${l^2} + {m^2} + {n^2} = 1$$<br><br>$$ \therefore $$ $$2{n^2} = 1 $$ ($$ \because $$ l<sup>2</sup> + m<sup>2</sup> $$-$$ n<sup>2</sup> = 0)
<br><br>$$\Rightarrow n = \pm {1 \over {\sqrt 2 }}$$<br><br>$$ \therefore $$ $${l^2} + {m^2} = {1 \over 2}$$ & $$l + m = {1 \over {\sqrt 2 }}$$<br><br>$$ \Rightarrow {1 \over 2} - 2lm = {1 \over 2}$$<br><br>$$ \Rightarrow lm = 0$$ or $$m = 0$$<br><br>$$ \therefore $$ $$l = 0,m = {1 \over {\sqrt 2 }}$$ or $$l = {1 \over {\sqrt 2 }}$$<br><br>$$ < 0,{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }} > $$ or $$ < {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }} > $$<br><br>$$ \therefore $$ $$\cos \alpha = 0 + 0 + {1 \over 2} = {1 \over 2}$$<br><br>$$ \therefore $$ $${\sin ^4}\alpha + {\cos ^4}\alpha = 1 - {1 \over 2}{\sin ^2}(2\alpha ) = 1 - {1 \over 2}.{3 \over 4} = {5 \over 8}$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
1ktfvxn17 | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m $$-$$ n = 0 and mn + nl + lm = 0, is : | [{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$$\\pi - {\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right)$$"}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{8 \\over 9}} \\right)$$"}, {"identifier": "D", "content": "$${\\pi \\over 3}$$"}] | ["A"] | null | n = 2 (l + m)<br><br>lm + n(l + m) = 0<br><br>lm + 2(l + m)<sup>2</sup> = 0<br><br>2l<sup>2</sup> + 2m<sup>2</sup> + 5ml = 0<br><br>$$2{\left( {{l \over m}} \right)^2} + 2 + 5\left( {{l \over m}} \right) = 0$$<br><br>2t<sup>2</sup> + 5t + 2 = 0<br><br>(t + 2)(2t + 1) = 0<br><br>$$ \Rightarrow t = - 2; - {1 \over 2}$$<br><br>(i) $${l \over m} = - 2$$<br><br>$${n \over m} = - 2$$<br><br>($$-$$2m, m, $$-$$2m)<br><br>($$-$$2, 1, $$-$$2)<br><br>(ii) $${l \over m} = - {1 \over 2}$$<br><br>n = $$-$$2l<br><br>(l, $$-$$2l, $$-$$2l)<br><br>(1, $$-$$2, $$-$$2)<br><br>$$\cos \theta = {{ - 2 - 2 + 4} \over {\sqrt 9 \sqrt 9 }} = 0 \Rightarrow 0 = {\pi \over 2}$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1l57oj6hy | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | <p>If two straight lines whose direction cosines are given by the relations $$l + m - n = 0$$, $$3{l^2} + {m^2} + cnl = 0$$ are parallel, then the positive value of c is :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}] | ["A"] | null | <p>Given that the direction cosines satisfy $l + m - n = 0$, we find that $n = l + m$.</p>
<p>The other equation is $3l^2 + m^2 + cnl = 0$, and substituting $n = l + m$ gives $3l^2 + m^2 + cl(l + m) = 0$.</p>
<p>This simplifies to $(3 + c)l^2 + clm + m^2 = 0$.</p>
<p>As the lines are parallel, they share the same direction ratios, so we can express $l$ in terms of $m$, say $l = km$. Substituting this into our equation gives $(3 + c)(km)^2 + ckm^2 + m^2 = 0$.</p>
<p>This simplifies to $m^2[k^2(3 + c) + kc + 1] = 0$.</p>
<p>Since $m \neq 0$, we must have $k^2(3 + c) + kc + 1 = 0$. Here, we consider the ratio $k = \frac{l}{m}$ to be constant, since the lines are parallel. The equation then becomes a quadratic equation in $k$.</p>
<p>As the lines are parallel, the discriminant of the quadratic equation must be equal to zero for the equation to have equal roots. Hence, the discriminant $D = (c^2 - 4(3 + c)) = 0$.</p>
<p>Solving this quadratic equation gives $c^2 - 4c - 12 = 0$. </p>
<p>Factoring this equation gives $(c - 6)(c + 2) = 0$.</p>
<p>Solving for $c$ gives $c = 6, -2$. </p>
<p>However, we are looking for the positive value of $c$, so $c = 6$.</p>
<p>Therefore, the correct answer is 6</p>
| mcq | jee-main-2022-online-27th-june-morning-shift |
1l6m6hocr | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | <p>Let $$\mathrm{P}(-2,-1,1)$$ and $$\mathrm{Q}\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$$ be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are $$\alpha,-1, \beta$$, where both $$\alpha$$ and $$\beta$$ are integers of minimum absolute values, then $$\alpha^{2}+\beta^{2}$$ is equal to ____________.</p> | [] | null | 450 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb45xm/29b2975e-d397-4ba9-941a-4cc0bd7b15f1/c362a6a0-2e7f-11ed-8702-156c00ced081/file-1l7rb45xn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rb45xm/29b2975e-d397-4ba9-941a-4cc0bd7b15f1/c362a6a0-2e7f-11ed-8702-156c00ced081/file-1l7rb45xn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Morning Shift Mathematics - 3D Geometry Question 121 English Explanation"></p>
<p>d.r's of $$RS = < \alpha , - 1,\beta > $$</p>
<p>d.r's of $$PQ = < {{90} \over {17}},{{60} \over {17}},{{94} \over {17}} > \, = \, < 45,30,47 > $$</p>
<p>as PQ and RS are diagonals of rhombus</p>
<p>$$\alpha (45) + 30( - 1) + 47(\beta ) = 0$$</p>
<p>$$ \Rightarrow 45\alpha + 47\beta = 30$$</p>
<p>i.e., $$\alpha = {{30 - 47\beta } \over {45}}$$</p>
<p>for minimum integral value $$\alpha = - 15$$ and $$\beta = 15$$</p>
<p>$$ \Rightarrow {\alpha ^2} + {\beta ^2} = 450$$.</p> | integer | jee-main-2022-online-28th-july-morning-shift |
lsam59tf | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | Consider a $\triangle A B C$ where $A(1,3,2), B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle B A C$ meets
the line $B C$ at $D$, then the length of the projection of the vector $\overrightarrow{A D}$ on the vector $\overrightarrow{A C}$ is : | [{"identifier": "A", "content": "$\\frac{37}{2 \\sqrt{38}}$"}, {"identifier": "B", "content": "$\\sqrt{19}$"}, {"identifier": "C", "content": "$\\frac{39}{2 \\sqrt{38}}$"}, {"identifier": "D", "content": "$\\frac{\\sqrt{38}}{2}$"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqeg6dq/195e875b-1d41-47ba-b7a7-6f56da027c9a/8f22c6e0-cdc0-11ee-9f50-677e7e372eae/file-6y3zli1lsqeg6dr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqeg6dq/195e875b-1d41-47ba-b7a7-6f56da027c9a/8f22c6e0-cdc0-11ee-9f50-677e7e372eae/file-6y3zli1lsqeg6dr.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Evening Shift Mathematics - 3D Geometry Question 43 English Explanation">
<br><br>$D$ divides $B C$ in ratio $1: 1$
<br><br>$$
D:\left(\frac{1}{2}, 7, \frac{7}{2}\right)
$$
<br><br>$\begin{aligned} & \overrightarrow{A D}=\left(\frac{1}{2}-1\right) \hat{i}+(7-3) \hat{j}+\left(\frac{7}{2}-2\right) \hat{k} \\\\ & =-\frac{1}{2} \hat{i}+4 \hat{j}+\frac{3}{2} \hat{k} \\\\ & \overrightarrow{A C}=2 \hat{i}+3 \hat{j}+5 \hat{k}\end{aligned}$
<br><br>Projection of $\overrightarrow{A D}$ on $\overrightarrow{A C}$
<br><br>$$
=\frac{-1+12+\frac{15}{2}}{\sqrt{4+9+25}}=\frac{37}{2 \sqrt{38}}
$$ | mcq | jee-main-2024-online-1st-february-evening-shift |
jaoe38c1lseyroz7 | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | <p>Let $$O$$ be the origin and the position vectors of $$A$$ and $$B$$ be $$2 \hat{i}+2 \hat{j}+\hat{k}$$ and $$2 \hat{i}+4 \hat{j}+4 \hat{k}$$ respectively. If the internal bisector of $$\angle \mathrm{AOB}$$ meets the line $$\mathrm{AB}$$ at $$\mathrm{C}$$, then the length of $$O C$$ is</p> | [{"identifier": "A", "content": "$$\\frac{3}{2} \\sqrt{34}$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{3} \\sqrt{31}$$\n"}, {"identifier": "C", "content": "$$\\frac{2}{3} \\sqrt{34}$$\n"}, {"identifier": "D", "content": "$$\\frac{3}{2} \\sqrt{31}$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt2yowkb/31073c70-5b0c-4f0f-80c0-62ea5ac61070/2035eab0-d4a9-11ee-bdd1-01c80c3e2d9a/file-1lt2yowkc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt2yowkb/31073c70-5b0c-4f0f-80c0-62ea5ac61070/2035eab0-d4a9-11ee-bdd1-01c80c3e2d9a/file-1lt2yowkc.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Morning Shift Mathematics - 3D Geometry Question 28 English Explanation"></p>
<p>$$\text { length of } O C=\frac{\sqrt{136}}{3}=\frac{2 \sqrt{34}}{3}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfkjy12 | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | <p>Let $$\mathrm{P}(3,2,3), \mathrm{Q}(4,6,2)$$ and $$\mathrm{R}(7,3,2)$$ be the vertices of $$\triangle \mathrm{PQR}$$. Then, the angle $$\angle \mathrm{QPR}$$ is</p> | [{"identifier": "A", "content": "$$\\cos ^{-1}\\left(\\frac{7}{18}\\right)$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{6}$$\n"}, {"identifier": "C", "content": "$$\\cos ^{-1}\\left(\\frac{1}{18}\\right)$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{3}$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8wgd3/6b24122f-fe39-4086-be4a-889687b965f8/a5aaee70-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8wgd4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr8wgd3/6b24122f-fe39-4086-be4a-889687b965f8/a5aaee70-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8wgd4.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Evening Shift Mathematics - 3D Geometry Question 27 English Explanation"></p>
<p>Direction ratio of $$\mathrm{PR}=(4,1,-1)$$</p>
<p>Direction ratio of $$\mathrm{PQ}=(1,4,-1)$$</p>
<p>Now, $$\cos \theta=\left|\frac{4+4+1}{\sqrt{18} \cdot \sqrt{18}}\right|$$</p>
<p>$$\theta=\frac{\pi}{3}$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
lv5grw9p | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | <p>Let $$P(x, y, z)$$ be a point in the first octant, whose projection in the $$x y$$-plane is the point $$Q$$. Let $$O P=\gamma$$; the angle between $$O Q$$ and the positive $$x$$-axis be $$\theta$$; and the angle between $$O P$$ and the positive $$z$$-axis be $$\phi$$, where $$O$$ is the origin. Then the distance of $$P$$ from the $$x$$-axis is</p> | [{"identifier": "A", "content": "$$\\gamma \\sqrt{1-\\sin ^2 \\phi \\cos ^2 \\theta}$$\n"}, {"identifier": "B", "content": "$$\\gamma \\sqrt{1+\\cos ^2 \\theta \\sin ^2 \\phi}$$\n"}, {"identifier": "C", "content": "$$\\gamma \\sqrt{1+\\cos ^2 \\phi \\sin ^2 \\theta}$$\n"}, {"identifier": "D", "content": "$$\\gamma \\sqrt{1-\\sin ^2 \\theta \\cos ^2 \\phi}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \\
& \overrightarrow{O Q}=x \hat{i}+y \hat{j} \\
& |O P|=\gamma=\sqrt{x^2+y^2+z^2} \\
& \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \\
& \cos \phi=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\gamma} \\
& \text { Distance of } P \text { from } x \text {-axis }=\sqrt{y^2+z^2} \\
& d=\sqrt{\gamma^2-x^2} \\
& \Rightarrow x^2=\gamma^2 \sin ^2 \phi \cos ^2 \theta \\
& \Rightarrow d=\sqrt{\gamma^2-\gamma^2 \sin ^2 \phi \cos ^2 \theta} \\
& =\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lv7v4fxr | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | <p>If the line $$\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$$ makes a right angle with the line $$\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$$, then $$4 \lambda+9 \mu$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "13"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["D"] | null | <p>$$\begin{aligned}
& L_1: \frac{x-2}{(-3)}=\frac{y-\frac{2}{3}}{\left(\frac{4 \lambda+1}{3}\right)}=\frac{}{(-1)} \\
& L_2: \frac{x+3}{3 \mu}=\frac{y-\frac{1}{2}}{-3}=\frac{z-5}{-7} \\
& \because L_1 \perp L_2 \\
& \Rightarrow(-3)(3 \mu)+\left(\frac{4 \lambda+1}{3}\right)(-3)+(-1)(-7)=0 \\
& -9 \mu-4 \lambda-1+7=0 \\
& \Rightarrow 4 \lambda+9 \mu=6 \\
&
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
hW9Uid5rVNiUGULz | maths | 3d-geometry | lines-and-plane | A plane which passes through the point $$(3,2,0)$$ and the line
<br/><br>$${{x - 4} \over 1} = {{y - 7} \over 5} = {{z - 4} \over 4}$$ is :</br> | [{"identifier": "A", "content": "$$x-y+z=1$$"}, {"identifier": "B", "content": "$$x+y+z=5$$ "}, {"identifier": "C", "content": "$$x+2y-z=1$$ "}, {"identifier": "D", "content": "$$2x-y+z=5$$"}] | ["A"] | null | As the point $$\left( {3,2,0} \right)$$ lies on the given line
<br><br>$${{x - 4} \over 1} = {{y - 7} \over 5} = {{z - 4} \over 4}$$
<br><br>$$\therefore$$ There can be infinite many planes passing through this line. But here out of the four options only first option is satisfied by the coordinates of both the points $$\left( {3,\,2,\,0} \right)$$ and $$\left( {4,\,7,\,4} \right)$$
<br><br>$$\therefore$$ $$x - y + z = 1$$ is the required plane. | mcq | aieee-2002 |
EbGhqY1JBAk9FShb | maths | 3d-geometry | lines-and-plane | If the angel $$\theta $$ between the line $${{x + 1} \over 1} = {{y - 1} \over 2} = {{z - 2} \over 2}$$ and
<br/><br/>the plane $$2x - y + \sqrt \lambda \,\,z + 4 = 0$$ is such that $$\sin \,\,\theta = {1 \over 3}$$ then value of $$\lambda $$ is : | [{"identifier": "A", "content": "$${5 \\over 3}$$"}, {"identifier": "B", "content": "$${-3 \\over 5}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "$${-4 \\over 3}$$"}] | ["A"] | null | If $$\theta $$ is the angle between line and plane then $$\left( {{\pi \over 2} - 0} \right)$$
<br><br>is the angle between line and normal to plane given by
<br><br>$$\cos \left( {{\pi \over 2} - 0} \right) = {{\left( {\widehat i + 2\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j + \sqrt \lambda \widehat k} \right)} \over {3\sqrt {4 + 1 + \lambda } }}$$
<br><br>$$\cos \left( {{\pi \over 2} - \theta } \right) = {{2 - 2 + 2\sqrt \lambda } \over {3 \times \sqrt 5 + \lambda }}$$
<br><br>$$ \Rightarrow \sin \theta = {{2\sqrt \lambda } \over {3\sqrt 5 + \lambda }} = {1 \over 3}$$
<br><br>$$ \Rightarrow 4\lambda = 5 + \lambda \Rightarrow \lambda = {5 \over 3}$$ | mcq | aieee-2005 |
nTl94Fj4X6KBIr3k | maths | 3d-geometry | lines-and-plane | If the plane $$2ax-3ay+4az+6=0$$ passes through the midpoint of the line joining the centres of the spheres
<br/><br>$${x^2} + {y^2} + {z^2} + 6x - 8y - 2z = 13$$ and
<br/><br>$${x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8$$ then a equals :</br></br> | [{"identifier": "A", "content": "$$-1$$"}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$-2$$ "}, {"identifier": "D", "content": "$$2$$"}] | ["C"] | null | Centers of given spheres are $$\left( { - 3,4,1} \right)$$ and $$\left( {5, - 2,1} \right).$$
<br><br>Mid point of centers is $$\left( {1,1,1} \right).$$
<br><br>Satisfying this in the equation of plane, we get
<br><br>$$2a - 3a + 4a + 6 = 0$$
<br/><br/>$$ \Rightarrow a = - 2$$ | mcq | aieee-2005 |
LrYeTFfhmmgV6ZGq | maths | 3d-geometry | lines-and-plane | The distance between the line
<br/><br>$$\overrightarrow r = 2\widehat i - 2\widehat j + 3\widehat k + \lambda \left( {i - j + 4k} \right),$$ and the plane
<br/><br>$$\overrightarrow r .\left( {\widehat i + 5\widehat j + \widehat k} \right) = 5$$ is </br></br> | [{"identifier": "A", "content": "$${{10} \\over 9}$$ "}, {"identifier": "B", "content": "$${{10} \\over {3\\sqrt 3 }}$$ "}, {"identifier": "C", "content": "$${{3} \\over 10}$$"}, {"identifier": "D", "content": "$${{10} \\over 3}$$"}] | ["B"] | null | A point on lines is $$\left( {2, - 2,3} \right)$$ its perpendicular distance
<br><br>from the plane $$x + 5y + z - 5 = 0$$ is
<br><br>$$ = \left| {{{2 - 10 + 3 - 5} \over {\sqrt {1 + 25 + 1} }}} \right| = {{10} \over {3\sqrt 3 }}$$ | mcq | aieee-2005 |
JeLCqr5IqeCB7frm | maths | 3d-geometry | lines-and-plane | The image of the point $$(-1, 3,4)$$ in the plane $$x-2y=0$$ is : | [{"identifier": "A", "content": "$$\\left( { - {{17} \\over 3}, - {{19} \\over 3},4} \\right)$$ "}, {"identifier": "B", "content": "$$(15,11,4)$$ "}, {"identifier": "C", "content": "$$\\left( { - {{17} \\over 3}, - {{19} \\over 3},1} \\right)$$"}, {"identifier": "D", "content": "None of these "}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265452/exam_images/ibx2kf4tgkt5ewlrdikt.webp" loading="lazy" alt="AIEEE 2006 Mathematics - 3D Geometry Question 309 English Explanation">
<br><br>$$E{q^n}\,\,\,\,$$ of $$\,\,\,\,PN: $$-
<br><br>$${{x + 1} \over 1} = {{y - 3} \over { - 2}} = {{z - 4} \over 0} = \lambda $$
<br><br>$$N\left( {\lambda - 1, - 2\lambda + 3 - 4} \right)$$
<br><br>It lies on $$x-2y=0$$
<br><br>$$ \Rightarrow \lambda - 1 + 4\lambda - 6 = 0$$
<br><br>$$ \Rightarrow \lambda = 7/5$$
<br><br>$$N\left( {{2 \over 5},{1 \over 5},4} \right)$$
<br><br>$$N$$ is mid point of $$PP'$$
<br><br>$$\therefore$$ $$\alpha - 1 = {4 \over 5},\beta + 3 = {2 \over 5},r + 4 = 8$$
<br><br>$$ \Rightarrow \alpha = {9 \over 5},\beta = {{ - 13} \over 5},r = 4$$
<br><br>$$\therefore$$ Image is $$\left( {{9 \over 5},{{ - 13} \over 5},4} \right)$$ | mcq | aieee-2006 |
UvC5NyuvIjBty2fQ | maths | 3d-geometry | lines-and-plane | Let the line $$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$$ lie in the plane $$\,\,\,\,\,$$ $$x + 3y - \alpha z + \beta = 0.$$ Then $$\left( {\alpha ,\beta } \right)$$ equals | [{"identifier": "A", "content": "$$(-6,7)$$ "}, {"identifier": "B", "content": "$$(5,-15)$$ "}, {"identifier": "C", "content": "$$(-5,5)$$ "}, {"identifier": "D", "content": "$$(6, -17)$$ "}] | ["A"] | null | As the line $${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$$ lie in the plane
<br><br>$$x + 3y - \alpha z + \beta = 0$$
<br><br>$$\therefore$$ $$Pt\left( {2,1, - 2} \right)$$ lies on the plane
<br><br>i.e. $$2 + 3 + 2\alpha + \beta = 0$$
<br><br>or $$\,\,\,\,2\alpha + \beta + 5 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$
<br><br>Also normal to plane will be perpendicular to line,
<br><br>$$\therefore$$ $$\,\,\,3 \times 1 - 5 \times 3 + 2 \times \left( { - \alpha } \right) = 0$$
<br><br>$$ \Rightarrow \alpha = - 6$$
<br><br>From equation $$(i)$$ then, $$\beta = 7$$
<br><br>$$\therefore$$ $$\left( {\alpha ,\beta } \right) = \left( { - 6,7} \right)$$ | mcq | aieee-2009 |
eeoIaVQuoj4uNE1A | maths | 3d-geometry | lines-and-plane | If the angle between the line $$x = {{y - 1} \over 2} = {{z - 3} \over \lambda }$$ and the plane
<br/><br>$$x+2y+3z=4$$ is $${\cos ^{ - 1}}\left( {\sqrt {{5 \over {14}}} } \right),$$ then $$\lambda $$ equals :</br> | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${2 \\over 5}$$"}, {"identifier": "C", "content": "$${5 \\over 3}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["D"] | null | If $$\theta $$ be the angle between the given line and plane, then
<br><br>$$\sin \theta = {{1 \times 1 + 2 \times 2 + \lambda \times 3} \over {\sqrt {{1^2} + {2^2} + {\lambda ^2}} .\sqrt {{1^2} + {2^2} + {3^2}} }}$$
<br><br>$$ = {{5 + 3\lambda } \over {\sqrt {14} .\sqrt {5 + {\lambda ^2}} }}$$
<br><br>But it is given that
<br><br>$$\theta = {\cos ^{ - 1}}\sqrt {{5 \over {14}}} \Rightarrow \sin \theta = {3 \over {\sqrt {14} }}$$
<br><br>$$\therefore$$ $${{5 + 3\lambda } \over {\sqrt {14} \sqrt {5 + {\lambda ^2}} }} = {3 \over {\sqrt {14} }} \Rightarrow \lambda = {2 \over 3}$$ | mcq | aieee-2011 |
hFijuvylYi3GaK1R | maths | 3d-geometry | lines-and-plane | The image of the line $${{x - 1} \over 3} = {{y - 3} \over 1} = {{z - 4} \over { - 5}}\,$$ in the plane $$2x-y+z+3=0$$ is the line : | [{"identifier": "A", "content": "$${{x - 3} \\over 3} = {{y + 5} \\over 1} = {{z - 2} \\over { - 5}}$$ "}, {"identifier": "B", "content": "$${{x - 3} \\over { - 3}} = {{y + 5} \\over { - 1}} = {{z - 2} \\over 5}\\,$$ "}, {"identifier": "C", "content": "$${{x + 3} \\over 3} = {{y - 5} \\over 1} = {{z - 2} \\over { - 5}}\\,$$ "}, {"identifier": "D", "content": "$${{x + 3} \\over { - 3}} = {{y - 5} \\over { - 1}} = {{z + 2} \\over 5}$$ "}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266411/exam_images/lc7tycmcmkjqgjziprzg.webp" loading="lazy" alt="JEE Main 2014 (Offline) Mathematics - 3D Geometry Question 295 English Explanation">
<br><br>$${{a - 1} \over 2} = {{b - 3} \over { - 1}} = {{c - 4} \over 1} = \lambda \left( {let} \right)$$
<br><br>$$ \Rightarrow a = 2\lambda + 1,\,\,b = 3 - \lambda ,\,\,c = 4 + \lambda $$
<br><br>$$p = \left( {{{a + 1} \over 2},{{b + 3} \over 2},{{c + 4} \over 2}} \right)$$
<br><br>$$ = \left( {\lambda + 1,{{6 - \lambda } \over 2},{{\lambda + 8} \over 2}} \right)$$
<br><br>$$\therefore$$ $$2\left( {\lambda + 1} \right) - {{6 - \lambda } \over 2} + {{\lambda + 8} \over 2} + 3 = 0$$
<br><br>$$3\lambda + 6 = 0 \Rightarrow \lambda = - 2$$
<br><br>$$a = - 3,b = 5,c = 2$$
<br><br>Required line is $${{x + 3} \over 3} = {{y - 5} \over 1} = {{z - 2} \over { - 5}}$$ | mcq | jee-main-2014-offline |
pTa4dhxh07jnpFBx | maths | 3d-geometry | lines-and-plane | The equation of the plane containing the line $$2x-5y+z=3; x+y+4z=5,$$ and parallel to the plane, $$x+3y+6z=1,$$ is : | [{"identifier": "A", "content": "$$x+3y+6z=7$$"}, {"identifier": "B", "content": "$$2x+6y+12z=-13$$"}, {"identifier": "C", "content": "$$2x+6y+12z=13$$ "}, {"identifier": "D", "content": "$$x+3y+6z=-7$$ "}] | ["A"] | null | Equation of the plane containing the lines
<br><br>$$2x - 5y + z = 3$$ and $$x + y + 4z = 5$$ is
<br><br>$$2x - 5y + z - 3 + \lambda \left( {x + y + 4z - 5} \right) = 0$$
<br><br>$$ \Rightarrow \left( {2 + \lambda } \right)x + \left( { - 5 + \lambda } \right)y + \left( {1 + 4\lambda } \right)z + $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 - 5\lambda } \right) = 0....\left( i \right)$$
<br><br>Since the plane $$(i)$$ parallel to the given plane
<br><br>$$x + 3y + 6z = 1$$
<br><br>$$\therefore$$ $$\,\,\,{{2 + \lambda } \over 1} = {{ - 5 + \lambda } \over 3} = {{1 + 4\lambda } \over 6}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \lambda = - {{11} \over 2}$$
<br><br>Hence equation of the required plane is
<br><br>$$\left( {2 - {{11} \over 2}} \right)x + \left( { - 5 - {{11} \over 2}} \right)y + \left( {1 - {{44} \over 2}} \right)z + $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 + {{55} \over 2}} \right) = 0$$
<br><br>$$ \Rightarrow x + 3y + 6z = 7$$ | mcq | jee-main-2015-offline |
3n9s47wXddiAH8vV | maths | 3d-geometry | lines-and-plane | The distance of the point $$(1, 0, 2)$$ from the point of intersection of the line $${{x - 2} \over 3} = {{y + 1} \over 4} = {{z - 2} \over {12}}$$ and the plane $$x - y + z = 16,$$ is : | [{"identifier": "A", "content": "$$3\\sqrt {21} $$ "}, {"identifier": "B", "content": "$$13$$"}, {"identifier": "C", "content": "$$2\\sqrt {14} $$"}, {"identifier": "D", "content": "$$8$$"}] | ["B"] | null | General point on given line $$ \equiv P\left( {3r + 2,4r - 1,12r + 2} \right)$$
<br><br>Point $$P$$ must satisfy equation of plane
<br><br>$$\left( {3r + 2} \right) - \left( {4r - 1} \right) + \left( {12r + 2} \right) = 16$$
<br><br>$$11r + 5 = 16$$
<br><br>$$r=1$$
<br><br>$$P\left( {3 \times 1 + 2,4 \times 1 - 1,12 \times 1 + 2} \right) = P\left( {5,3,14} \right)$$
<br><br>Distance between $$P$$ and $$(1,0,2)$$
<br><br>$$D = \sqrt {{{\left( {5 - 1} \right)}^2} + {3^2} + {{\left( {14 - 2} \right)}^2}} = 13$$ | mcq | jee-main-2015-offline |
hxfsFTp3VBUn9E8D | maths | 3d-geometry | lines-and-plane | If the line, $${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z + 4} \over 3}\,$$ lies in the planes, $$lx+my-z=9,$$ then $${l^2} + {m^2}$$ is equal to : | [{"identifier": "A", "content": "$$5$$ "}, {"identifier": "B", "content": "$$2$$ "}, {"identifier": "C", "content": "$$26$$"}, {"identifier": "D", "content": "$$18$$"}] | ["B"] | null | Line lies in the plane $$ \Rightarrow \left( {3, - 2, - 4} \right)$$ lie in the plane
<br><br>$$ \Rightarrow 3\ell - 2m + 4 = 9$$ or $$3\ell - 2m = 5....\left( 1 \right)$$
<br><br>Also, $$\ell ,$$ $$m, - 1$$ are dr's of line perpendicular to plane
<br><br>and $$2, - 1,3$$ are dr's of line lying in the plane
<br><br>$$ \Rightarrow 2\ell - m - 3 = 0\,\,\,$$ or $$\,\,\,2\ell - m = 3....\left( 2 \right)$$
<br><br>Solving $$(1)$$ and $$(2)$$ we get $$\ell = 1$$ and $$m=-1$$
<br><br>$$ \Rightarrow {\ell ^2} + {m^2} = 2.$$ | mcq | jee-main-2016-offline |
rvY5wvQqGsfmIoNI | maths | 3d-geometry | lines-and-plane | The distance of the point $$(1,-5,9)$$ from the plane $$x-y+z=5$$ measured along the line $$x=y=z$$ is : | [{"identifier": "A", "content": "$${{10} \\over {\\sqrt 3 }}$$ "}, {"identifier": "B", "content": "$${20 \\over 3}$$"}, {"identifier": "C", "content": "$$3\\sqrt {10} $$"}, {"identifier": "D", "content": "$$10\\sqrt {3} $$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264547/exam_images/fmk5f20pf8f2dnokx8xv.webp" loading="lazy" alt="JEE Main 2016 (Offline) Mathematics - 3D Geometry Question 288 English Explanation">
<br><br>$$e{q^n}\,\,$$ of $$\,\,PO:\,\,$$ $${{x - 1} \over 1} = {{y + 5} \over 1} = {{z - 9} \over 1} = \lambda $$
<br><br>$$ \Rightarrow x = \lambda + 1;\,\,y = \lambda - 5;\,\,z = \lambda + 9$$
<br><br>Putting these in $$e{q^n}$$ of plane :-
<br><br>$$\lambda + 1 - \lambda + 5 + \lambda + 9 = 5$$
<br><br>$$ \Rightarrow \lambda = - 10$$
<br><br>$$ \Rightarrow O$$ is $$\left( { - 9, - 15, - 1} \right)$$
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ Distance $$OP = 10\sqrt 3 .$$ | mcq | jee-main-2016-offline |
QThv7lwI00EXfX0gLpjI7 | maths | 3d-geometry | lines-and-plane | The number of distinct real values of $$\lambda $$ for which the lines
<br/><br/>$${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over {{\lambda ^2}}}$$ and $${{x - 3} \over 1} = {{y - 2} \over {{\lambda ^2}}} = {{z - 1} \over 2}$$ are coplanar is :
| [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["D"] | null | As lines are coplanar, so
<br><br>$$\left| {\matrix{
{3 - 1} & {2 - 2} & {1 - \left( { - 3} \right)} \cr
1 & 2 & {{\lambda ^2}} \cr
1 & {{\lambda ^2}} & 2 \cr
} } \right| $$ = 0
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
2 & 0 & 4 \cr
1 & 2 & {{\lambda ^2}} \cr
1 & {{\lambda ^2}} & 2 \cr
} } \right| $$ = 0
<br><br>$$ \Rightarrow $$ 2(4 $$-$$ $$\lambda $$<sup>4</sup>) + 4($$\lambda $$<sup>2</sup> $$-$$ 2) = 0
<br><br>$$ \Rightarrow $$ 4 $$-$$ $$\lambda $$<sup>4</sup> + 2$$\lambda $$<sup>2</sup> $$-$$ 4 = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$<sup>2</sup>($$\lambda $$<sup>2</sup> $$-$$ 2) = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 0, $$\sqrt 2 , - \sqrt 2 $$
<br><br>$$ \therefore $$ 3 distinct real values are possible. | mcq | jee-main-2016-online-10th-april-morning-slot |
u2D3Iv4j5JXG4jCJ | maths | 3d-geometry | lines-and-plane | If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,
<br/><br/>$${x \over 1} = {y \over 4} = {z \over 5}$$ is Q, then PQ is equal to:
| [{"identifier": "A", "content": "$$2\\sqrt {42} $$"}, {"identifier": "B", "content": "$$\\sqrt {42} $$"}, {"identifier": "C", "content": "$$6\\sqrt 5 $$"}, {"identifier": "D", "content": "$$3\\sqrt 5 $$"}] | ["A"] | null | Equation of line PQ is $${{x - 1} \over 1} = {{y + 2} \over 4} = {{z - 3} \over 5}$$
<br><br>Let F be ($$\lambda $$ + 1, 4$$\lambda $$ $$-$$ 2, 5$$\lambda $$ + 3)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265539/exam_images/j1l2cyuk0ctnijsktmb6.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Offline) Mathematics - 3D Geometry Question 284 English Explanation">
<br>Since F lies on the plane
<br><br>$$ \therefore $$ 2($$\lambda $$ + 1) + 3(4$$\lambda $$ $$-$$ 2) $$-$$ 4(5$$\lambda $$ + 3) + 22 $$=$$ 0
<br><br>$$ \Rightarrow $$ $$-$$ 6$$\lambda $$ + 6 = 0 $$
\Rightarrow $$ $$\lambda $$ = 1
<br><br>$$ \therefore $$ F is (2, 2, 8)
<br><br>PQ = 2 PF = 2$$\sqrt {{1^2} + {4^2} + {5^2}} $$ = 2$$\sqrt {42} $$
| mcq | jee-main-2017-offline |
Xs9FqdMNdzzpW5zC | maths | 3d-geometry | lines-and-plane | The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, – 1), having normal
perpendicular to both the lines
<br/><br/>$${{x - 1} \over 1} = {{y + 2} \over { - 2}} = {{z - 4} \over 3}$$
<br/><br>and
<br/><br/>$${{x - 2} \over 2} = {{y + 1} \over { - 1}} = {{z + 7} \over { - 1}}$$ is :</br> | [{"identifier": "A", "content": "$${{10} \\over {\\sqrt {83} }}$$"}, {"identifier": "B", "content": "$${{5} \\over {\\sqrt {83} }}$$"}, {"identifier": "C", "content": "$${{10} \\over {\\sqrt {74} }}$$"}, {"identifier": "D", "content": "$${{20} \\over {\\sqrt {74} }}$$"}] | ["A"] | null | Let the plane be
<br><br>a(x $$-$$ 1) + b(y + 1) + c (z + 1) = 0
<br><br>Normal vector
<br><br>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 2} & 3 \cr
2 & { - 1} & { - 1} \cr
} } \right| = 5\widehat i + 7\widehat j + 3\widehat k$$
<br><br>So plane is 5(x $$-$$ 1) + 7(y + 1) + 3(z + 1) = 0
<br><br>$$ \Rightarrow $$ 5x + 7y + 3z + 5 = 0
<br><br>Distance of point (1, 3, $$-$$ 7) from the plane is
<br><br>$${{5 + 21 - 21 + 5} \over {\sqrt {25 + 49 + 9} }} = {{10} \over {\sqrt {83} }}$$ | mcq | jee-main-2017-offline |
eOHV4R4qf5Fr8Dwl8tYJH | maths | 3d-geometry | lines-and-plane | The coordinates of the foot of the perpendicular from the point (1, $$-$$2, 1) on the plane containing the lines, $${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$$ and $${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7},$$ is : | [{"identifier": "A", "content": "(2, $$-$$4, 2)"}, {"identifier": "B", "content": "($$-$$ 1, 2, $$-$$1)"}, {"identifier": "C", "content": "(0, 0, 0)"}, {"identifier": "D", "content": "(1, 1, 1)"}] | ["C"] | null | $$\overrightarrow n $$ = $$\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $$ = $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
6 & 7 & 8 \cr
3 & 5 & 7 \cr
} } \right|$$
<br><br>= (9, $$-$$ 18, 9)
<br><br>= (1, $$-$$2, 1)
<br><br>$$ \therefore $$ Equation of plane is
<br><br>1(x + 1) $$-$$ 2(y $$-$$ 1) + (z $$-$$ 3) = 0
<br><br>$$ \Rightarrow $$ x $$-$$ 2y + z = 0
<br><br>foot to z
<br><br>$${{x - 1} \over 1}$$ = $${{y + 2} \over { - 2}}$$ = $${{z - 1} \over 1}$$ = $$ - {{\left[ {1 + 4 + 1} \right]} \over 6}$$
<br><br><b>x = 0,</b> <b>y = 0,</b> <b>z = 0</b> | mcq | jee-main-2017-online-8th-april-morning-slot |
qeI2oXKiRz35ggv6LQjaF | maths | 3d-geometry | lines-and-plane | The line of intersection of the planes $$\overrightarrow r .\left( {3\widehat i - \widehat j + \widehat k} \right) = 1\,\,$$ and
<br/>$$\overrightarrow r .\left( {\widehat i + 4\widehat j - 2\widehat k} \right) = 2,$$ is : | [{"identifier": "A", "content": "$${{x - {4 \\over 7}} \\over { - 2}} = {y \\over 7} = {{z - {5 \\over 7}} \\over {13}}$$"}, {"identifier": "B", "content": "$${{x - {4 \\over 7}} \\over 2} = {y \\over { - 7}} = {{z + {5 \\over 7}} \\over {13}}$$"}, {"identifier": "C", "content": "$${{x - {6 \\over {13}}} \\over 2} = {{y - {5 \\over {13}}} \\over { - 7}} = {z \\over { - 13}}$$"}, {"identifier": "D", "content": "$${{x - {6 \\over {13}}} \\over 2} = {{y - {5 \\over {13}}} \\over 7} = {z \\over { - 13}}$$"}] | ["C"] | null | $$\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $$
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & { - 1} & 1 \cr
1 & 4 & { - 2} \cr
} } \right| = \widehat i\left( { - 2} \right) - \widehat j\left( { - 7} \right) + \widehat k\left( {13} \right)$$
<br><br>$$\overrightarrow n = - 2\widehat i + 7\widehat j + 13\widehat k$$
<br><br>Now,
<br><br> $$3x - y + z = 1$$
<br><br> $$x + 4y - 2z = 2$$
<br><br>but z $$=$$ 0 & solving the given
<br><br> x $$=$$ 6/13 & y = 5/13
<br><br>$$ \therefore $$ required equation of a line is
<br><br>$${{x - 6/13} \over 2} = {{y - 5/13} \over { - 7}} = {z \over { - 13}}$$ | mcq | jee-main-2017-online-8th-april-morning-slot |
TvlgxK9cJHYSO6FsebHnW | maths | 3d-geometry | lines-and-plane | If the line, $${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}}$$ lies in the plane, 2x−4y+3z=2, then the shortest distance between this line and the line, $${{x - 1} \over {12}} = {y \over 9} = {z \over 4}$$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "3"}] | ["C"] | null | Point (3, $$-$$ 2, $$-$$ $$\lambda $$) on p line 2x $$-$$ 4y + 3z $$-$$ 2 $$=$$ 0
<br><br>$$=$$ 6 + 8 $$-$$ 3$$\lambda $$ $$-$$ 2 = 0
<br><br>$$=$$ 3$$\lambda $$ $$=$$ 12
<br><br><b>$$\lambda $$ $$=$$ 4</b>
<br><br>Now,
<br><br>$${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + 4} \over { - 2}} = {k_1}$$ . . .(i)
<br><br>$${{x - 1} \over {12}} = {y \over 9} = {z \over 4} = {k_2}$$ . . .(ii)
<br><br>Point on equation (i) P (k<sub>1</sub> + 3, $$-$$ k<sub>1</sub> $$-$$ 2, $$-$$ 2k<sub>1</sub> $$-$$ 4)
<br><br>Point on equation (ii) Q(12k<sub>2</sub> + 1, 9k<sub>2</sub>, 4k<sub>2</sub>)
<br><br>k<sub>1</sub> + 3 $$=$$ 12k<sub>2</sub> + 1 $$\left| { - {k_1} - 2 = 9{k_2}} \right|$$ $$-$$ 2k<sub>1</sub> $$-$$ 4 $$=$$ 4k<sub>2</sub>
<br><br>k<sub>2</sub> $$=$$ 0
<br><br>k<sub>1</sub> $$=$$ $$-$$ 2
<br><br>p (1, 0, 0) lie on equation of a line 1
<br><br>gives shortest distance $$=$$ 0 | mcq | jee-main-2017-online-9th-april-morning-slot |
Hby8suwc7u2mMH1O | maths | 3d-geometry | lines-and-plane | The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane,
x + y + z = 7 is : | [{"identifier": "A", "content": "$$\\sqrt {{2 \\over 3}} $$"}, {"identifier": "B", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266605/exam_images/sruywneq0jjxnahol5cq.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - 3D Geometry Question 286 English Explanation">
<br><br>PQ is the projection of line segment AB on the plane x + y + z = 7
<br><br>P and Q are called foot of perpendicular on the plane x + y + z = 7
<br><br>Let P = (x, y, z) then
<br><br>$${{x - 5} \over 1} = {{y + 1} \over 1} = {{z - 4} \over 1} = {{ - \left( {5 - 1 + 4} \right)} \over {{1^2} + {1^2} + {1^2}}}$$
<br><br>$$ \Rightarrow \,\,\,\,x - 5 = y + 1 = z - 4 = - {8 \over 3}$$
<br><br>$$\therefore\,\,\,$$ x = $${7 \over 3}$$ , y = $$-$$ $${{11} \over 3},$$ z = $${4 \over 3}$$
<br><br>$$\therefore\,\,\,$$ Point P = $$\left( {{7 \over 3}, - {{11} \over 3},{4 \over 3}} \right)$$
<br><br>Let Q = (x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>) then
<br><br>$${{{a_1} - 4} \over 1} = {{{y_1} + 1} \over 1} = {{{z_1} - 3} \over 1} = {{ - \left( {4 - 1 + 3} \right)} \over {{1^2} + {1^2} + {1^2}}}$$
<br><br>$$ \Rightarrow \,\,\,\,$$ x<sub>1</sub> $$-$$ 4 = y<sub>1</sub> +1= z<sub>1</sub> $$-$$ 3 = $$-$$ 2
<br><br>$$\therefore\,\,\,$$ x = 2, y = $$-$$ 3, z = 1
<br><br>$$\therefore\,\,\,$$ Point Q = (2, $$-$$ 3, 1)
<br><br>Now length of PQ is
<br><br>$$\sqrt {{{\left( {{7 \over 3} - 2} \right)}^2} + {{\left( { - {{11} \over 3} + 3} \right)}^2} + {{\left( {{4 \over 3} - 1} \right)}^2}} $$
<br><br>$$ = \sqrt {{1 \over 9} + {4 \over 9} + {1 \over 9}} $$
<br><br>$$ = \sqrt {{6 \over 9}} $$
<br><br>$$ = \sqrt {{2 \over 3}} $$ | mcq | jee-main-2018-offline |
IHWRS21qK99uEo6wdYXzx | maths | 3d-geometry | lines-and-plane | An angle between the plane, x + y + z = 5 and the line of intersection of the planes, 3x + 4y + z $$-$$ 1 = 0 and 5x + 8y + 2z + 14 =0, is : | [{"identifier": "A", "content": "$${\\sin ^{ - 1}}\\left( {\\sqrt {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} } \\right)$$ "}, {"identifier": "B", "content": "$${\\cos ^{ - 1}}\\left( {\\sqrt {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} } \\right)$$"}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} \\right)$$"}, {"identifier": "D", "content": "$${\\sin ^{ - 1}}\\left( {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} \\right)$$"}] | ["A"] | null | Normal to $$3x + 4y + z = 1$$ is $$3\widehat i + 4\widehat j + \widehat k$$
<br><br>Normal to $$5x + 8y + 2z =$$ $$ - 14$$ is $$5\widehat i + 8\widehat j + 2\widehat k$$
<br><br>The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are directly proportional to the cross product of the normal vectors.
<br><br>Therefore the direction ratios of the line is $$ - \widehat j + 4\widehat k$$
<br><br>Hence the angle between the plane x + y + z + 5 = 0
<br><br>and the intersection line is $${\sin ^{ - 1}}\left( {{{ - 1 + 4} \over {\sqrt {17} \sqrt 3 }}} \right) = {\sin ^{ - 1}}\left( {\sqrt {{3 \over {17}}} } \right)$$ | mcq | jee-main-2018-online-15th-april-morning-slot |
KhXrS1fo8yYoLcyUzbXUs | maths | 3d-geometry | lines-and-plane | The equation of the line passing through (–4, 3, 1), parallel
<br/><br>to the plane x + 2y – z – 5 = 0 and intersecting
<br/><br>the line $${{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}}$$ is :</br></br> | [{"identifier": "A", "content": "$${{x + 4} \\over 3} = {{y - 3} \\over {-1}} = {{z - 1} \\over 1}$$"}, {"identifier": "B", "content": "$${{x + 4} \\over 1} = {{y - 3} \\over {1}} = {{z - 1} \\over 3}$$"}, {"identifier": "C", "content": "$${{x + 4} \\over -1} = {{y - 3} \\over {1}} = {{z - 1} \\over 1}$$"}, {"identifier": "D", "content": "$${{x - 4} \\over 2} = {{y + 3} \\over {1}} = {{z + 1} \\over 4}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264981/exam_images/ehep4l3pw7lbyi0lsu7d.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - 3D Geometry Question 268 English Explanation">
<br><br>The line L is parallel to the plane P and intersect with line 4 at point R.
<br><br>Let the coordinate of point R is, (x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>) and it passes through L<sub>2</sub>.
<br><br>$${{{x_1} + 1} \over { - 3}} = {{{y_1} - 3} \over 2} = {{{z_1} - 2} \over { - 1}} = t$$
<br><br>$$ \therefore $$ x<sub>1</sub> = $$-$$1 $$-$$ 3t, y<sub>1</sub> = 3 + 2t, z<sub>1</sub> = 2 $$-$$ t
<br><br>$$\overrightarrow {AR} = \left( {3 - 3t} \right)\widehat i + \left( {2t} \right)\widehat j + \left( {1 - t} \right)\widehat k$$
<br><br>$$\overrightarrow n = \widehat i + 2\widehat j - \widehat k$$
<br><br>As $$\overrightarrow {AR} $$ and $$\overrightarrow n $$ are perpendicular to each other, So
<br><br>$$\overrightarrow {AR} $$ $$ \cdot $$ $$\overrightarrow n $$ = 0
<br><br>$$ \Rightarrow $$ (3 $$-$$ 3t) 1 + (2t)2 + (1 $$-$$ t) ($$-$$ 1) = 0
<br><br>$$ \Rightarrow $$ 3 $$-$$ 3t + 4t $$-$$ 1 + t = 0
<br><br>$$ \Rightarrow $$ 2 + 2t = 0
<br><br>$$ \Rightarrow $$ t = $$-$$ 1
<br><br>$$ \therefore $$ point R = (2, 1, 3)
<br><br>$$ \therefore $$ DR of line L is
<br><br>= (2 $$-$$ ($$-$$ 4), $$1$$ $$-$$ 3, 3 $$-$$ 1)
<br><br>= (6, $$-$$ 2, 2)
<br><br>$$ \therefore $$ Equation of line is
<br><br>$${{x + 4} \over 6} = {{y - 3} \over { - 2}} = {{z - 1} \over 2}$$
<br><br>or $${{x + 4} \over 3} = {{y - 3} \over { - 1}} = {{z - 1} \over 1}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
xFbNYocl8Ychr6fBoL3rsa0w2w9jxayhpcy | maths | 3d-geometry | lines-and-plane | The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines
<br/>$$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$$ and $$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$$ is : | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$${\\sqrt 3 }$$"}] | ["D"] | null | Vector of the plane is <br><br>
$$\left| {\matrix{
{\hat i} & {\hat j} & {\hat k} \cr
1 & 2 & { - 1} \cr
{ - 1} & 1 & { - 2} \cr
} } \right| = - 3\hat i + 3\hat j + 3\hat k$$<br><br>
Now equation of plane is <br><br>
$$ - 3x + 3y + 3z = c$$<br><br>
(1, 1, 0) will satisfy the plane<br><br>
$$ \Rightarrow - 3 + 3 + 0 = c$$<br><br>
$$ \Rightarrow $$ c = 0<br><br>
$$ - 3x + 3y + 3z = 0$$<br><br>
distance from (2, 1, 4) is<br><br>
$$ \Rightarrow \left| {{{ - 6 + 3 + 12} \over {\sqrt {27} }}} \right| = \left| {{9 \over {3\sqrt 3 }}} \right| = \sqrt 3 \,\,units$$ | mcq | jee-main-2019-online-12th-april-evening-slot |
klTd3ShmDde4vluBEB3rsa0w2w9jx6gsul6 | maths | 3d-geometry | lines-and-plane | If the line $${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}}$$
intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane
3x + y + 4z = 16 at a point Q, then PQ is equal to : | [{"identifier": "A", "content": "$$2\\sqrt 7 $$"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "$$2\\sqrt {14} $$"}, {"identifier": "D", "content": "$$\\sqrt {14} $$"}] | ["C"] | null | $${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}} = \lambda $$<br><br>
$$A(3\lambda + 2,2\lambda - 1, - \lambda + 1)$$ line on 2x + 3y -z + 13 = 0<br><br>
$$ \Rightarrow 2(3\lambda + 2) + 3(2\lambda - 1) - ( - \lambda + 1) + 13 = 0$$<br><br>
$$ \Rightarrow 13\lambda + 13 = 0 \Rightarrow \lambda = - 1$$<br><br>
Now point P(-1, -3, 2) lie on 3x + y + 4z = 16<br><br>
$$ \Rightarrow 3(3\lambda + 2) + (3\lambda + 2) + 4(3\lambda + 2) = 16$$<br><br>
$$ \Rightarrow 9\lambda + 6 + 2\lambda - 4\lambda - 1 + 4 = 16$$<br><br>
$$ \Rightarrow 7\lambda = 7 \Rightarrow \lambda = 1$$<br><br>
$$ \Rightarrow $$ Q(5, 1, 0)<br><br>
$$ \therefore $$ PQ = $$\sqrt {36 + 16 + 4} = \sqrt {56} = 2\sqrt {14} $$ | mcq | jee-main-2019-online-12th-april-morning-slot |
kvFqafTbB4SGPDJlq03rsa0w2w9jx2h1x8t | maths | 3d-geometry | lines-and-plane | A perpendicular is drawn from a point on the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1}$$ to the plane x + y + z = 3 such that the
foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are : | [{"identifier": "A", "content": "(4, 0, \u2013 1)"}, {"identifier": "B", "content": "(2, 0, 1)"}, {"identifier": "C", "content": "(1, 0, 2)"}, {"identifier": "D", "content": "(\u2013 1, 0, 4)"}] | ["B"] | null | $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1} = \lambda $$<br><br>
Let a point P on the line is <br><br>
(2$$\lambda $$ + 1, – $$\lambda $$ –1, + $$\lambda $$)<br><br>
Foot of $${ \bot ^r}Q$$ is given by<br><br>
$${{x - 2\lambda - 1} \over 1} = {{y + \lambda + 1} \over 1} = {{z - \lambda } \over 1} = - {{\left( {2\lambda - 3} \right)} \over 3}$$<br><br>
$$ \therefore $$ Q lies on x + y + z = 3 & x – y + z = 3<br><br>
$$ \Rightarrow $$ x + z = 3 & y = 0<br><br>
$$ \therefore $$ $$y = 0 \Rightarrow \lambda + 1 = {{ - 2\lambda + 3} \over 3} \Rightarrow \lambda = 0$$<br><br>
$$ \therefore $$ Q is (2, 0, 1)
| mcq | jee-main-2019-online-10th-april-evening-slot |
CdW7hr9rKHc4IjEEzC18hoxe66ijvwp23ys | maths | 3d-geometry | lines-and-plane | If the line, $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}$$ meets the plane,
x + 2y + 3z = 15 at a point P, then the distance of P from the origin is : | [{"identifier": "A", "content": "$${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "B", "content": "2$$\\sqrt 5$$"}, {"identifier": "C", "content": "9/2"}, {"identifier": "D", "content": "7/2"}] | ["C"] | null | Let $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}$$ = $$\lambda $$
<br><br>Any arbitary point on the line is P( 2$$\lambda $$ + 1, 3$$\lambda $$ - 1, 4$$\lambda $$ + 2).
<br><br>This point also lies on the plane x + 2y + 3z = 15.
<br><br>$$ \therefore $$ (2$$\lambda $$ + 1) + 2(3$$\lambda $$ - 1) + 3(4$$\lambda $$ + 2) = 15
<br><br>$$ \Rightarrow $$ 20$$\lambda $$ = 10
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${1 \over 2}$$
<br><br>So point P is (2, $${1 \over 2}$$, 4).
<br><br>Distance from the origin(O) of point P(2, $${1 \over 2}$$, 4) is
<br><br>OP = $$\sqrt {{{\left( 2 \right)}^2} + {{\left( {{1 \over 2}} \right)}^2} + {{\left( 4 \right)}^2}} $$
<br><br>= $$\sqrt {4 + {1 \over 4} + 16} $$
<br><br>= $$\sqrt {{{81} \over 4}} $$
<br><br>= $${9 \over 2}$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
sG8EkXlzDW1GqEPMdLzTF | maths | 3d-geometry | lines-and-plane | The equation of a plane containing the line of
intersection of the planes 2x – y – 4 = 0 and
y + 2z – 4 = 0 and passing through the point
(1, 1, 0) is : | [{"identifier": "A", "content": "x \u2013 3y \u2013 2z = \u20132"}, {"identifier": "B", "content": "2x \u2013 z = 2"}, {"identifier": "C", "content": "x \u2013 y \u2013 z = 0"}, {"identifier": "D", "content": "x + 3y + z = 4"}] | ["C"] | null | The equation of any plane passing through the intersection of the planes 2x – y – 4 = 0 and
y + 2z – 4 = 0 is :
<br><br>(2x – y – 4) + $$\lambda $$(y + 2z – 4) = 0 ........(1)
<br><br>As this plane passes through (1, 1, 0) then this point satisfy the equation (1).
<br><br>$$ \therefore $$ (2 – 1 – 4) + $$\lambda $$(1 + 0 – 4) = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = -1
<br><br>Equation of required plane will be
<br><br>(2x – y – 4) – (y + 2z – 4) = 0
<br><br>$$ \Rightarrow $$ 2x – 2y – 2z = 0
<br><br>$$ \Rightarrow $$ x – y – z = 0 | mcq | jee-main-2019-online-8th-april-morning-slot |
XPWSJ5jLrtku5PYAmM6MV | maths | 3d-geometry | lines-and-plane | The vector equation of the plane through the line
of intersection of the planes x + y + z = 1 and 2x
+ 3y+ 4z = 5 which is perpendicular to the plane
x – y + z = 0 is : | [{"identifier": "A", "content": "$$\\mathop r\\limits^ \\to \\times \\left( {\\mathop i\\limits^ \\wedge - \\mathop k\\limits^ \\wedge } \\right) - 2 = 0$$"}, {"identifier": "B", "content": "$$\\mathop r\\limits^ \\to . \\left( {\\mathop i\\limits^ \\wedge + \\mathop k\\limits^ \\wedge } \\right) + 2 = 0$$"}, {"identifier": "C", "content": "$$\\mathop r\\limits^ \\to . \\left( {\\mathop i\\limits^ \\wedge - \\mathop k\\limits^ \\wedge } \\right) + 2 = 0$$"}, {"identifier": "D", "content": "$$\\mathop r\\limits^ \\to \\times \\left( {\\mathop i\\limits^ \\wedge - \\mathop k\\limits^ \\wedge } \\right) + 2 = 0$$"}] | ["C"] | null | Given,
<br/><br/>P<sub>1</sub> : x + y + z = 1
<br><br>P<sub>1</sub> : 2x
+ 3y + 4z = 5
<br><br>Equation of the plane passing through the line
of intersection of the plane P<sub>1</sub> and P<sub>2</sub> is :
<br><br>P<sub>1</sub> + $$\lambda $$P<sub>2</sub> = 0
<br><br>$$ \Rightarrow $$ (x + y + z –1) + $$\lambda $$(2x + 3y + 4z – 5) = 0
<br><br>$$ \Rightarrow $$ x(1 + 2$$\lambda $$) + y(1 + 3$$\lambda $$) + z(1 + 4$$\lambda $$) - 5$$\lambda $$ - 1 = 0 .....(1)
<br><br>Direction Ratio (D.R) of this plane = (1 + 2$$\lambda $$, 1 + 3$$\lambda $$, 1 + 4$$\lambda $$)
<br><br>Plane (1) is perpendicular to x - y + z = 0, whose D.R = (1, -1, 1)
<br><br>As they are perpendicular so dot product of D.R = 0
<br><br>$$ \therefore $$ (1) (1 + 2$$\lambda $$) + (–1) (1 + 3$$\lambda $$) + (1) (1 + 4$$\lambda $$) = 0
<br><br>$$ \Rightarrow $$ 1 + 2$$\lambda $$ –1 – 3$$\lambda $$ + 1 + 4$$\lambda $$ = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $$ - {1 \over 3}$$
<br><br>Putting the value of $$\lambda $$ in equation (1), we get
<br><br>$$ \Rightarrow $$ $${x \over 3} - {z \over 3} + {2 \over 3} = 0$$
<br><br>$$ \Rightarrow $$ x - z + 2 = 0
<br><br>Vector form of this plane,
<br><br>$$\mathop r\limits^ \to . \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) + 2 = 0$$ | mcq | jee-main-2019-online-8th-april-evening-slot |
INBHKvgfzgvl3QVAjDdYB | maths | 3d-geometry | lines-and-plane | If an angle between the line, $${{x + 1} \over 2} = {{y - 2} \over 1} = {{z - 3} \over { - 2}}$$ and the plane, $$x - 2y - kz = 3$$ is $${\cos ^{ - 1}}\left( {{{2\sqrt 2 } \over 3}} \right),$$ then a value of k is : | [{"identifier": "A", "content": "$$\\sqrt {{3 \\over 5}} $$"}, {"identifier": "B", "content": "$$ - {5 \\over 2}$$"}, {"identifier": "C", "content": "$$ - {3 \\over 2}$$"}, {"identifier": "D", "content": "$$\\sqrt {{5 \\over 3}} $$"}] | ["D"] | null | DR's of line are 2, 1, $$-$$2
<br><br>normal vector of plane is $$\widehat i$$ $$-$$ 2$$\widehat j$$ $$-$$ k$$\widehat k$$
<br><br>sin$$\alpha $$ = $${{\left( {2\widehat i + \widehat j - 2\widehat k} \right).\left( {\widehat i - 2\widehat j - k\widehat k} \right)} \over {3\sqrt {1 + 4 + {k^2}} }}$$
<br><br>sin $$\alpha $$ = $${{2k} \over {3\sqrt {{k^2} + 5} }}$$ . . . . . . (1)
<br><br>cos $$\alpha $$ =$${{2\sqrt 2 } \over 3}$$ . . . . .. . (2)
<br><br>(1)<sup>2</sup> + (2)<sup>2</sup> = 1 $$ \Rightarrow $$ k<sup>2</sup> = $${5 \over 3}$$ | mcq | jee-main-2019-online-12th-january-evening-slot |
Yf8eyIaJz6x0Vp1iYcFpM | maths | 3d-geometry | lines-and-plane | Two lines $${{x - 3} \over 1} = {{y + 1} \over 3} = {{z - 6} \over { - 1}}$$ and $${{x + 5} \over 7} = {{y - 2} \over { - 6}} = {{z - 3} \over 4}$$ intersect at the point R. The reflection of R in the xy-plane has coordinates :
| [{"identifier": "A", "content": "(2, 4, 7)"}, {"identifier": "B", "content": "(2, $$-$$ 4, $$-$$7)"}, {"identifier": "C", "content": "(2, $$-$$ 4, 7)"}, {"identifier": "D", "content": "($$-$$ 2, 4, 7)"}] | ["B"] | null | Point on L<sub>1</sub> ($$\lambda $$ + 3, 3$$\lambda $$ $$-$$ 1, $$-$$$$\lambda $$ + 6)
<br><br>Point on L<sub>2</sub> (7$$\mu $$ $$-$$ 5, $$-$$6$$\mu $$ + 2, 4$$\mu $$ + 3
<br><br>$$ \Rightarrow $$ $$\lambda $$ + 3 = 7$$\mu $$ $$-$$ 5 . . . . (i)
<br><br>3$$\lambda $$ $$-$$ 1 = $$-$$6$$\mu $$ + 2 . . . .(ii)
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $$-$$1, $$\mu $$ = 1
<br><br>point R(2, $$-$$ 4, 7)
<br><br>Reflection is (2, $$-$$4, $$-$$ 7) | mcq | jee-main-2019-online-11th-january-evening-slot |
j3rxTjeh5YBNG4T1B1C5X | maths | 3d-geometry | lines-and-plane | The plane containing the line $${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z - 1} \over 3}$$ and also containing its projection on the plane 2x + 3y $$-$$ z = 5, contains which one of the following points ? | [{"identifier": "A", "content": "($$-$$ 2, 2, 2)"}, {"identifier": "B", "content": "(2, 2, 0)"}, {"identifier": "C", "content": "(2, 0, $$-$$ 2)"}, {"identifier": "D", "content": "(0, $$-$$ 2, 2)"}] | ["C"] | null | The normal vector of required plane
<br><br>$$ = \left( {2\widehat i - \widehat j + 3\widehat k} \right) \times \left( {2\widehat i + 3\widehat j - \widehat k} \right)$$
<br><br>$$ = - 8\widehat i + 8\widehat j + 8\widehat k$$
<br><br>So, direction ratio of normal is $$\left( { - 1,1,1} \right)$$
<br><br>So required plane is
<br><br>$$ - \left( {x - 3} \right) + \left( {y + 2} \right) + \left( {z - 1} \right) = 0$$
<br><br>$$ \Rightarrow - x + y + z + 4 = 0$$
<br><br>Which is satisfied by $$\left( {2,0, - 2} \right)$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
kGwIvanvC36NRfACAH6nl | maths | 3d-geometry | lines-and-plane | On which of the following lines lies the point of intersection of the line, $${{x - 4} \over 2} = {{y - 5} \over 2} = {{z - 3} \over 1}$$ and the plane,
x + y + z = 2 ? | [{"identifier": "A", "content": "$${{x - 4} \\over 1} = {{y - 5} \\over 1} = {{z - 5} \\over { - 1}}$$"}, {"identifier": "B", "content": "$${{x - 2} \\over 2} = {{y - 3} \\over 2} = {{z + 3} \\over 3}$$"}, {"identifier": "C", "content": "$${{x - 1} \\over 1} = {{y - 3} \\over 2} = {{z + 4} \\over { - 5}}$$"}, {"identifier": "D", "content": "$${{x + 3} \\over 3} = {{4 - y} \\over 3} = {{z + 1} \\over { - 2}}$$"}] | ["C"] | null | General point on the given line is
<br><br>x = 2$$\lambda $$ + 4
<br><br>y = 2$$\lambda $$ + 5
<br><br>z = $$\lambda $$ + 3
<br><br>Solving with plane,
<br><br>2$$\lambda $$ + 4 + 2$$\lambda $$ + 5 + $$\lambda $$ + 3 = 2
<br><br>5$$\lambda $$ + 12 = 2
<br><br>5$$\lambda $$ = $$-$$ 10
<br><br>$$\lambda $$ = $$-$$ 2 | mcq | jee-main-2019-online-10th-january-evening-slot |
8CSQaPFHmG0GHfXGEg7dU | maths | 3d-geometry | lines-and-plane | The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis
also passes through the point : | [{"identifier": "A", "content": "(\u20133, 0, -1) "}, {"identifier": "B", "content": "(3, 2, 1) "}, {"identifier": "C", "content": "(3, 3, -1)"}, {"identifier": "D", "content": "(\u20133, 1, 1)"}] | ["B"] | null | The equation of plane
<br><br>(x + y + z $$-$$ 1) + $$\lambda $$ (2x + 3y $$-$$ z + 4) = 0
<br><br>$$ \Rightarrow $$ (1 + 2$$\lambda $$)x + (1 + 3$$\lambda $$)y + (1 $$-$$ $$\lambda $$)z + 4$$\lambda $$ $$-$$ 1 = 0
<br><br>As plane is parallel to y axis so the normal vector of plane and dot product of $$\widehat j$$ is zero.
<br><br>$$ \therefore $$ 1 + 3$$\lambda $$ = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $$-$$ $${1 \over 3}$$
<br><br>$$ \therefore $$ So the equation of the plane is
<br><br>x(1 $$-$$ $${2 \over 3}$$) + (1 $$-$$ $${3 \over 3}$$) y + (1 + $${1 \over 3}$$) $$-$$ $${4 \over 3}$$ $$-$$ 1 = 0
<br><br>$$ \Rightarrow $$ x ($${1 \over 3}$$) + z($${4 \over 3}$$) $$-$$ $${7 \over 3}$$ = 0
<br><br>$$ \Rightarrow $$ x + 4z $$-$$ 7 = 0
<br><br>By checking each options you can see only point (3, 2, 1) lies on the plane. | mcq | jee-main-2019-online-9th-january-morning-slot |
X9jCkBVS5S2zXHqiFo7k9k2k5ki7wil | maths | 3d-geometry | lines-and-plane | If the distance between the plane,
23x – 10y – 2z + 48 = 0 and the plane<br/><br/>
containing the lines
$${{x + 1} \over 2} = {{y - 3} \over 4} = {{z + 1} \over 3}$$<br/><br/> and
$${{x + 3} \over 2} = {{y + 2} \over 6} = {{z - 1} \over \lambda }\left( {\lambda \in R} \right)$$<br/><br/> is equal to
$${k \over {\sqrt {633} }}$$, then k is equal to ______. | [] | null | 3 | Required distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)
<br><br>$$ \Rightarrow $$ $$\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$$ = $${k \over {\sqrt {633} }}$$
<br><br>$$ \Rightarrow $$ $$\left| {{3 \over {\sqrt {633} }}} \right|$$ = $${k \over {\sqrt {633} }}$$
<br><br>$$ \Rightarrow $$ k = 3 | integer | jee-main-2020-online-9th-january-evening-slot |
z3jkKUxU8nDZA2wXVQjgy2xukg3b9f5y | maths | 3d-geometry | lines-and-plane | A plane P meets the coordinate axes at A, B
and C respectively. The centroid of $$\Delta $$ABC is
given to be (1, 1, 2). Then the equation of the
line through this centroid and perpendicular to
the plane P is : | [{"identifier": "A", "content": "$${{x - 1} \\over 1} = {{y - 1} \\over 1} = {{z - 2} \\over 2}$$"}, {"identifier": "B", "content": "$${{x - 1} \\over 2} = {{y - 1} \\over 1} = {{z - 2} \\over 1}$$"}, {"identifier": "C", "content": "$${{x - 1} \\over 2} = {{y - 1} \\over 2} = {{z - 2} \\over 1}$$"}, {"identifier": "D", "content": "$${{x - 1} \\over 1} = {{y - 1} \\over 2} = {{z - 2} \\over 2}$$"}] | ["C"] | null | Let, Equation of plane is
<br><br>$${x \over a} + {y \over b} + {z \over c}$$ = 1
<br><br>A = ($$a$$, 0, 0) B
= (0, b, 0), C
= (0, 0, c)
<br><br>$$ \therefore $$ Centroid = $$\left( {{a \over 3},{b \over 3},{c \over 3}} \right)$$ = (1, 1, 2)
<br><br>$$ \Rightarrow $$ $$a$$ = 3, b = 3, c = 6
<br><br>$$ \therefore $$ Plane : $${x \over 3} + {y \over 3} + {z \over 6}$$ = 1
<br><br>$$ \Rightarrow $$ 2x + 2y + z = 6
<br><br>The equation of the
line through this centroid (1, 1, 2) and <br>perpendicular to
the plane 2x + 2y + z = 6 is :
<br><br>$${{x - 1} \over 2} = {{y - 1} \over 2} = {{z - 2} \over 1}$$ | mcq | jee-main-2020-online-6th-september-evening-slot |
Qojv10KpxgJcV2JxWHjgy2xukfuuxqrg | maths | 3d-geometry | lines-and-plane | The shortest distance between the lines
<br/><br>$${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$$ <br/><br>and x + y + z + 1 = 0, 2x – y + z
+ 3 = 0 is :</br></br> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 3 }}$$"}] | ["D"] | null | Plane through line of intersection is
<br><br>x + y + z + 1 + $$\lambda $$ (2x –y + z + 3) = 0
<br><br>It should be parallel to given line $${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$$
<br><br>$$ \therefore $$ 0(1 + 2$$\lambda $$) - 1(1 - $$\lambda $$) + 1(1 + $$\lambda $$) = 0 $$ \Rightarrow $$ $$\lambda $$ = 0
<br><br>$$ \therefore $$ Required Plane : x + y + z + 1 = 0
<br><br>Shortest distance of (1, –1, 0) from this plane
<br><br>= $${{\left| {1 - 1 + 0 + 1} \right|} \over {\sqrt {{1^2} + {1^2} + {1^2}} }}$$ = $${1 \over {\sqrt 3 }}$$ | mcq | jee-main-2020-online-6th-september-morning-slot |
u8By5wccby20xeA0tgjgy2xukfqch09n | maths | 3d-geometry | lines-and-plane | If for some $$\alpha $$ $$ \in $$ R, the lines
<br/><br/>L<sub>1</sub> : $${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$$ and
<br/><br>L<sub>2</sub> : $${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$$ are coplanar,
<br/><br/>then the line L<sub>2</sub>
passes through the point :</br> | [{"identifier": "A", "content": "(10, 2, 2)"}, {"identifier": "B", "content": "(2, \u201310, \u20132)"}, {"identifier": "C", "content": "(10, \u20132, \u20132)"}, {"identifier": "D", "content": "(\u20132, 10, 2)"}] | ["B"] | null | L<sub>1</sub> : $${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$$ and
<br><br>L<sub>2</sub> : $${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$$ are coplanar.
<br><br>$$ \therefore $$ $$\left| {\matrix{
1 & 3 & 2 \cr
2 & { - 1} & 1 \cr
\alpha & {5 - \alpha } & 1 \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ –1(–1 + $$\alpha $$ - 5) + 3(2 - $$\alpha $$) - 2(10 - 2$$\alpha $$ + $$\alpha $$) = 0
<br><br>$$ \Rightarrow $$ 6 - $$\alpha $$ + 6 - 3$$\alpha $$ + 2$$\alpha $$ - 20 = 0
<br><br>$$ \Rightarrow $$ –8 –2$$\alpha $$ = 0
<br><br>$$ \Rightarrow $$ $$\alpha $$ = -4
<br><br>$$ \therefore $$ Equation of L<sub>2</sub> : $${{x + 2} \over { - 4}} = {{y + 1} \over 9} = {{z + 1} \over 1}$$
<br><br>Check options (2, –10, –2) lies on L<sub>2</sub>. | mcq | jee-main-2020-online-5th-september-evening-slot |
b7yr7A2nAMnX9Hf8Fojgy2xukfagymx7 | maths | 3d-geometry | lines-and-plane | The distance of the point (1, –2, 3) from<br/><br> the plane x – y + z = 5 measured parallel to <br/><br>the line $${x \over 2} = {y \over 3} = {z \over { - 6}}$$ is :</br></br> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${1 \\over 7}$$"}, {"identifier": "D", "content": "$${7 \\over 5}$$"}] | ["B"] | null | Equation of line parallel to $${x \over 2} = {y \over 3} = {z \over { - 6}}$$ passes through $$(1, - 2,3)$$ is<br><br>$${{x - 1} \over 2} = {{y + 2} \over 3} = {{z - 3} \over { - 6}} = r$$<br><br>$$x = 2r + 1$$<br><br>$$y = 3r - 2$$, <br><br>$$z = - 6r + 3$$
<br><br>A point on whole line = (2r + 1, 3r – 2, – 6r + 3).
<br><br>This point lies on plane x – y + 2 = 5
<br><br>so, $$2r + 1 - 3r + 2 - 6r + 3 = 5$$<br><br>$$ \Rightarrow $$ $$r = {1 \over 7}$$<br><br>$$ \therefore $$ $$x = {9 \over 7}$$, $$y = {{ - 11} \over 7}$$, $$z = {{15} \over 7}$$<br><br>Distance is = $$\sqrt {{{\left( {{9 \over 7} - 1} \right)}^2} + {{\left( {2 - {{11} \over 7}} \right)}^2} + {{\left( {3 - {{15} \over 7}} \right)}^2}} $$<br><br>$$ = \sqrt {{{\left( {{2 \over 7}} \right)}^2} + {{\left( {{3 \over 7}} \right)}^2} + {{\left( {{6 \over 7}} \right)}^2}} $$<br><br>$$ = {1 \over 7}\sqrt {4 + 9 + 36} $$ = 1 | mcq | jee-main-2020-online-4th-september-evening-slot |
ZyHrTWGrvXewytqQX0jgy2xukezm71el | maths | 3d-geometry | lines-and-plane | The foot of the perpendicular drawn from the
point (4, 2, 3) to the line joining the points
(1, –2, 3) and (1, 1, 0) lies on the plane : | [{"identifier": "A", "content": "x \u2013 2y + z = 1"}, {"identifier": "B", "content": "x + 2y \u2013 z = 1"}, {"identifier": "C", "content": "x \u2013 y \u2013 2y = 1"}, {"identifier": "D", "content": "2x + y \u2013 z = 1"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267567/exam_images/staijxnqv8p4honzsnr6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Mathematics - 3D Geometry Question 228 English Explanation">
<br><br>Equation of AB,
<br><br>$${{x - 1} \over 0} = {{y + 2} \over 3} = {{z - 3} \over { - 3}} = \lambda $$
<br><br>$$ \therefore $$ Coordinates of any point on the line (M) = $$\left( { - 3,3\lambda - 2, - 3\lambda } \right)$$
<br><br>$$\overrightarrow {PM} = - 3\widehat i + \left( {3\lambda - 4} \right)\widehat j - 3\lambda \widehat k$$
<br><br>$$\overrightarrow {AB} = 3\widehat j - 3\widehat k$$
<br><br>As $$\overrightarrow {PM} \bot \overrightarrow {AB} $$
<br><br>$$ \therefore $$ $$\overrightarrow {PM} .\overrightarrow {AB} = 0$$
<br><br>$$ \Rightarrow $$ $$\left( { - 3} \right).0 + \left( {3\lambda - 4} \right)\left( 3 \right) + \left( { - 3\lambda } \right)\left( { - 3} \right)$$ = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${2 \over 3}$$
<br><br>$$ \therefore $$ M = (1, 0, 1)
<br><br>By checking each options we can see M lies on 2x + y – z = 1. | mcq | jee-main-2020-online-3rd-september-morning-slot |
Qarw3NWSaAza5ekKgYjgy2xukezbvldh | maths | 3d-geometry | lines-and-plane | A plane passing through the point (3, 1, 1)
contains two lines whose direction ratios are 1,
–2, 2 and 2, 3, –1 respectively. If this plane also
passes through the point ($$\alpha $$, –3, 5), then
$$\alpha $$ is
equal to: | [{"identifier": "A", "content": "-10"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "-5"}] | ["C"] | null | As normal is perpendicular to both the lines so normal vector to the plane is<br><br>
$$\overrightarrow n = \left( {\widehat i - 2\widehat j + 2\widehat k} \right) \times \left( {2\widehat i + 3\widehat j - \widehat k} \right)$$<br><br>
$$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 2} & 2 \cr
2 & 3 & { - 1} \cr
} } \right|$$<br><br>
$$\overrightarrow n = \left( {2 - 6} \right)\widehat i - \left( { - 1 - 4} \right)\widehat j + \left( {3 + 4} \right)\widehat k$$<br><br>
$$\overrightarrow n = - 4\widehat i + 5\widehat j + 7\widehat k$$<br><br>
Now equation of plane passing through (3,1,1) is<br><br>
$$ \Rightarrow $$ –4(x – 3) + 5(y – 1) + 7(z – 1) = 0<br><br>
$$ \Rightarrow $$ –4x + 12 + 5y – 5 + 7z – 7 = 0<br><br>
$$ \Rightarrow $$ –4x + 5y + 7z = 0 ...(1)<br><br>
Plane is also passing through ($$\alpha $$, –3, 5) so this point satisfies the equation of plane so put in equation (1)<br><br>
–4$$\alpha $$ + 5 × (–3) + 7 × (5) = 0<br><br>
$$ \Rightarrow $$ –4$$\alpha $$ – 15 + 35 = 0<br><br>
$$ \Rightarrow\alpha $$ = 5
| mcq | jee-main-2020-online-2nd-september-evening-slot |
mKqODJ2TViU4wWfYdSjgy2xukewrehxd | maths | 3d-geometry | lines-and-plane | The plane passing through the points (1, 2, 1),
<br/>(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
<br/>also passes through the point : | [{"identifier": "A", "content": "(0, 6, \u20132)"}, {"identifier": "B", "content": "(\u20132, 0, 1)"}, {"identifier": "C", "content": "(0, \u20136, 2)"}, {"identifier": "D", "content": "(2, 0 \u20131)"}] | ["B"] | null | Equation of plane passing through (2, 1, 2)<br><br>a(x $$-$$ 2) + b(y $$-$$ 1) + c(z $$-$$ 2) = 0 ......(1)<br><br>As point (1, 2, 1) also passes through the plane, so it satisfy the equation, <br><br>a(1 $$-$$ 2) + b(2 $$-$$ 1) + c(1 $$-$$ 2) = 0<br><br>$$ \Rightarrow $$ $$-$$a + b $$-$$ c = 0 ....(2)<br><br>Given line 2x = 3y and z = 1,<br><br>So, symmetric form of the line<br><br>$${x \over 3} = {y \over 2} = {{z - 1} \over 0}$$<br><br>$$ \therefore $$ Direction ratio of this line is (3, 2, 0) and Direction ration of plane = (a, b, c)<br><br>As plane is parallel to the line so the normal of the plane is perpendicular to the line.<br><br>$$ \therefore $$ Dot product of direction ratio = 0<br><br>3a + 2b + 0(c) = 0 .....(3)<br><br>Equation of plane, <br><br>$$\left| {\matrix{
{x - 2} & {y - 1} & {z - 2} \cr
{ - 1} & 1 & { - 1} \cr
3 & 2 & 0 \cr
} } \right| = 0$$<br><br>$$ \Rightarrow 3(1 - y + 2 - z) - 2( - x + 2 + z - 2) = 0$$<br><br>$$ \Rightarrow 9 - 3y - 3z + 2x - 2z = 0$$<br><br>$$ \Rightarrow 2x - 3y - 5z + 9 = 0$$<br><br>By checking all options you can see ($$-$$2, 0, 1) satisfy the equation. | mcq | jee-main-2020-online-2nd-september-morning-slot |
1t1lV3VVbYNVWnJEu8jgy2xukf49l5v7 | maths | 3d-geometry | lines-and-plane | Let a plane P contain two lines
<br/>$$\overrightarrow r = \widehat i + \lambda \left( {\widehat i + \widehat j} \right)$$, $$\lambda \in R$$ and
<br/>$$\overrightarrow r = - \widehat j + \mu \left( {\widehat j - \widehat k} \right)$$, $$\mu \in R$$
<br/>If Q($$\alpha $$, $$\beta $$, $$\gamma $$) is the foot of the perpendicular
drawn from the point M(1, 0, 1) to P, then
3($$\alpha $$ + $$\beta $$ + $$\gamma $$) equals _______. | [] | null | 5 | Given lines,<br><br>$$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$$ parallel to $$(\widehat i + \widehat j)$$<br><br>Let, $$\overrightarrow {{n_1}} = (\widehat i + \widehat j)$$<br><br>and $$\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$$ parallel to $$(\widehat j - \widehat k)$$<br><br>Let, $$\overrightarrow {{n_2}} = (\widehat j - \widehat k)$$<br><br>$$ \therefore $$ Normal of plane, $$\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $$<br><br>$$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & 1 & 0 \cr
0 & 1 & { - 1} \cr
} } \right|$$<br><br>$$ = - \widehat i + \widehat j + \widehat k$$<br><br>Line $$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.<br><br>$$ \therefore $$ Equation of the plane, <br><br>$$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$$<br><br>$$ \Rightarrow x - y - z - 1 = 0$$<br><br>Foot of perpendicular from (x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>) on the plane,<br><br>$${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$$<br><br>Here foot of perpendicular is drawn from M(1, 0, 1),<br><br>$$ \therefore $$ $${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$$<br><br>$$ \therefore $$ $$x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$$<br><br>$${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$$<br><br>$${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$$<br><br>According to the question,<br><br>$$x = \alpha $$, $$y = \beta $$, $$z = \gamma $$<br><b $$$$\beta="-" {1="" \over="" 3}$$<br=""><br>$$ \therefore $$ $$\alpha = {4 \over 3}$$, $$\beta = - {1 \over 3}$$, $$\gamma = {2 \over 3}$$<br><br>$$ \therefore $$ $$3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$$</b> | integer | jee-main-2020-online-3rd-september-evening-slot |
a2z3tAA7ldpoqgXxjx1klrep90f | maths | 3d-geometry | lines-and-plane | The distance of the point (1, 1, 9) from the point of intersection of the line
$${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}$$
and the plane x + y + z = 17 is : | [{"identifier": "A", "content": "$$19\\sqrt 2 $$"}, {"identifier": "B", "content": "$$2\\sqrt {19} $$"}, {"identifier": "C", "content": "38"}, {"identifier": "D", "content": "$$\\sqrt {38} $$"}] | ["D"] | null | Given, P(1, 1, 9).<br/><br/>Equation of plane x + y + z = 17<br/><br/>Equation of line $$\Rightarrow$$ $${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}$$<br/><br/>$$\Rightarrow$$ $${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2} = \lambda $$ (let)<br/><br/>$$\Rightarrow$$ x = $$\lambda$$ + 3; y = 2$$\lambda$$ + 4; z = 2$$\lambda$$ + 5<br/><br/>$$\therefore$$ The point we have is ($$\lambda$$ + 3, 2$$\lambda$$ + 4, 2$$\lambda$$ + 5).<br/><br/>$$\because$$ This point lies on the plane x + y + z = 17.<br/><br/>$$\therefore$$ $$\lambda$$ + 3 + 2$$\lambda$$ + 4 + 2$$\lambda$$ + 5 = 17<br/><br/>$$\Rightarrow$$ $$\lambda$$ = 1<br/><br/>$$\therefore$$ The coordinate of point is (4, 6, 7)<br/><br/>$$\therefore$$ Required distance between (1, 1, 9) and (4, 6, 7) is<br/><br/>$$ = \sqrt {{{(4 - 1)}^2} + {{(6 - 1)}^2} + {{(7 - 9)}^2}} $$<br/><br/>$$ = \sqrt {9 + 25 + 4} = \sqrt {38} $$ | mcq | jee-main-2021-online-24th-february-morning-slot |
wENOHw7UshvkPbCF2x1klrlu7yu | maths | 3d-geometry | lines-and-plane | The vector equation of the plane passing through the intersection<br/><br/> of the planes $$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$$ and $$\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$$, and the point (1, 0, 2) is : | [{"identifier": "A", "content": "$$\\overrightarrow r .\\left( {\\widehat i + 7\\widehat j + 3\\widehat k} \\right) = {7 \\over 3}$$"}, {"identifier": "B", "content": "$$\\overrightarrow r .\\left( {\\widehat i + 7\\widehat j + 3\\widehat k} \\right) = 7$$"}, {"identifier": "C", "content": "$$\\overrightarrow r .\\left( {3\\widehat i + 7\\widehat j + 3\\widehat k} \\right) = 7$$"}, {"identifier": "D", "content": "$$\\overrightarrow r .\\left( {\\widehat i - 7\\widehat j + 3\\widehat k} \\right) = {7 \\over 3}$$"}] | ["B"] | null | Given, point (1, 0, 2)<br/><br/>Equation of plane = <br/><br/>$$\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) = 1$$ and $$\overrightarrow r\,.\,(\widehat i - 2\widehat j) = - 2$$<br/><br/>Equation of plane passing through the intersection of given planes is<br/><br/>$$[\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) - 1] + \lambda [\overrightarrow r\,.\,(\widehat i - 2\widehat j) + 2] = 0$$<br/><br/>$$\because$$ This plane passes through point (1, 0, 2) i.e., <br/><br/>vector $$(\widehat i + 2\widehat k)$$<br/><br/>$$\therefore$$ $$[(\widehat i + 2\widehat k)\,.\,(\widehat i + \widehat j + \widehat k) - 1] + \lambda [(\widehat i + 2\widehat k)\,.\,(\widehat i - 2\widehat j) + 2] = 0$$<br/><br/>$$ \Rightarrow (3 - 1) + \lambda (1 + 2) = 0$$<br/><br/>$$ \Rightarrow 2 + \lambda \times 3 = 0$$<br/><br/>$$ \Rightarrow \lambda = - 2/3$$<br/><br/>Hence, equation of required plane is<br/><br/>$$[\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) - 1] + \left( {{{ - 2} \over 3}} \right)[\overrightarrow r\,.\,(\widehat i - 2\widehat j) + 2] = 0$$<br/><br/>$$ \Rightarrow $$ $$3[\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) - 1] - 2[\overrightarrow r\,.\,(\widehat i - 2\widehat j) + 2] = 0$$<br/><br/>$$ \Rightarrow $$ $$\overrightarrow r\,.\,(\widehat i + 7\widehat j + 3\widehat k) = 7$$ | mcq | jee-main-2021-online-24th-february-evening-slot |
oAhZZg0ACHs4c6hAhW1kluhipw0 | maths | 3d-geometry | lines-and-plane | Let ($$\lambda$$, 2, 1) be a point on the plane which passes through the point (4, $$-$$2, 2). If the plane is perpendicular to the line joining the points ($$-$$2, $$-$$21, 29) and ($$-$$1, $$-$$16, 23), then $${\left( {{\lambda \over {11}}} \right)^2} - {{4\lambda } \over {11}} - 4$$ is equal to __________. | [] | null | 8 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265776/exam_images/vghtupk7erbaxkhau8et.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Morning Shift Mathematics - 3D Geometry Question 209 English Explanation"><br><br>$$\overrightarrow {AB} \bot \overrightarrow {PQ} $$<br><br>$$\left[ {(4 - \lambda )\widehat i - 4\widehat j + \widehat k} \right].\left[ { + \widehat i + 5\widehat j - 6\widehat k} \right] = 0$$<br><br>$$4 - \lambda - 20 - 6 = 0$$<br><br>$$ \Rightarrow $$ $$\lambda $$ = -22<br><br>Now, $${\lambda \over {11}} = - 2$$<br><br>$$ \Rightarrow {\left( {{\lambda \over {11}}} \right)^2} - {{4\lambda } \over {11}} - 4$$<br><br>$$ \Rightarrow 4 + 8 - 4 = 8$$ | integer | jee-main-2021-online-26th-february-morning-slot |
0syG6wUdgkgrwz9vew1kluxx2hr | maths | 3d-geometry | lines-and-plane | Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P($$\alpha$$, $$\beta$$, $$\gamma$$) is the foot of perpendicular from (3, 2, 1) on L, then the <br/>value of 21($$\alpha$$ + $$\beta$$ + $$\gamma$$) equals : | [{"identifier": "A", "content": "102"}, {"identifier": "B", "content": "142"}, {"identifier": "C", "content": "136"}, {"identifier": "D", "content": "68"}] | ["A"] | null | Dr's of line $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 1 \cr
0 & 1 & 2 \cr
} } \right| = 3\widehat i - 2\widehat j + \widehat k$$<br><br>Dr/s : - (3, $$-$$2, 1)<br><br>Points on the line ($$-$$2, 4, 0)<br><br>Equation of the line $${{x + 2} \over 3} = {{y - 4} \over { - 2}} = {z \over 1} = \lambda $$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266970/exam_images/wlg6fjohqqr0lnvjjp15.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - 3D Geometry Question 207 English Explanation"><br><br>Dr's of PQ : $$3\lambda - 5, - 2\lambda + 2,\lambda - 1$$<br><br>Dr's of y lines are (3, $$-$$2, 1)<br><br>Since $$PQ \bot $$ line<br><br>$$3(3\lambda - 5) - 2( - 2\lambda + 2) + 1(\lambda - 1) = 0$$<br><br>$$\lambda = {{10} \over 7}$$<br><br>$$P\left( {{{16} \over 7},{8 \over 7},{{10} \over 7}} \right)$$<br><br>$$21(\alpha + \beta + \gamma ) = 21\left( {{{34} \over 7}} \right) = 102$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
CkOToyCiyoKHvpwLR91kmhx32x5 | maths | 3d-geometry | lines-and-plane | Let P be a plane lx + my + nz = 0 containing <br/><br/>the line, $${{1 - x} \over 1} = {{y + 4} \over 2} = {{z + 2} \over 3}$$. If plane P divides the line segment AB joining <br/><br/>points A($$-$$3, $$-$$6, 1) and B(2, 4, $$-$$3) in ratio k : 1 then the value of k is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1.5"}, {"identifier": "D", "content": "4"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264908/exam_images/td5ojeuzdtmnzkretbhj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Mathematics - 3D Geometry Question 205 English Explanation">
<br>Line lies on plane<br><br>$$ - l + 2m + 3n = 0$$ ..... (1)<br><br>Point on line (1, $$-$$4, $$-$$2) lies on plane<br><br>$$l - 4m - 2n = 0$$ .... (2)<br><br>from (1) & (2)<br><br>$$ - 2m + n = 0 \Rightarrow 2m = n$$<br><br>$$l = 3n + 2m \Rightarrow l = 4n$$<br><br>$$l:m:n::4n:{n \over 2}:n$$<br><br>$$l:m:n::8n:n:2n$$<br><br>$$l:m:n::8:1:2$$<br><br>Now equation of plane is 8x + y + 2z = 0<br><br>R divide AB is ratio k : 1<br><br>$$R:\left( {{{ - 3 + 2k} \over {k + 1}},{{ - 6 + 4k} \over {k + 1}},{{1 - 3k} \over {k + 1}}} \right)$$ lies on plane<br><br>$$8\left( {{{ - 3 + 2k} \over {k + 1}}} \right) + \left( {{{ - 6 + 4k} \over {k + 1}}} \right) + 2\left( {{{1 - 3k} \over {k + 1}}} \right) = 0$$<br><br>$$ - 24 + 16k - 6 + 4k + 2 - 6k = 0$$<br><br>$$ - 28 + 14k = 0$$<br><br>$$k = 2$$ | mcq | jee-main-2021-online-16th-march-morning-shift |
jTuoAqCOlT0ueBNhrI1kmizayck | maths | 3d-geometry | lines-and-plane | If the distance of the point (1, $$-$$2, 3) from the plane x + 2y $$-$$ 3z + 10 = 0 measured parallel to the line, $${{x - 1} \over 3} = {{2 - y} \over m} = {{z + 3} \over 1}$$ is $$\sqrt {{7 \over 2}} $$, then the value of |m| is equal to _________. | [] | null | 2 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267379/exam_images/vmb11b1oodts8ory1wgg.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - 3D Geometry Question 202 English Explanation">
<br>Given line L,
<br><br>$${{x - 1} \over 3} = {{2 - y} \over m} = {{z + 3} \over 1}$$
<br><br>$$ \Rightarrow $$ $${{x - 1} \over 3} = {{y - 2} \over -m} = {{z + 3} \over 1}$$
<br><br>$$ \therefore $$ D.R of line = <3, -m, 1>
<br><br>D.R of parallel line PQ will also be same.
<br><br>$$ \therefore $$ Equation of line PQ,
<br><br>$${{x - 1} \over 3} = {{y + 2} \over { - m}} = {{z - 3} \over 1} = \lambda $$<br><br>Pt. $$Q(3\lambda + 1, - m\lambda - 2,\lambda + 3)$$ lie on plane<br><br>$$(3\lambda + 1) + 2( - m\lambda - 2) - 3(\lambda + 3) + 10 = 0$$<br><br>$$ \Rightarrow 3\lambda - 2m\lambda - 3\lambda + 1 - 4 - 9 + 10 = 0$$<br><br>$$ \Rightarrow - 2m\lambda = 2$$<br><br>$$m\lambda = - 1 \Rightarrow \lambda = - {1 \over m}$$<br><br>$$Q\left[ { - {3 \over m} + 1, - 1, - {1 \over m} + 3} \right]$$<br><br>Given, $$PQ = \sqrt {{7 \over 2}} $$<br><br>$$ \Rightarrow $$ $$\sqrt {{{\left( { - {3 \over m}} \right)}^2} + 1 + {{\left( { - {1 \over m}} \right)}^2}} = \sqrt {{7 \over 2}} $$<br><br>$$ \Rightarrow {{10 + {m^2}} \over {{m^2}}} = {7 \over 2}$$<br><br>$$ \Rightarrow 20 + 2{m^2} = 7{m^2}$$<br><br>$$ \Rightarrow $$ $${m^2} = 4 \Rightarrow |m| = 2$$ | integer | jee-main-2021-online-16th-march-evening-shift |
oMM8M7sXfZX0uLWycA1kmjbpapm | maths | 3d-geometry | lines-and-plane | If the equation of the plane passing through the line of intersection of the planes 2x $$-$$ 7y + 4z $$-$$ 3 = 0, 3x $$-$$ 5y + 4z + 11 = 0 and the point ($$-$$2, 1, 3) is ax + by + cz $$-$$ 7 = 0, then the value of 2a + b + c $$-$$ 7 is ____________. | [] | null | 4 | Equation of plane can be written using family of planes : P<sub>1</sub> + $$\lambda$$P<sub>2</sub> = 0<br><br>(2x $$-$$ 7y + 4z $$-$$ 3) + $$\lambda$$ (3x $$-$$ 5y + 4z + 11) = 0<br><br>It passes through ($$-$$2, 1, 3)<br><br>$$ \therefore $$ ($$-$$4 + 7 + 12 $$-$$ 3) + $$\lambda$$ ($$-$$6 $$-$$ 5 + 12 + 11) = 0<br><br>$$-$$2 + $$\lambda$$ (12) = 0<br><br>$$\lambda$$ = $${1 \over 6}$$.<br><br>$$ \therefore $$ 12x $$-$$ 42y + 24z $$-$$ 18 + 3x $$-$$ 5y + 4z + 11 = 0<br><br>15x $$-$$ 47y + 28z $$-$$ 7 = 0<br><br>$$ \therefore $$ a = 15, b = $$-$$47, c = 28<br><br>$$ \therefore $$ 2a + b + c $$-$$ 7 = 30 $$-$$ 47 + 28 $$-$$ 7 = 4 | integer | jee-main-2021-online-17th-march-morning-shift |
hYzxuFS1bzdbL0JjTy1kmkm69ew | maths | 3d-geometry | lines-and-plane | If the equation of plane passing through the mirror image of a point (2, 3, 1) with respect to line $${{x + 1} \over 2} = {{y - 3} \over 1} = {{z + 2} \over { - 1}}$$ and containing the line $${{x - 2} \over 3} = {{1 - y} \over 2} = {{z + 1} \over 1}$$ is $$\alpha$$x + $$\beta$$y + $$\gamma$$z = 24, then $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to : | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "19"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "20"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267417/exam_images/rfcxq7voccwbfmoextkg.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267008/exam_images/neowpsgzybeeqozyfefn.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267715/exam_images/itpc8wihkk6ay7ozb0q7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Mathematics - 3D Geometry Question 196 English Explanation 1"></picture><br><br>Let point M is (2$$\lambda$$ $$-$$ 1, $$\lambda$$ + 3, $$-$$ $$\lambda$$ $$-$$ 2)<br><br>D.R.'s of AM line are < 2$$\lambda$$ $$-$$ 1 $$-$$ 2, $$\lambda$$ + 3 $$-$$ 3, $$-$$$$\lambda$$ $$-$$ 2 $$-$$ 1>
<br><br>= < 2$$\lambda$$ $$-$$ 3, $$\lambda$$, $$-$$$$\lambda$$ $$-$$3 ><br><br>AM $$ \bot $$ line L<sub>1</sub><br><br>$$ \therefore $$ $$2(2\lambda - 3) + 1(\lambda ) - 1( - \lambda - 3) = 0$$<br><br>$$6\lambda = 3,\lambda = {1 \over 2}$$ $$ \therefore $$ $$M \equiv \left( {0,{7 \over 2},{{ - 5} \over 2}} \right)$$<br><br>M is mid-point of A & B<br><br>$$M = {{A + B} \over 2}$$<br><br>B = 2M $$-$$ A<br><br>B $$ \equiv $$ ($$-$$2, 4, $$-$$6)<br><br>Now we have to find equation of plane passing through B($$-$$2, 4, $$-$$6) & also containing the line<br><br>$${{x - 2} \over 3} = {{1 - y} \over 2} = {{z + 1} \over 1}$$ ....... (1)<br><br>$${{x - 2} \over 3} = {{y - 1} \over { - 2}} = {{z + 1} \over 1}$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267697/exam_images/pcxamvzsbmuou2a9eitt.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Mathematics - 3D Geometry Question 196 English Explanation 2"><br><br>Point P on line is (2, 1, $$-$$1)<br><br>$${\overrightarrow b _2}$$ of line L<sub>2</sub> is 3, $$-$$2, 1<br><br>$$\overrightarrow n ||({\overrightarrow b _2} \times \overrightarrow {PB} )$$<br><br>$${\overrightarrow b _2} = 3\widehat i - 2\widehat j + \widehat k$$<br><br>$$\overrightarrow {PB} = - 4\widehat i + 3\widehat j - 5\widehat k$$<br><br>$$\overrightarrow n = 7\widehat i + 11\widehat j + \widehat k$$<br><br>$$ \therefore $$ equation of plane is $$\overrightarrow r \,.\,\overrightarrow n = \overrightarrow a \,.\,\overrightarrow n $$<br><br>$$\overrightarrow r \,.\,(7\widehat i + 11\widehat j + \widehat k) = ( - 2\widehat i + 4\widehat j - 6\widehat k).(7\widehat i + 11\widehat j + \widehat k)$$<br><br>7x + 11y + z = $$-$$14 + 44 $$-$$6<br><br>7x + 11y + z = 24<br><br>$$ \therefore $$ $$\alpha$$ = 7<br><br>$$\beta$$ = 11<br><br>$$\gamma$$ = 1<br><br>$$ \therefore $$ $$\alpha$$ + $$\beta$$ + $$\gamma$$ = 19 | mcq | jee-main-2021-online-17th-march-evening-shift |
Y1ictxRYxt6QuvFxjI1kmm42n93 | maths | 3d-geometry | lines-and-plane | Let P be a plane containing the line $${{x - 1} \over 3} = {{y + 6} \over 4} = {{z + 5} \over 2}$$ and parallel to the line $${{x - 1} \over 4} = {{y - 2} \over { - 3}} = {{z + 5} \over 7}$$. If the point (1, $$-$$1, $$\alpha$$) lies on the plane P, then the value of |5$$\alpha$$| is equal to ____________. | [] | null | 38 | <p>Equation of required plane is $$\left| {\matrix{
{x - 1} & {y + 6} & {z + 5} \cr
3 & 4 & 2 \cr
4 & { - 3} & 7 \cr
} } \right| = 0$$</p>
<p>Since, (1, $$-$$1, $$\alpha$$) lies on it,</p>
<p>So, replace x by 1, y by ($$-$$1) and z and $$\alpha$$.</p>
<p>$$\left| {\matrix{
0 & 5 & {\alpha + 5} \cr
3 & 4 & 2 \cr
4 & { - 3} & 7 \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow 5\alpha + 38 = 0 \Rightarrow 5\alpha = - 38$$</p>
<p>$$\therefore$$ $$\left| {5\alpha } \right| = \left| { - 38} \right| = 38$$</p> | integer | jee-main-2021-online-18th-march-evening-shift |
1krq0kpif | maths | 3d-geometry | lines-and-plane | Let P be a plane passing through the points (1, 0, 1), (1, $$-$$2, 1) and (0, 1, $$-$$2). Let a vector $$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$$ be such that $$\overrightarrow a $$ is parallel to the plane P, perpendicular to $$(\widehat i + 2\widehat j + 3\widehat k)$$ and $$\overrightarrow a \,.\,(\widehat i + \widehat j + 2\widehat k) = 2$$, then $${(\alpha - \beta + \gamma )^2}$$ equals ____________. | [] | null | 81 | Equation of plane :<br><br>$$\left| {\matrix{
{x - 1} & {y - 0} & {z - 1} \cr
{1 - 1} & 2 & {1 - 1} \cr
{1 - 0} & {0 - 1} & {1 + 2} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow 3x - z - 2 = 0$$<br><br>$$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$$ || to 3x $$-$$ z $$-$$ 2 = 0<br><br>$$ \Rightarrow 3\alpha - 8 = 0$$ ..... (1)<br><br>$$\overrightarrow a \bot \widehat i + \widehat j + 3\widehat k$$<br><br>$$ \Rightarrow \alpha + 2\beta + 3\gamma = 0$$ ...... (2)<br><br>$$\overrightarrow a .(\widehat i + \widehat j + 2\widehat k) = 0$$<br><br>$$\Rightarrow$$ $$\alpha$$ + $$\beta$$ + 2$\gamma$ = 2 ........ (3)<br><br>On solving 1, 2 & 3<br><br>$$\alpha$$ = 1, $$\beta$$ = $$-$$5, $\gamma$ = 3<br><br>So, ($$\alpha$$ $$-$$ $$\beta$$ + $\gamma$)<sup>2</sup> = 81 | integer | jee-main-2021-online-20th-july-morning-shift |
1krrtutnt | maths | 3d-geometry | lines-and-plane | Consider the line L given by the equation <br/><br/>$${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$$. <br/><br/>Let Q be the mirror image of the point (2, 3, $$-$$1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P? | [{"identifier": "A", "content": "($$-$$1, 1, 2)"}, {"identifier": "B", "content": "(1, 1, 1)"}, {"identifier": "C", "content": "(1, 1, 2)"}, {"identifier": "D", "content": "(1, 2, 2)"}] | ["D"] | null | Plane p is $${ \bot ^r}$$ to line $${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$$ & passes through pt. (2, 3) equation of plane p <br><br>2(x $$-$$ 2) + 1(y $$-$$ 3) + 1 (z + 1) = 0<br><br>2x + y + z $$-$$ 6 = 0<br><br>Point (1, 2, 2) satisfies above equation | mcq | jee-main-2021-online-20th-july-evening-shift |
1krtbltik | maths | 3d-geometry | lines-and-plane | Let L be the line of intersection of planes $$\overrightarrow r .(\widehat i - \widehat j + 2\widehat k) = 2$$ and $$\overrightarrow r .(2\widehat i + \widehat j - \widehat k) = 2$$. If $$P(\alpha ,\beta ,\gamma )$$ is the foot of perpendicular on L from the point (1, 2, 0), then the value of $$35(\alpha + \beta + \gamma )$$ is equal to : | [{"identifier": "A", "content": "101"}, {"identifier": "B", "content": "119"}, {"identifier": "C", "content": "143"}, {"identifier": "D", "content": "134"}] | ["B"] | null | $${P_1}:x - y + 2z = 2$$<br><br>$${P_2}:2x + y - 3 = 2$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265409/exam_images/gzyjdvs460f1b9cskjhd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 22th July Evening Shift Mathematics - 3D Geometry Question 187 English Explanation"><br>Let line of Intersection of planes P<sub>1</sub> and P<sub>2</sub> cuts xy plane in point Q.<br><br>$$\Rightarrow$$ z-coordinate of point Q is zero<br><br>$$ \Rightarrow \left. {\matrix{
{x - y = 2} \cr
{and\,2x + y = 2} \cr
} } \right\} \Rightarrow x = {4 \over 3},y = {{ - 2} \over 3}$$<br><br>$$ \Rightarrow Q\left( {{4 \over 3},{{ - 2} \over 3},0} \right)$$<br><br>Vector parallel to the line of intersection<br><br>$$\overrightarrow a = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 1} & 2 \cr
2 & 1 & { - 1} \cr
} } \right| = - \widehat i + 5\widehat j + 3\widehat k$$<br><br>Equation of Line of intersection<br><br>$${{x - {4 \over 3}} \over { - 1}} = {{y + {2 \over 3}} \over 5} = {{z - 0} \over 3} = \lambda $$ (say)<br><br>Let coordinates of foot of perpendicular be <br><br>$$F\left( { - \lambda + {4 \over 3},5\lambda - {2 \over 3},3\lambda } \right)$$<br><br>$$\overrightarrow {PF} = \left( { - \lambda + {1 \over 3}} \right)\widehat i + \left( {5\lambda - {8 \over 3}} \right)\widehat j + (3\lambda )\widehat k$$<br><br>$$\overrightarrow {PF} .\overrightarrow a = 0$$<br><br>$$ \Rightarrow \lambda - {1 \over 3} + 25\lambda {{ - 40} \over 3} + 9\lambda = 0$$<br><br>$$ \Rightarrow 35\lambda = {{41} \over 3} \Rightarrow \lambda = {{41} \over {105}}$$<br><br>Now, $$\alpha = - \lambda + {4 \over 3},\beta = 5\lambda - {2 \over 3},\gamma = 3\lambda $$<br><br>$$ \Rightarrow \alpha + \beta + \gamma = 7\lambda + {2 \over 3}$$<br><br>$$ = 7\left( {{{41} \over {105}}} \right) + {2 \over 3}$$<br><br>$$ = {{51} \over {15}}$$<br><br>$$ \Rightarrow 35(\alpha + \beta + \gamma ) = {{51} \over {15}} \times 35 = 119$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krw1mzhz | maths | 3d-geometry | lines-and-plane | Let the foot of perpendicular from a point P(1, 2, $$-$$1) to the straight line $$L:{x \over 1} = {y \over 0} = {z \over { - 1}}$$ be N. Let a line be drawn from P parallel to the plane x + y + 2z = 0 which meets L at point Q. If $$\alpha$$ is the acute angle between the lines PN and PQ, then cos$$\alpha$$ is equal to ________________. | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt 3 }}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264384/exam_images/ockxunbxex6ruzroagy0.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - 3D Geometry Question 186 English Explanation 1"><br>$$\overrightarrow {PN} .(\widehat i - \widehat k) = 0$$<br><br>$$\Rightarrow$$ N(1, 0, $$-$$1)<br><br>Now, <br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265917/exam_images/on0qybbkpgettztu34nb.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - 3D Geometry Question 186 English Explanation 2"><br>$$\overrightarrow {PQ} .(\widehat i + \widehat j + 2\widehat k) = 0$$<br><br>$$\Rightarrow$$ $$\mu$$ = $$-$$ 1<br><br>$$\Rightarrow$$ Q ($$-$$1, 0, 1)<br><br>$$\overrightarrow {PN} $$ = 2$$\widehat j$$ and $$\overrightarrow {PQ} $$ = $$2\widehat i + 2\widehat j - 2\widehat k$$<br><br>$$ \Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1krxgs6qs | maths | 3d-geometry | lines-and-plane | For real numbers $$\alpha$$ and $$\beta$$ $$\ne$$ 0, if the point of intersection of the straight lines<br/><br/>$${{x - \alpha } \over 1} = {{y - 1} \over 2} = {{z - 1} \over 3}$$ and $${{x - 4} \over \beta } = {{y - 6} \over 3} = {{z - 7} \over 3}$$, lies on the plane x + 2y $$-$$ z = 8, then $$\alpha$$ $$-$$ $$\beta$$ is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "7"}] | ["D"] | null | First line is ($$\phi$$ + $$\alpha$$, 2$$\phi$$ + 1, 3$$\phi$$ + 1)<br><br>and second line is (q$$\beta$$ + 4, 3q + 6, 3q + 7)<br><br>For intersection<br><br>$$\phi$$ + $$\alpha$$ = q$$\beta$$ + 4 ...... (i)<br><br>2$$\phi$$ + 1 = 3q + 6 .... (ii)<br><br>3$$\phi$$ + 1 = 3q + 7 ...... (iii)<br><br>for (ii) & (iii) $$\phi$$ = 1, q = $$-$$1<br><br>So, from (i) $$\alpha$$ + $$\beta$$ = 3<br><br>Now, point of intersection is ($$\alpha$$ + 1, 3, 4)<br><br>It lies on the plane.<br><br>Hence, $$\alpha$$ = 5 & $$\beta$$ = $$-$$2 | mcq | jee-main-2021-online-27th-july-evening-shift |
1kryflta1 | maths | 3d-geometry | lines-and-plane | The distance of the point P(3, 4, 4) from the point of intersection of the line joining the points. Q(3, $$-$$4, $$-$$5) and R(2, $$-$$3, 1) and the plane 2x + y + z = 7, is equal to ______________. | [] | null | 7 | $$\overrightarrow {QR} : - {{x - 3} \over 1} = {{y + 4} \over { - 1}} = {{z + 5} \over { - 6}} = r$$<br><br>$$ \Rightarrow (x,y,z) \equiv (r + 3, - r - 4, - 6r - 5)$$<br><br>Now, satisfying it in the given plane.<br><br>We get r = $$-$$2<br><br>so, required point of intersection is T(1, $$-$$2, 7).<br><br>Hence, PT = 7. | integer | jee-main-2021-online-27th-july-evening-shift |
1krzrobee | maths | 3d-geometry | lines-and-plane | If the lines $${{x - k} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$ and <br/>$${{x + 1} \over 3} = {{y + 2} \over 2} = {{z + 3} \over 1}$$ are co-planar, then the value of k is _____________. | [] | null | 1 | $$\left| {\matrix{
{k + 1} & 4 & 6 \cr
1 & 2 & 3 \cr
3 & 2 & 1 \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ $$(k + 1)[2 - 6] - 4[1 - 9] + 6[2 - 6] = 0$$<br><br>$$ \Rightarrow $$ $$k = 1$$ | integer | jee-main-2021-online-25th-july-evening-shift |
1ks0cxriw | maths | 3d-geometry | lines-and-plane | Let a plane P pass through the point (3, 7, $$-$$7) and contain the line, $${{x - 2} \over { - 3}} = {{y - 3} \over 2} = {{z + 2} \over 1}$$. If distance of the plane P from the origin is d, then d<sup>2</sup> is equal to ______________. | [] | null | 3 | $$\overrightarrow {BA} = (\widehat i + 4\widehat j - 5\widehat k)$$<br><br>$$\overrightarrow {BA} \times \overrightarrow l = \overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
{ - 3} & 2 & 1 \cr
1 & 4 & { - 5} \cr
} } \right|$$<br><br>$$a\widehat i + b\widehat j + c\widehat k = - 14\widehat i - \widehat j(14) + \widehat k( - 14)$$<br><br>a = 1, b = 1, c = 1<br><br>Plane is (x $$-$$ 2) + (y $$-$$ 3) + (z + 2) = 0<br><br>$$ \Rightarrow $$ x + y + z $$-$$ 3 = 0<br><br>$$ \therefore $$ d = $$\sqrt 3 $$ $$\Rightarrow$$ d<sup>2</sup> = 3 | integer | jee-main-2021-online-27th-july-morning-shift |
1ktbf6smu | maths | 3d-geometry | lines-and-plane | A plane P contains the line $$x + 2y + 3z + 1 = 0 = x - y - z - 6$$, and is perpendicular to the plane $$ - 2x + y + z + 8 = 0$$. Then which of the following points lies on P? | [{"identifier": "A", "content": "($$-$$1, 1, 2)"}, {"identifier": "B", "content": "(0, 1, 1)"}, {"identifier": "C", "content": "(1, 0, 1)"}, {"identifier": "D", "content": "(2, $$-$$1, 1)"}] | ["B"] | null | Equation of plane P can be assumed as<br><br>P : x + 2y + 3z + 1 + $$\lambda$$ (x $$-$$ y $$-$$ z $$-$$ 6) = 0<br><br>$$\Rightarrow$$ P : (1 + $$\lambda$$)x + (2 $$-$$ $$\lambda$$)y + (3 $$-$$ $$\lambda$$)z + 1 $$-$$ 6$$\lambda$$ = 0<br><br>$$ \Rightarrow {\overrightarrow n _1} = (1 + \lambda )\widehat i + (2 - \lambda )\widehat j + (3 - \lambda )\widehat k$$<br><br>$$\therefore$$ $${\overrightarrow n _1}\,.\,{\overrightarrow n _2} = 0$$<br><br>$$\Rightarrow$$ 2(1 + $$\lambda$$) $$-$$ (2 $$-$$ $$\lambda$$) $$-$$ (3 $$-$$ $$\lambda$$) = 0<br><br>$$\Rightarrow$$ 2 + 2$$\lambda$$ $$-$$ 2 + $$\lambda$$ $$-$$ 3 + $$\lambda$$ = 0 $$\Rightarrow$$ $$\lambda$$ = $${3 \over 4}$$<br><br>$$\Rightarrow$$ $$P:{{7x} \over 4} + {5 \over 4}y + {{9z} \over 4} - {{14} \over 4} = 0$$<br><br>$$\Rightarrow$$ 7x + 5y + 9z = 14<br><br>(0, 1, 1) lies on P. | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktbi6k7h | maths | 3d-geometry | lines-and-plane | Let the line L be the projection of the line $${{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 4} \over 2}$$ in the plane x $$-$$ 2y $$-$$ z = 3. If d is the distance of the point (0, 0, 6) from L, then d<sup>2</sup> is equal to _______________. | [] | null | 26 | To find the projection let's find the foot of perpendicular from $(1,3$,
4) to plane $x-2 y-z=3$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lfg2ca35/20453aa3-5bf8-445c-837d-c8b6d848626f/8ecab100-c6b2-11ed-b4b3-b306e87ca523/file-1lfg2ca36.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lfg2ca35/20453aa3-5bf8-445c-837d-c8b6d848626f/8ecab100-c6b2-11ed-b4b3-b306e87ca523/file-1lfg2ca36.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 26th August Morning Shift Mathematics - 3D Geometry Question 179 English Explanation">
<br>$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y-3}{-2}=\frac{z-4}{-1}=\lambda_1 \\\\
& \left(\lambda_1+1\right)-2\left(-2 \lambda_1+3\right)-\left(-\lambda_1+4\right)=3 \\\\
& \Rightarrow 6 \lambda_1=12 \Rightarrow \lambda_1=2
\end{aligned}
$$
<br><br>So, foot of perpendicular from $(1,3,4)$ to plane $x-2 y-z=3$ is $A$ $(3,-1,2)$.
<br><br>Let us also find the intersection point of the plane and line
<br><br>$$
\begin{gathered}
\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}=\lambda_2 \\\\
\left(2 \lambda_2+1\right)-2\left(\lambda_2+3\right)-\left(2 \lambda_2+4\right)=3-2 \lambda_2=12 \Rightarrow \lambda_2=-6
\end{gathered}
$$
<br><br>The intersection point of the plane and line is $B(-11,-3,-8)$ Line passing through $A$ and $B$ is
<br><br>$$
\begin{aligned}
& \frac{x-3}{-14}=\frac{y+1}{-2}=\frac{z-2}{-10}=\mu \\\\
& \frac{x-3}{7}=\frac{y+1}{1}=\frac{z-2}{5}=\mu
\end{aligned}
$$
<br><br>Now, let's find the distance from $O(0,0,6)$ to this line $L$.
<br><br>Let's say $C(7 \mu+3, \mu-1,5 \mu+2)$ is any point on $L$. Then,
<br><br>$$
\begin{aligned}
& \{(7 \mu+3)-0\} \cdot 7+\{(\mu-1)-0\} \cdot 1+\{(5 \mu+2)-6\} \cdot 5=0 \\\\
& \Rightarrow 49 \mu+21+\mu-1+25 \mu-20=0 \Rightarrow \mu=0 \\\\
& \therefore C(3,-1,2) \\\\
& \text { Distance }=\sqrt{(3-0)^2+(-1-0)^2+(2-6)^2}=\sqrt{26} \\\\
& d^2=26
\end{aligned}
$$ | integer | jee-main-2021-online-26th-august-morning-shift |
1ktd44pm7 | maths | 3d-geometry | lines-and-plane | Let Q be the foot of the perpendicular from the point P(7, $$-$$2, 13) on the plane containing the lines $${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$$ and $${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$$. Then (PQ)<sup>2</sup>, is equal to ___________. | [] | null | 96 | Containing the line $$\left| {\matrix{
{x + 1} & {y - 1} & {z - 3} \cr
6 & 7 & 8 \cr
3 & 5 & 7 \cr
} } \right| = 0$$<br><br>$$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$$<br><br>$$x - 2y + z = 0$$<br><br>$$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6 $$<br><br>$$P{Q^2} = 96$$ | integer | jee-main-2021-online-26th-august-evening-shift |
1kteihgh9 | maths | 3d-geometry | lines-and-plane | The distance of the point (1, $$-$$2, 3) from the plane x $$-$$ y + z = 5 measured parallel to a line, whose direction ratios are 2, 3, $$-$$6 is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263282/exam_images/vaartuqi2uokdkrcurcw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift Mathematics - 3D Geometry Question 175 English Explanation"><br><br>$$(1 + 2\lambda ) + 2 - 3\lambda + 3 - 6\lambda = 5$$<br><br>$$ \Rightarrow 6 - 7\lambda = 5 \Rightarrow \lambda = {1 \over 7}$$<br><br>so, $$P = \left( {{9 \over 7}, - {{11} \over 7},{{15} \over 7}} \right)$$<br><br>$$AP = \sqrt {{{\left( {1 - {9 \over 7}} \right)}^2} + {{\left( { - 2 + {{11} \over 7}} \right)}^2} + {{\left( {3 - {{15} \over 7}} \right)}^2}} $$<br><br>$$AP = \sqrt {\left( {{4 \over {49}}} \right) + {9 \over {49}} + {{36} \over {49}}} = 1$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktekeym0 | maths | 3d-geometry | lines-and-plane | Equation of a plane at a distance $$\sqrt {{2 \over {21}}} $$ from the origin, which contains the line of intersection of the planes x $$-$$ y $$-$$ z $$-$$ 1 = 0 and 2x + y $$-$$ 3z + 4 = 0, is : | [{"identifier": "A", "content": "$$3x - y - 5z + 2 = 0$$"}, {"identifier": "B", "content": "$$3x - 4z + 3 = 0$$"}, {"identifier": "C", "content": "$$ - x + 2y + 2z - 3 = 0$$"}, {"identifier": "D", "content": "$$4x - y - 5z + 2 = 0$$"}] | ["D"] | null | Required equation of plane<br><br>$${P_1} + \lambda {P_2} = 0$$<br><br>$$(x - y - z - 1) + \lambda (2x + y - 3z + 4) = 0$$<br><br>Given that its dist. From origin is $${2 \over {\sqrt {21} }}$$<br><br>Thus, $${{|4\lambda - 1|} \over {\sqrt {{{(2\lambda + 1)}^2} + {{(\lambda - 1)}^2} + {{( - 3\lambda - 1)}^2}} }} = {{\sqrt 2 } \over {\sqrt {21} }}$$<br><br>$$ \Rightarrow 21{(4\lambda - 1)^2} = 2(14{\lambda ^2} + 8\lambda + 3)$$<br><br>$$ \Rightarrow 336{\lambda ^2} - 168\lambda + 21 = 28{\lambda ^2} + 16\lambda + 6$$<br><br>$$ \Rightarrow 308{\lambda ^2} - 184\lambda + 15 = 0$$<br><br>$$ \Rightarrow 308{\lambda ^2} - 154\lambda - 30\lambda + 15 = 0$$<br><br>$$ \Rightarrow (2\lambda - 1)(154\lambda - 15) = 0$$<br><br>$$ \Rightarrow \lambda = {1 \over 2}$$ or $${{15} \over {154}}$$<br><br>for $$\lambda = {1 \over 2}$$ reqd. plane is $$4x - y - 5z + 2 = 0$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktfzryxs | maths | 3d-geometry | lines-and-plane | The equation of the plane passing through the line of intersection of the planes $$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$$ and $$\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0$$ and parallel to the x-axis is : | [{"identifier": "A", "content": "$$\\overrightarrow r .\\left( {\\widehat j - 3\\widehat k} \\right) + 6 = 0$$"}, {"identifier": "B", "content": "$$\\overrightarrow r .\\left( {\\widehat i + 3\\widehat k} \\right) + 6 = 0$$"}, {"identifier": "C", "content": "$$\\overrightarrow r .\\left( {\\widehat i - 3\\widehat k} \\right) + 6 = 0$$"}, {"identifier": "D", "content": "$$\\overrightarrow r .\\left( {\\widehat j - 3\\widehat k} \\right) - 6 = 0$$"}] | ["A"] | null | Equation of planes are<br><br>$$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) - 1 = 0 \Rightarrow x + y + z - 1 = 0$$<br><br>and $$\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0 \Rightarrow 2x + 3y - z + 4 = 0$$<br><br>equation of planes through line of intersection of these planes is :-<br><br>$$(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0$$<br><br>$$ \Rightarrow (1 + 2\lambda )x + (1 + 3\lambda )y + (1 - \lambda )z - 1 + 4\lambda = 0$$<br><br>But this plane is parallel to x-axis whose direction are (1, 0, 0)<br><br>$$\therefore$$ $$(1 + 2\lambda )1 + (1 + 3\lambda )0 + (1 - \lambda )0 = 0$$<br><br>$$\lambda = - {1 \over 2}$$<br><br>$$\therefore$$ Required plane is <br><br>$$0x + \left( {1 - {3 \over 2}} \right)y + \left( {1 + {1 \over 2}} \right)z - 1 + 4\left( {{{ - 1} \over 2}} \right) = 0$$<br><br>$$ \Rightarrow {{ - y} \over 2} + {3 \over 2}z - 3 = 0$$<br><br>$$ \Rightarrow y - 3z + 6 = 0$$<br><br>$$ \Rightarrow \overrightarrow r .\left( {\widehat j - 3\widehat k} \right) + 6 = 0$$ Ans. | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktgobl7e | maths | 3d-geometry | lines-and-plane | Let S be the mirror image of the point Q(1, 3, 4) with respect to the plane 2x $$-$$ y + z + 3 = 0 and let R(3, 5, $$\gamma$$) be a point of this plane. Then the square of the length of the line segment SR is ___________. | [] | null | 72 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266720/exam_images/h71pebht6whdybpiqe1o.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - 3D Geometry Question 173 English Explanation"> <br><br>Since R(3, 5, $$\gamma$$) lies on the plane 2x $$-$$ y + z + 3 = 0. <br><br>Therefore, 6 $$-$$ 5 + $$\gamma$$ + 3 = 0<br><br>$$\Rightarrow$$ $$\gamma$$ = $$-$$4<br><br>Now, <br><br>dr's of line QS are 2, $$-$$1, 1<br><br>equation of line QS is <br><br>$${{x - 1} \over 2} = {{y - 3} \over { - 1}} = {{z - 4} \over 1} = \lambda $$ (say)<br><br>$$ \Rightarrow F(2\lambda + 1, - \lambda + 3,\lambda + 4)$$<br><br>F lies in the plane<br><br>$$ \Rightarrow 2(2\lambda + 1) - ( - \lambda + 3) + (\lambda + 4) + 3$$ = 0<br><br>$$ \Rightarrow 4\lambda + 2 + \lambda - 3 + \lambda + 7 = 0$$<br><br>$$ \Rightarrow 6\lambda + 6 = 0 \Rightarrow \lambda = - 1$$<br><br>$$\Rightarrow$$ F($$-$$1, 4, 3)<br><br>Since, F is mid-point of QS.<br><br>Therefore, coordinated of S are ($$-$$3, 5, 2).<br><br>So, SR = $$\sqrt {36 + 0 + 36} = \sqrt {72} $$<br><br>SR<sup>2</sup> = 72. | integer | jee-main-2021-online-27th-august-evening-shift |
1ktiom3vk | maths | 3d-geometry | lines-and-plane | Let the equation of the plane, that passes through the point (1, 4, $$-$$3) and contains the line of intersection of the <br/>planes 3x $$-$$ 2y + 4z $$-$$ 7 = 0 <br/>and x + 5y $$-$$ 2z + 9 = 0, be <br/>$$\alpha$$x + $$\beta$$y + $$\gamma$$z + 3 = 0, then $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to : | [{"identifier": "A", "content": "$$-$$23"}, {"identifier": "B", "content": "$$-$$15"}, {"identifier": "C", "content": "23"}, {"identifier": "D", "content": "15"}] | ["A"] | null | 3x $$-$$ 2y + 4z $$-$$ 7 + $$\lambda$$(x + 5y $$-$$ 2z + 9) = 0<br><br>(3 + $$\lambda$$)x + (5$$\lambda$$ $$-$$ 2)y + (4 $$-$$ 2$$\lambda$$)z + 9$$\lambda$$ $$-$$ 7 = 0<br><br>passing through (1, 4, $$-$$3)<br><br>$$\Rightarrow$$ 3 + $$\lambda$$ + 20$$\lambda$$ $$-$$ 8 $$-$$ 12 + 6$$\lambda$$ + 9$$\lambda$$ $$-$$ 7 = 0<br><br>$$\Rightarrow$$ $$\lambda$$ = $${2 \over 3}$$<br><br>$$\Rightarrow$$ equation of plane is<br><br>$$-$$11x $$-$$ 4y $$-$$ 8z + 3 = 0<br><br>$$\Rightarrow$$ $$\alpha$$ + $$\beta$$ + $$\gamma$$ = $$-$$23 | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktis4alt | maths | 3d-geometry | lines-and-plane | The square of the distance of the point of intersection <br/><br/>of the line $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$$ and the plane $$2x - y + z = 6$$ from the point ($$-$$1, $$-$$1, 2) is __________. | [] | null | 61 | $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6} = \lambda $$<br><br>$$x = 2\lambda + 1,y = 3\lambda + 2,z = 6\lambda - 1$$<br><br>for point of intersection of line & plane<br><br>$$2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$$<br><br>$$7\lambda = 7 \Rightarrow \lambda = 1$$<br><br>point : (3, 5, 5)<br><br>(distance)<sup>2</sup> = $${(3 + 1)^2} + {(5 + 1)^2} + {(5 - 2)^2}$$<br><br>$$ = 16 + 36 + 9 = 61$$ | integer | jee-main-2021-online-31st-august-morning-shift |
1ktk5cco4 | maths | 3d-geometry | lines-and-plane | The distance of the point ($$-$$1, 2, $$-$$2) from the line of intersection of the planes 2x + 3y + 2z = 0 and x $$-$$ 2y + z = 0 is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt {42} } \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt {34} } \\over 2}$$"}] | ["D"] | null | P<sub>1</sub> : 2x + 3y + 2z = 0<br><br>$$\Rightarrow$$ $${\overrightarrow n _1} = 2\widehat i + 3\widehat j + 2\widehat k$$<br><br>P<sub>2</sub> : x $$-$$ 2y + z = 0<br><br>$$\Rightarrow$$ $${\overrightarrow n _2} = \widehat i - 2\widehat j + \widehat k$$<br><br>Direction vector of line L which is line of intersection of P<sub>1</sub> & P<sub>2</sub><br><br>$$\overrightarrow r = {\overrightarrow n _1} \times {\overrightarrow n _2} = 7\widehat i - 7\widehat k$$<br><br>DR's of L are (1, 0, $$-$$1)<br><br>$$\Rightarrow$$ Equation of L : $${x \over 1} = {y \over 0} = {z \over { - 1}} = \lambda $$<br><br> <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266901/exam_images/kkhzoknkbvrgsubbusfx.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Evening Shift Mathematics - 3D Geometry Question 166 English Explanation"> <br><br>DR's of $$\overrightarrow {PQ} $$ = ($$\lambda$$ + 1, $$-$$2, 2 $$-$$ $$\lambda$$)<br><br>$$\because$$ $$\overrightarrow {PQ} \bot \overrightarrow r $$<br><br>$$ \Rightarrow (\lambda + 1)(1) + ( - 2)(0) + (2 - \lambda )( - 1) = 0$$<br><br>$$ \Rightarrow \lambda = {1 \over 2} \Rightarrow Q\left( {{1 \over 2},0,{{ - 1} \over 2}} \right)$$<br><br>$$ \Rightarrow PQ = {{\sqrt {34} } \over 2}$$ | mcq | jee-main-2021-online-31st-august-evening-shift |
1ktkdi4ed | maths | 3d-geometry | lines-and-plane | Suppose, the line $${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$$ lies on the plane $$x + 3y - 2z + \beta = 0$$. Then $$(\alpha + \beta )$$ is equal to _______. | [] | null | 7 | <p>Given equation of line</p>
<p>$${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$$ ...... (i)</p>
<p>and plane x + 3y $$-$$ 2z + $$\beta$$ = 0 ...... (ii)</p>
<p>Line (i) passes through (2, 2, $$-$$2)</p>
<p>which lies on plane (ii).</p>
<p>$$\therefore$$ 2 + 6 + 4 + $$\beta$$ = 0 $$\Rightarrow$$ $$\beta$$ = $$-$$ 12</p>
<p>Also, given line is perpendicular to normal of the plane</p>
<p>$$\alpha$$(1) $$-$$ 5(3) + 2($$-$$2) = 0 $$\Rightarrow$$ $$\alpha$$ = 19</p>
<p>$$\therefore$$ $$\alpha$$ + $$\beta$$ = 19 + (-12) = 19 - 12 = 7</p> | integer | jee-main-2021-online-31st-august-evening-shift |
1l546e82f | maths | 3d-geometry | lines-and-plane | <p>Let $${P_1}:\overrightarrow r \,.\,\left( {2\widehat i + \widehat j - 3\widehat k} \right) = 4$$ be a plane. Let P<sub>2</sub> be another plane which passes through the points (2, $$-$$3, 2), (2, $$-$$2, $$-$$3) and (1, $$-$$4, 2). If the direction ratios of the line of intersection of P<sub>1</sub> and P<sub>2</sub> be 16, $$\alpha$$, $$\beta$$, then the value of $$\alpha$$ + $$\beta$$ is equal to ________________.</p> | [] | null | 28 | <p>Direction ratio of normal to $${P_1} \equiv < 2,1, - 3 > $$</p>
<p>and that of $${P_2} \equiv \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 1 & { - 5} \cr
{ - 1} & { - 2} & 5 \cr
} } \right| = - 5\widehat i - \widehat j( - 5) + \widehat k(1)$$</p>
<p>i.e. $$ < - 5,5,1 > $$</p>
<p>d.r's of line of intersection are along vector</p>
<p>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & 1 & { - 3} \cr
{ - 5} & 5 & 1 \cr
} } \right| = \widehat i(16) - \widehat j( - 13) + \widehat k(15)$$</p>
<p>i.e. $$ < 16,13,15 > $$</p>
<p>$$\therefore$$ $$\alpha + \beta = 13 + 15 = 28$$</p> | integer | jee-main-2022-online-29th-june-morning-shift |
1l54bcrlo | maths | 3d-geometry | lines-and-plane | <p>Let $${{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}}$$ lie on the plane $$px - qy + z = 5$$, for some p, q $$\in$$ R. The shortest distance of the plane from the origin is :</p> | [{"identifier": "A", "content": "$$\\sqrt {{3 \\over {109}}} $$"}, {"identifier": "B", "content": "$$\\sqrt {{5 \\over {142}}} $$"}, {"identifier": "C", "content": "$${5 \\over {\\sqrt {71} }}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt {142} }}$$"}] | ["B"] | null | $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}=\lambda$
<br/><br/>
$(3 \lambda+2,-2 \lambda-1,-\lambda-3)$ lies on plane $p x-q y+z=5$ <br/><br/>$p(3 \lambda+2)-q(-2 \lambda-1)+(-\lambda-3)=5$
<br/><br/>
$\lambda(3 p+2 q-1)+(2 p+q-8)=0$
<br/><br/>
$3 p+2 q-1=0\} p=15$
<br/><br/>
$2 p+q-8=0\} q=-22$
<br/><br/>
Equation of plane $15 x+22 y+z-5=0$
<br/><br/>
Shortest distance from origin $=\frac{|0+0+0-5|}{\sqrt{15^{2}+22^{2}+1}}$
<br/><br/>
$=\frac{5}{\sqrt{710}}$
<br/><br/>
$=\sqrt{\frac{5}{142}}$ | mcq | jee-main-2022-online-29th-june-evening-shift |
1l54t8ew5 | maths | 3d-geometry | lines-and-plane | <p>Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line $$\overrightarrow r = - \widehat k + \lambda \left( {\widehat i + \widehat j + 2\widehat k} \right),\,\lambda \in R$$. Then, which of the following points lies on T?</p> | [{"identifier": "A", "content": "(2, 1, 0)"}, {"identifier": "B", "content": "(1, 2, 1)"}, {"identifier": "C", "content": "(1, 2, 2)"}, {"identifier": "D", "content": "(1, 3, 2)"}] | ["B"] | null | $P(1,2,1)$ image in plane $x+2 y+2 z=16$
<br><br>
$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2(1+2 \times 2+2 \times 1-16)}{1^{2}+2^{2}+2^{2}} \\\\
& \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=2 \\\\
& Q(3,6,5) \\\\
& \vec{r}=-\hat{k}+\lambda(\hat{i}+\hat{j}+2 \hat{k})
\end{aligned}
$$
<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc5qpgrl/6590d524-93d7-4f0f-8cfb-f2167895c70e/934a1200-85a0-11ed-95b6-d9b2225f1c8e/file-1lc5qpgrm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc5qpgrl/6590d524-93d7-4f0f-8cfb-f2167895c70e/934a1200-85a0-11ed-95b6-d9b2225f1c8e/file-1lc5qpgrm.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - 3D Geometry Question 161 English Explanation"><br>
$A Q=3 \hat{i}+6 \hat{j}+6 \hat{k}$
<br><br>
$$
\begin{aligned}
& =3(\hat{i}+2 \hat{j}+2 \hat{k}) \\\\
& \vec{n}=(\hat{i}+2 \hat{j}+2 \hat{k}) \times(\hat{i}+\hat{j}+2 \hat{k}) \\\\
& \left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\1 & 2 & 2 \\1 & 1 & 2\end{array}\right| \\\\
& =2 \hat{i}-0 \hat{j}-\hat{k}
\end{aligned}
$$<br><br>
Equation of plane $\equiv 2(x-0)+0(y-0)-1(z+1)$ $=0$
<br><br>
$2 x-z=1$
<br><br>
Point lying on plane from the option is $(1,2,1)$ i.e., option (B) | mcq | jee-main-2022-online-29th-june-evening-shift |
1l566xtof | maths | 3d-geometry | lines-and-plane | <p>If two distinct point Q, R lie on the line of intersection of the planes $$ - x + 2y - z = 0$$ and $$3x - 5y + 2z = 0$$ and $$PQ = PR = \sqrt {18} $$ where the point P is (1, $$-$$2, 3), then the area of the triangle PQR is equal to :</p> | [{"identifier": "A", "content": "$${2 \\over 3}\\sqrt {38} $$"}, {"identifier": "B", "content": "$${4 \\over 3}\\sqrt {38} $$"}, {"identifier": "C", "content": "$${8 \\over 3}\\sqrt {38} $$"}, {"identifier": "D", "content": "$$\\sqrt {{{152} \\over 3}} $$"}] | ["B"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5obyhm8/67807093-6d4d-453b-a541-6801368b15df/45fa9b00-0544-11ed-987f-3938cfc0f7f1/file-1l5obyhm9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5obyhm8/67807093-6d4d-453b-a541-6801368b15df/45fa9b00-0544-11ed-987f-3938cfc0f7f1/file-1l5obyhm9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Morning Shift Mathematics - 3D Geometry Question 158 English Explanation"> </p>
<p>Line L is x = y = z</p>
<p>$$\overrightarrow {PQ} .\,(\widehat i + \widehat j + \widehat k) = 0$$</p>
<p>$$ \Rightarrow (\alpha - 3) + \alpha + 2 + \alpha - 1 = 0$$</p>
<p>$$ \Rightarrow \alpha = {2 \over 3}$$ so, $$T = \left( {{2 \over 3},{2 \over 3},{2 \over 3}} \right)$$</p>
<p>$$PT = \sqrt {{{38} \over 3}} $$</p>
<p>$$ \Rightarrow QT = {4 \over {\sqrt 3 }}$$</p>
<p>So, Area $$ = \left( {{1 \over 2} \times {4 \over {\sqrt 3 }} \times {{\sqrt {38} } \over {\sqrt 3 }}} \right).\,2$$</p>
<p>$$ = {{4\sqrt {38} } \over 3}$$ sq. units</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l5672y46 | maths | 3d-geometry | lines-and-plane | <p>Let the plane $$P:\overrightarrow r \,.\,\overrightarrow a = d$$ contain the line of intersection of two planes $$\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 6$$ and $$\overrightarrow r \,.\,\left( { - 6\widehat i + 5\widehat j - \widehat k} \right) = 7$$. If the plane P passes through the point $$\left( {2,3,{1 \over 2}} \right)$$, then the value of $${{|13\overrightarrow a {|^2}} \over {{d^2}}}$$ is equal to :</p> | [{"identifier": "A", "content": "90"}, {"identifier": "B", "content": "93"}, {"identifier": "C", "content": "95"}, {"identifier": "D", "content": "97"}] | ["B"] | null | <p>$${P_1}:x + 3y - z = 6$$</p>
<p>$${P_2}: - 6x + 5y - z = 7$$</p>
<p>Family of planes passing through line of intersection of P<sub>1</sub> and P<sub>2</sub> is given by $$x(1 - 6\lambda ) + y(3 + 5\lambda ) + z( - 1 - \lambda ) - (6 + 7\lambda ) = 0$$</p>
<p>It passes through $$\left( {2,3,{1 \over 2}} \right)$$</p>
<p>So, $$2(1 - 6\lambda ) + 3(3 + 5\lambda ) + {1 \over 2}( - 1 - \lambda ) - (6 + 7\lambda ) = 0$$</p>
<p>$$ \Rightarrow 2 - 12\lambda + 9 + 15\lambda - {1 \over 2} - {\lambda \over 2} - 6 - 7\lambda = 0$$</p>
<p>$$ \Rightarrow {9 \over 2} - {{9\lambda } \over 2} = 0 \Rightarrow \lambda = 1$$</p>
<p>Required plane is</p>
<p>$$ - 5x + 8y - 2z - 13 = 0$$</p>
<p>Or $$\overrightarrow r .\,( - 5\widehat i + 8\widehat j - 2\widehat k) = 13$$</p>
<p>$${{|13\overrightarrow a {|^2}} \over {|d{|^2}}} = {{{{13}^2}} \over {{{(13)}^2}}}.\,|\overrightarrow a {|^2} = 93$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l56r99j8 | maths | 3d-geometry | lines-and-plane | <p>Let the foot of the perpendicular from the point (1, 2, 4) on the line $${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$$ be P. Then the distance of P from the plane $$3x + 4y + 12z + 23 = 0$$ is :</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "$${{50} \\over {13}}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$${{63} \\over {13}}$$"}] | ["A"] | null | <p>$$L:{{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3} = t$$</p>
<p>Let P = (4t $$-$$ 2, 2t + 1, 3t $$-$$ 1)</p>
<p>$$\because$$ P is the foot of perpendicular of (1, 2, 4)</p>
<p>$$\therefore$$ $$4(4t - 3) + 2(2t - 1) + 3(3t - 5) = 0$$</p>
<p>$$ \Rightarrow 29t = 29 \Rightarrow t = 1$$</p>
<p>$$\therefore$$ P = (2, 3, 2)</p>
<p>Now, distance of P from the plane</p>
<p>$$3x + 4y + 12z + 23 = 0$$, is</p>
<p>$$\left| {{{6 + 12 + 24 + 23} \over {\sqrt {9 + 16 + 144} }}} \right| = {{65} \over {13}} = 5$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l58a1l78 | maths | 3d-geometry | lines-and-plane | <p>Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x $$-$$ 3y + 5z = 8. If the mirror image of the point $$\left( {2, - {1 \over 2},2} \right)$$ in the rotated plane is B(a, b, c), then :</p> | [{"identifier": "A", "content": "$${a \\over 8} = {b \\over 5} = {c \\over { - 4}}$$"}, {"identifier": "B", "content": "$${a \\over 4} = {b \\over 5} = {c \\over { - 2}}$$"}, {"identifier": "C", "content": "$${a \\over 8} = {b \\over { - 5}} = {c \\over 4}$$"}, {"identifier": "D", "content": "$${a \\over 4} = {b \\over 5} = {c \\over 2}$$"}] | ["A"] | null | <p>Consider the equation of plane,</p>
<p>$$P:(2x + 3y + z + 20) + \lambda (x - 3y + 5z - 8) = 0$$</p>
<p>$$P:(2 + \lambda )x + 3(3 - 3\lambda )y + 1(1 + 5\lambda )z + (20 - 8\lambda ) = 0$$</p>
<p>$$\because$$ Plane P is perpendicular to $$2x + 3y + z + 20 = 0$$</p>
<p>So, $$4 + 2\lambda + 9 - 9\lambda + 1 + 5\lambda = 0$$</p>
<p>$$ \Rightarrow \lambda = 7$$</p>
<p>$$P:9x - 18y + 36z - 36 = 0$$</p>
<p>or $$P:x - 2y + 4z = 4$$</p>
<p>If image of $$\left( {2, - {1 \over 2},2} \right)$$ in plane P is (a, b, c) then</p>
<p>$${{a - 2} \over 1} = {{b + {1 \over 2}} \over { - 2}} = {{c - 2} \over 4}$$</p>
<p>and $$\left( {{{a + 2} \over 2}} \right) - 2\left( {{{b - {1 \over 2}} \over 2}} \right) + 4\left( {{{c + 2} \over 2}} \right) = 4$$</p>
<p>Clearly $$a = {4 \over 3}$$, $$b = {5 \over 6}$$ and $$c = - {2 \over 3}$$</p>
<p>So, $$a:b:c = 8:5: - 4$$</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58g7um7 | maths | 3d-geometry | lines-and-plane | <p>If the lines $$\overrightarrow r = \left( {\widehat i - \widehat j + \widehat k} \right) + \lambda \left( {3\widehat j - \widehat k} \right)$$ and $$\overrightarrow r = \left( {\alpha \widehat i - \widehat j} \right) + \mu \left( {2\widehat i - 3\widehat k} \right)$$ are co-planar, then the distance of the plane containing these two lines from the point ($$\alpha$$, 0, 0) is :</p> | [{"identifier": "A", "content": "$${2 \\over 9}$$"}, {"identifier": "B", "content": "$${2 \\over 11}$$"}, {"identifier": "C", "content": "$${4 \\over 11}$$"}, {"identifier": "D", "content": "2"}] | ["B"] | null | <p>$$\because$$ Both lines are coplanar, so</p>
<p>$$\left| {\matrix{
{\alpha - 1} & 0 & { - 1} \cr
0 & 3 & { - 1} \cr
2 & 0 & { - 3} \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow \alpha = {5 \over 3}$$</p>
<p>Equation of plane containing both lines</p>
<p>$$\left| {\matrix{
{x - 1} & {y + 1} & {z - 1} \cr
0 & 3 & { - 1} \cr
2 & 0 & { - 3} \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow 9x + 2y + 6z = 13$$</p>
<p>So, distance of $$\left( {{5 \over 3},0,0} \right)$$ from this plane</p>
<p>$$ = {2 \over {\sqrt {81 + 4 + 36} }} = {2 \over {11}}$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l5aiu9gk | maths | 3d-geometry | lines-and-plane | <p>Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, $$-$$1, $$-$$1), parallel to the line PQ meets the plane S at R, then QR<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "11"}] | ["B"] | null | <p>As L is parallel to PQ d.r.s of S is <1, 1, 1></p>
<p>$$\therefore$$ $$L \equiv {{x - 1} \over 1} = {{y + 1} \over 1} = {{z + 1} \over 1}$$</p>
<p>Point of intersection of L and S be $$\lambda$$</p>
<p>$$ \Rightarrow (\lambda + 1) + (\lambda - 1) + (\lambda - 1) = S$$</p>
<p>$$ \Rightarrow \lambda = 2$$</p>
<p>$$\therefore$$ $$R \equiv (3,1,1)$$</p>
<p>Let $$Q(\alpha ,\beta ,\gamma )$$</p>
<p>$$ \Rightarrow {{\alpha - 1} \over 1} = {\beta \over 1} = {{\gamma - 1} \over 1} = {{ - 2( - 3)} \over 3}$$</p>
<p>$$ \Rightarrow \alpha = 3,\,\beta = 2,\,\gamma = 3$$</p>
<p>$$ \Rightarrow Q \equiv (3,2,3)$$</p>
<p>$${(QR)^2} = {0^2} + {(1)^2} + {(2)^2} = 5$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5ajvyuf | maths | 3d-geometry | lines-and-plane | <p>Let the lines</p>
<p>$${L_1}:\overrightarrow r = \lambda \left( {\widehat i + 2\widehat j + 3\widehat k} \right),\,\lambda \in R$$</p>
<p>$${L_2}:\overrightarrow r = \left( {\widehat i + 3\widehat j + \widehat k} \right) + \mu \left( {\widehat i + \widehat j + 5\widehat k} \right);\,\mu \in R$$,</p>
<p>intersect at the point S. If a plane ax + by $$-$$ z + d = 0 passes through S and is parallel to both the lines L<sub>1</sub> and L<sub>2</sub>, then the value of a + b + d is equal to ____________.</p> | [] | null | 5 | <p>As plane is parallel to both the lines we have d.r's of normal to the plane as <7, $$-$$2, $$-$$1></p>
<p>$$\left( {from\,\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 3 \cr
1 & 1 & 5 \cr
} } \right| = 7\widehat i - \widehat j(2) + \widehat k( - 1)} \right)$$</p>
<p>Also point of intersection of lines is $$2\widehat i + 4\widehat j + 6\widehat k$$</p>
<p>$$\therefore$$ Equation of plane is</p>
<p>$$7(x - 2) - 2(y - 4) - 1(z - 6) = 0$$</p>
<p>$$ \Rightarrow 7x - 2y - z = 0$$</p>
<p>$$a + b + d = 7 - 2 + 0 = 5$$</p> | integer | jee-main-2022-online-25th-june-morning-shift |
1l5w09pb2 | maths | 3d-geometry | lines-and-plane | <p>The distance of the point (3, 2, $$-$$1) from the plane $$3x - y + 4z + 1 = 0$$ along the line $${{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}$$ is equal to :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l65uoj8c/397fdd39-a2f5-49e7-9863-c285adbe3668/d1a0d9c0-0ee6-11ed-a7de-eff776fdb55c/file-1l65uoj8d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l65uoj8c/397fdd39-a2f5-49e7-9863-c285adbe3668/d1a0d9c0-0ee6-11ed-a7de-eff776fdb55c/file-1l65uoj8d.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - 3D Geometry Question 138 English Explanation"></p>
<p>Line PQ is parallel to line $${{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}$$</p>
<p>$$\therefore$$ DR of PQ = DR of line = <$$-$$2, 2, 1></p>
<p>$$\therefore$$ Equation of line PQ passing through P(3, 2, $$-$$1) and DR = <$$-$$2, 2, 1> is</p>
<p>$${{x - 3} \over { - 2}} = {{y - 2} \over 2} = {{z + 1} \over 1}$$</p>
<p>Any General point on line PQ = (x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>)</p>
<p>$$\therefore$$ $${{{x_1} - 3} \over { - 2}} = {{{y_1} - 2} \over 2} = {{{z_1} + 1} \over 1} = \lambda $$</p>
<p>$$ \Rightarrow {x_1} = - 2\lambda + 3$$</p>
<p>$${y_1} = 2\lambda + 2$$</p>
<p>$${z_1} = \lambda - 1$$</p>
<p>$$\therefore$$ Point Q = ($$-$$2$$\lambda$$ + 3, 2$$\lambda$$ + 2, $$\lambda$$ $$-$$ 1)</p>
<p>Point Q lies on the plane $$3x - y + 4z + 1 = 0$$. So point Q satisfy the equation.</p>
<p>$$3( - 2\lambda + 3) - (2\lambda + 2) + 4(\lambda - 1) + 1 = 0$$</p>
<p>$$ \Rightarrow - 6\lambda + 9 - 2\lambda - 2 + 4\lambda - 4 + 1 = 0$$</p>
<p>$$ \Rightarrow - 4\lambda + 4 = 0$$</p>
<p>$$ \Rightarrow \lambda = 1$$</p>
<p>$$\therefore$$ Point Q = ($$-$$2 $$\times$$ 1 + 3, 2 $$\times$$ 1 + 2, 1 $$-$$ 1)</p>
<p>= (1, 4, 0)</p>
<p>$$\therefore$$ Distance of the point P(3, 2, $$-$$1) from the plane = Length of PQ</p>
<p>$$ = \sqrt {{{(3 - 1)}^2} + {{(2 - 4)}^2} + {{( - 1 - 0)}^2}} $$</p>
<p>$$ = \sqrt {{2^2} + {{( - 2)}^2} + {{( - 1)}^2}} $$</p>
<p>$$ = \sqrt {4 + 4 + 1} $$</p>
<p>$$ = \sqrt 9 $$</p>
<p>$$ = 3$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
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