datavorous/entrance-exam-dataset huggingface dataset:
The distance between the two charges $25 \mu \mathrm{C}$ and $36 \mu\mathrm{C}$ is $11 \mathrm{~cm}$ At what point on the line joining the two, the intensity will be zero

Options:
A At a distance of $5 \mathrm{~cm}$ from $25 \mu \mathrm{C}$
B At a distance of $5 \mathrm{~cm}$ from $36 \mu \mathrm{C}$
C At a distance of $10 \mathrm{~cm}$ from $25 \mu \mathrm{C}$
D At a distance of $11 \mathrm{~cm}$ from $36 \mu \mathrm{C}$

The correct answer is:
At a distance of $5 \mathrm{~cm}$ from $25 \mu \mathrm{C}$

Solution:
Let the point porn line whence intensity is zeno.
Then intensity between $25\mu c$ and $36 \mu c$
So distance b/w $P$ and 36 be $x$ and distance b/w p and 25 be $(11-x)$
$\frac{k 36}{x^2}=\frac{k(25)}{(11-x)^2}$
$\frac{11-x}{x}=\frac{5}{6}$
$x=6$
and $11-x=5$

Topic: Physics, Electrostatics, JEE Main
The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $\mathrm{C}$ is its initial capacitance, its final capacitance is equal to:

Options:
A $2 \mathrm{C}$
B $C / 2$
C $4 \mathrm{C}$
D $\mathrm{C} / 4$

The correct answer is:
$\mathrm{C} / 4$

Solution:
The capacitance of parallel plate capacitor,
$\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{d}$
(where, the symbols have their usual meanings.)
Given, $\quad d^{\prime}=2 \mathrm{~d} \& \mathrm{~A}^{\prime}=\frac{\mathrm{A}}{2}$
Then, $\mathrm{C}^{\prime}=\frac{\varepsilon_0\left(\frac{\mathrm{A}}{2}\right)}{2 d}=\frac{\varepsilon_0 \mathrm{~A}}{4 d}=\frac{\mathrm{C}}{4}$

Topic: Physics, Capacitance, JEE Main
The distance between the two points A and A' which lie on y=2 such that both the line segments AB and A'B (where B is the point 2,3) subtend angle π4 at the origin, is equal to

Options:
A 10
B 485
C 525
D 3

The correct answer is:
525

Solution:
Let the slope of OB be m1=32

and the slope of OA be m2=2x

Now angle between two lines, tanπ4=32-2x1+62x

i.e. 32-2x1+62x=1⇒3x-42x+6=1

3x-4=±2x+6

⇒x1=10, x2=-25

⇒AA'=10+25=525

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The distance between the virtual image from its real object is 60 cm in case of a concave mirror and the image formed is 3 times magnified. What is the magnitude of focal length (in cm) of the mirror?

Options:

The correct answer is:
22.5

Solution:
x + 3x=60 ⇒x=15 cm
u=-15 cm
m=+ 3
∴ m=ff-u⇒ + 3=ff-(-15)
∴ 3f+ 45=f ⇒f=- 22.5 cm

f=22.5 cm

Topic: Physics, Ray Optics, JEE Main
The distance between two particles is decreasing at the rate of $6 \mathrm{~m} / \mathrm{sec}$. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of $4 \mathrm{~m} / \mathrm{sec}$. The particles have speeds as

Options:
A $5 \mathrm{~m} / \sec : 1 \mathrm{~m} / \mathrm{sec}$
B $4 \mathrm{~m} / \sec : 1 \mathrm{~m} / \mathrm{sec}$
C $4 \mathrm{~m} / \sec : 2 \mathrm{~m} / \mathrm{sec}$
D $5 \mathrm{~m} / \sec : 2 \mathrm{~m} / \mathrm{sec}$

The correct answer is:
$5 \mathrm{~m} / \sec : 1 \mathrm{~m} / \mathrm{sec}$

Solution:
When two particles moves towards each other then $v_1+v_2=6$....(i)
When these particles moves in the same direction then $v_1-v_2=4$....(ii) By solving $v_1=5$ and $v_2=1 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In Two Dimensions, JEE Main
The distance between two plates of a capacitor is d and its capacitance is C1, when air is the medium between the plates. If a metal sheet of thickness 2d3 and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes C2 . The ratio C2C1 is

Options:
A 3 : 1
B 2 : 1
C 4 : 1
D 1 : 1

The correct answer is:
3 : 1

Solution:
The formula of capacitance is C1=ε0Ad

The dielectric constant of metal sheet is Kmetal=∞

The new value of capacitance is

C2=ε0Ad-t+tKmetal=ε0Ad-2d3+0=3C1

Clearly, the ratio,

C2C1=3C1C1=3:1

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The distance from the origin, of the normal to the curve, x=2cos⁡t+2tsin⁡t, y=2sin⁡t-2tcos⁡t at t=π4, is :

Options:
A 2
B 22
C 4
D 2

The correct answer is:
2

Solution:
At t= π 4 ,
x=212+2π4.12=2+π22=4+π22
y=212-2π4 ⋅ 12 = 2- π22=4-π22
dydx=2cos⁡t-2 cos⁡t+t (-sin⁡t)=2tsin⁡t
dxdt= -2sin⁡t+2 sin⁡t+t⋅cos⁡t=2tcos⁡t
dydx=tan⁡t at t=π4 ⇒ dydx=tan ⁡π4=1
dydx=1 .

Slope of tangent is 1 & therefore slope of normal would be -1.
Equation of normal y- 4-π22= -1 x- 4+π22
x+y=4+π22+ 4-π22
⇒x+y=822

Distance from origin = -8222=2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2015 (10 Apr Online)
The distance moved by the screw of a screw gauge is 2 mm in four rotations and there are 50 divisions on its cap. When nothing is put between its jaws, 30th division of circular scale coincides with the reference line, with zero of circular scale lying above the reference line. When a plate is placed between the jaws, the main scale reads 2 divisions and the circular scale reads 20 divisions. The thickness of the plate is

Options:
A 1.5 mm
B 1.2 mm
C 1.4 mm
D 1.6 mm

The correct answer is:
1.5 mm

Solution:
Pitch = 1 2 mm
least count = 0.5 mm 50 =0.01 mm
Reading =[ 2× 1 2 +20×0.01 −( −30×0.01 ) ]=1.5 mm

Topic: Physics, Experimental Physics, NEET
The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum $\mathrm{P}$ is $\mathrm{d}$. The distance of closest approach of alpha particle to nucleus, if the linear momentum of the alpha particle is $1.5 \mathrm{P}$

Options:
A $\frac{2 \mathrm{~d}}{3}$
B $\frac{3 \mathrm{~d}}{2}$
C $\frac{4 d}{9}$
D $\frac{9 \mathrm{~d}}{4}$

The correct answer is:
$\frac{4 d}{9}$

Solution:
At sufficient distance, electric potential energy is zero. So, $\mathrm{U}_1=0$
at (1) $\mathrm{K}_1=\frac{1}{2} \mathrm{mv}^2=\frac{\mathrm{P}^2}{2 \mathrm{~m}}$
at (2) $\mathrm{K}_2=0$
$\mathrm{U}_2=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_\alpha \mathrm{q}_{\mathrm{n}}}{\mathrm{d}_{\mathrm{c}}}$
By energy conservation
$\begin{aligned} & \mathrm{U}_1+\mathrm{K}_1=\mathrm{U}_2+\mathrm{K}_2 \\ & 0+\frac{\mathrm{p}^2}{2 \mathrm{~m}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_\alpha \mathrm{q}_{\mathrm{n}}}{\mathrm{d}_{\mathrm{c}}}+0\end{aligned}$
Distance of closest approach, $d_c \propto \frac{1}{p^2}$
$\begin{aligned} & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{p}_1^2}{\mathrm{p}_2^2} \\ & =\mathrm{d} \frac{\mathrm{p}^2}{(1.5 \mathrm{p})^2}=\frac{4 \mathrm{~d}}{9}\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
The distance of line 3y-2z-1=0=3x-z+4 from the point (2,-1,6) is :

Options:
A 25
B 26
C 26
D 42

The correct answer is:
26

Solution:
Given equation of the line 3y-2z-1=0=3x-z+4

⇒3y-12=z-01=3x+41

⇒x+4313=y-1323=z-01

Here, Point on the line is P(-43,13,0) and the vector in the direction of line is 13i^+23j^+k^ then

from the above diagram PR=13(2+43)+23(-1-13)+1(6-0)19+49+1

PR=4149

QR2=PQ2-PR2

=1009+169+36-2249

QR=24=26

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The distance of the centres of moon and earth is D. The mass of earth is 81 times the mass of the moon. At what distance from the centre of the earth, the gravitational force will be zero?

Options:
A $\frac{D}{2}$
B $\frac{2 D}{3}$
C $\frac{4 D}{3}$
D $\frac{9 D}{10}$

The correct answer is:
$\frac{9 D}{10}$

Solution:
$$
\begin{aligned}
& \text { ( } \frac{G m_e}{x^2}=\frac{G m_m}{(D-x)^2} \\
& \text { or } \frac{G(81 m)}{x^2}=\frac{m}{(D-x)^2} \\
& \therefore \quad x=\frac{9 D}{10} .
\end{aligned}
$$

Topic: Physics, Gravitation, JEE Main
The distance of the origin from the centroid of the triangle whose two sides have the equations x-2y+1=0 and 2x-y-1=0 and whose orthocenter is 73, 73 is:

Options:
A 2
B 2
C 22
D 4

The correct answer is:
22

Solution:
Let AB≡x-2y+1=0 and AC≡2x-y-1=0

So, A1, 1

Now, altitude from B which is perpendicular to line 2x-y-1=0 and which passes through 73,73 is BH=x+2y-7=0,

Now intersection of BH & AB will give point B3, 2

Similarly, altitude from C is CH=2x+y-7=0⇒C2, 3

Centroid of △ABC=E2, 2 & OE=22

where O is origin.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The distance of the plane x+2y-z=2 from the point 2,-1,3, measured in the direction with the direction ratios 2,2,1 is

Options:
A 1 units
B 2 units
C 3 units
D 4 units

The correct answer is:
3 units

Solution:
Direction ratios given are 2,2,1
The line is x-22=y+12=z-31=r
Any general point on this line is 2r+2,2r-1,r+3
If it lies on the plane x+2y-z=2, then
2r+2+22r-1-r+3=2
⇒2r+4r-r+2-2-3=2
⇒5r=5
⇒r=1
The point on the plane is 4,1,4 and its distance from 2,-1,3 is 4-22+1+12+4-32=4+4+1=3

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The distance of the point 1, 0, 2 from the point of intersection of the line x-23=y+14=z-212 and the plane x-y+z=16, is

Options:
A 1 3
B 2 1 4
C 8
D 3 2 1

The correct answer is:
1 3

Solution:
Let x-23=y+14=z-212=t
General points is (2+3t,- 1+4t,2+12t)
It lies on the plane x-y+z=16.

⇒2+3t+1-4t+2+12t=16

⇒t=1
Hence, the point of intersection will be
(2+3(1),-1+4(1),2+12(1))=5,3,14
Distance from (1,0,2) = 5 - 1 2 + 3 - 0 2 + 1 4 - 2 2

= 4 2 + 3 2 + 1 2 2

= 1 3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2015 (04 Apr)
The distance of the point 1,1,9 from the point of intersection of the line x-31=y-42=z-52 and the plane x+y+z=17 is:

Options:
A 192
B 219
C 38
D 38

The correct answer is:
38

Solution:
Let x-31=y-42=z-52=t

⇒ x=3+t, y=2t+4, z=2t+5

For a point of intersection with x+y+z=17, then it satisfies the equation of a plane.

⇒3+t+2 t+4+2 t+5=17

⇒ 5t=5⇒t=1

⇒ The point of intersection is 4,6,7.

Then, distance between 1,1,9 and 4,6,7 is 9+25+4=38.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The distance of the point â1,â1 from the line 12 x + 6 =5 yâ2 is

Options:

The correct answer is:
5

Solution:
The given line is 12 x + 6 = 5 yâ2
â12xâ5y+82=0
The perpendicular distance from x 1 , y 1 to the line axâ+by+c=0 is a x 1 + b y 1 + c a 2 + b 2
The point x 1 , y 1 is â1,â1 , therefore, pependicular distance from â1,â1 to the line 12xââ5y+82=0 is
= - 1 2 - 5 + 8 2 1 2 2 + - 5 2 = 6 5 1 4 4 + 2 5 = 6 5 1 6 9
= 6 5 1 3 = 5

Topic: Mathematics, Straight Lines, JEE Main
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 :

Options:
A 12
B 422
C 52
D 342

The correct answer is:
342

Solution:
P1: 2x+3y+2z=0

⇒ n1→=2i^+3j^+2k^

P2: x-2y+z=0

⇒ n2→=i^-2j^+k

Let, line L is the line of intersection of P1 & P2

The direction vector of the line is

r→=n→1×n→2=7i^-7k^

DR's of L are 1,0,-1 and origin is a common point on both the planes

⇒ Equation of L: x1=y0=z-1=λ

Now, DR's of PQ→=λ+1, -2, 2-λ

∵ PQ→⊥r→

⇒ λ+11+-20+2-λ-1=0

⇒ λ=12

⇒ Q12,0,-12

Now, PQ=-1-122+2-02+-2+122

⇒ PQ=342

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The distance of the point 1,2,3 from the plane x+y-z=5 measured along the straight line x=y=z is

Options:
A 53 units
B 103 units
C 33 units
D 35 units

The correct answer is:
53 units

Solution:
Equation of line parallel to the line x=y=z and passing through (1, 2, 3) is x-1=y-2=z-3
Coordinates of a general point on this line is
Î»+1,Î»+2,Î»+3 .
It meets the given plane x+y-z=5
âÎ»=5
So, the point of intersection is 6,7,8
Hence, distance =53

Topic: Mathematics, Three Dimensional Geometry, JEE Main
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

Options:
A 2
B 5
C 3
D 1

The correct answer is:
1

Solution:
Given, equation of plane x-y+z=5,

Point is 1,-2,3

And the direction ratio of plane parallel to the given plane is 2,3,-6

Therefore, Equation of plane when D.R of parallel plane is given is,x-12=y+23=z-3-6=λ

⇒x=2λ+1 , y=3λ-2 , z=-6λ+3

since, x-y+z=5

⇒2λ+1-3λ+2-6λ+3=5

⇒λ=17

⇒x=97 , y=-117 , z=157

Distance between points 1,-2,3&97,-117,157=272+372+-672=1 unit

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The distance of the point (1,-2,3) from the plane x-y+z=5 measured parallel to the line x2=y3=z-6 is :

Options:
A 75
B 1
C 17
D 7

The correct answer is:
1

Solution:
Equation PQ x-12=y+23=z-3-6=λ
Let Q≡(2λ+1,3λ-2,-6λ+3)
Q lies on x-y+z=5
⇒ (2λ+1)-(3λ-2)+(-6λ+3)=5⇒λ=17⇒Q≡97,-117,157
PQ=1-972+-2+1172+3-1572=1

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The distance of the point 1, -2, 4 from the plane passing through the point 1, 2, 2 and perpendicular to the planes x-y+2z=3 and 2x-2y+z+12=0, is :

Options:
A 2
B 2
C 22
D 12

The correct answer is:
22

Solution:
Let equation of plane be
ax-1+by-2+cz-2=0 .......i
Since (i) is perpendicular to the given planes, then

Dot product of the normal vectors of the perpendicular planes is 0.
a-b+2c=0
2a-2b+c=0
Solving above equations c=0 and a=b equation of plane i becomes x+y-3=0
Distance from 1, -2, 4 will be
D=1-2-31+1=42=22

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2016 (09 Apr Online)
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 x-11=y+2-2=z-43 and x-22=y+1-1=z+7-1 , is:

Options:
A 2074
B 1083
C 583
D 1074

The correct answer is:
1083

Solution:
Let l, m, n be the direction cosines of the line normal to the plane, then
l-2m+3n=0 and 2l-m-n=0
⇒l2+3=m6+1=n-1+4=λ
⇒l=5λ, m=7λ, n=3λ
∴ The equation of the plane is
5x+7y+3y+d=0
∵ it passes through 1,-1,-1
⇒5-7-3+d=0
⇒d=5
Hence, the equation of the plane is 5x+7y+3y+5=0
Now, PQ=5+21-21+552+72+32

⇒PQ=1025+49+9

⇒PQ=1083

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2017 (02 Apr)
The distance of the point 1,-5, 9 from the plane x-y+z=5 measured along the line x=y=z is

Options:
A 103
B 203
C 310
D 103

The correct answer is:
103

Solution:
Equation of line parallel to x=y=z through
1,-5, 9 is x-11=y+51=z-91=λ
If P λ+1, λ-5, λ+9 be point of intersection of line and plane.

⇒ λ+1-λ+5+λ+9=5
⇒ λ= -10
⇒ Coordinate points are -9, -15, -1
⇒ Required distance =λ12+12+12=103

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2016 (03 Apr)
The distance of the point -1,9,-16 from the plane 2x+3y-z=5 measure parallel to the line x+43=2-y4=z-312 is

Options:
A 132
B 31
C 26
D 203

The correct answer is:
26

Solution:
Equation of line passing through -1,9,-16 and parallel to x+43=y-2-4=z-312 is

x+13=y-9-4=z+1612=λ

Any point on this line is P3λ-1, -4λ+9, 12λ-16

For finding point of intersection of line and a plane, we have

6λ-2-12λ+27-12λ+16=5

⇒λ=2

So, P5,1,8

Required distance is

=5+12+1-92+8+162

= 36+64+576=26 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The distance of the point $(2,3,4)$ from the line $1-x=\frac{y}{2}=\frac{1}{3}(1+z)$ is

Options:
A $\frac{1}{7} \sqrt{35}$
B $\frac{4}{7} \sqrt{35}$
C $\frac{2}{7} \sqrt{35}$
D $\frac{3}{7} \sqrt{35}$

The correct answer is:
$\frac{3}{7} \sqrt{35}$

Solution:
Required distance $\begin{aligned} & =\sqrt{\sum\left(x_1-x_2\right)^2-\left[\sum l\left(x_1-x_2\right)\right]^2} \\ = & \frac{3}{7} \sqrt{35}\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The distance of the point (2,3) from the line 2x-3y+28=0, measured parallel to the line 3x-y+1=0, is equal to

Options:
A 42
B 63
C 3+42
D 4+63

The correct answer is:
4+63

Solution:
Let, L1:3x-y+1=0 and L2:2x-3y+28=0.

Slope, m1=3=tanθ

⇒θ=60°

⇒sinθ=32, cosθ=12

Now, equation of line passing through 2,3 and having slope 3 is,

x-212=y-332=r

⇒2x-3y+28=0

⇒4+r-9-332r+28=0

⇒2-332r+23=0

⇒r=4633-2×33+233+2

⇒r=4623×33+21=4+63

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The distance of the point 6,-22 from the common tangent y=mx+c,m>0, of the curves x=2y2 and x=1+y2 is

Options:
A 13
B 5
C 143
D 53

The correct answer is:
5

Solution:
Given,

Equation of the curves x=2y2 and x=1+y2

Now for parabola y2=x2, equation of tangent is given by y=mx+18m

Which is also tangent to parabola y2+1=x,

So, y=mx+18m will satisfy the curve y2+1=x,

Hence, mx+18m2+1=x

⇒m2x2+164m2+x4+1=x

⇒m2x2-3x4+1+164m2=0

Now making discriminant zero we get,

D=0⇒342-4×m2×1+164m2

⇒916-4m2+116=0

⇒m=122

Hence, the equation of tangent is x-22y+1=0

Now distance of point 6,-22 from the tangent will be, d=6+8+19=5

Topic: Mathematics, Parabola, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The distance, of the point (7,-2,11) from the line x-61=y-40=z-83 along the line x-52=y-1-3=z-56, is :

Options:
A 12
B 14
C 18
D 21

The correct answer is:
14

Solution:
Given lines are x-61=y-40=z-83 and x-52=y-1-3=z-56.

Let, AB be a line parallel to x-52=y-1-3=z-56 which intersects x-61=y-40=z-83 at point B, where A≡7,-2,11

So, equation of AB is,

x-72=y+2-3=z-116=λlet

⇒x=2λ+7, y=-3λ-2, z=6λ+11

⇒B≡(2λ+7,-3λ-2,6λ+11)

Also, the point B lies on x-61=y-40=z-83.

⇒2λ+7-61=-3λ-2-40=6λ+11-83

⇒-3λ-2-4=0

⇒-3λ=6

⇒λ=-2

⇒B≡(3,4,-1)

⇒AB=(7-3)2+(4+2)2+(11+1)2

⇒AB=16+36+144

⇒AB=196

⇒AB=14

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The distance of the point 7,-3,-4 from the plane containing the points 2,-3,1, -1,1,-2 and 3,-4,2 is equal to:

Options:
A 4
B 5
C 52
D 42

The correct answer is:
52

Solution:
Equation of plane passing through the points 2,-3,1, -1,1,-2 and 3,-4,2 is

x-2y+3z-1-34-31-11=0

⇒x-2-z+1=0

⇒x-z-1=0

Required distance is

=7+4-12

=102=52 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
The given line is \(12(x+6)=5(y-2)\)
\(\Rightarrow 12 x+72=5 y-10\)
or \(12 x-5 y+72+10=0 \Rightarrow 12 x-5 y+82=0\) The perpendicular distance from \(\left(x_1, y_1\right)\) to the line \(a x+b y+c=0\) is \(\frac{\left(\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}\right)}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\). The point \(\left(x_1, y_1\right)\) is \((-1,1)\), therefore, perpendicular distance from \((-1,1)\) to the line \(12 x-5 y+82=0\) is
\(=\frac{|-12-5+82|}{\sqrt{12^2+(-5)^2}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=5\)

Topic: Mathematics, Straight Lines, JEE Main
The distance of the point 2,3,2 from the plane 3x+4y+4z=23 measured parallel to the line x+31=y-6-2=z-11 is

Options:
A 108 units
B 12 units
C

54 units

D 236 units

The correct answer is:

54 units

Solution:
Equation of the line passing through 2,3,2 and parallel to the given line is x-21=y-3-2=z-21
Any general point on this line is Î»+2,-2Î»+3,Î»+2
This must satisfy the given equation of the plane
â3Î»+6-8Î»+12+4Î»+8=23
â-Î»+26=23âÎ»=3
The point on the plane is 5,-3,5

Hence, the required distance =9+36+9=54

units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The distance of the point having position vector -i^+2j^+6k^ from the straight line passing through the point 2, 3, -4 and parallel to the vector, 6i^+3j^-4k^ is

Options:
A 43
B 6
C 213
D 7

The correct answer is:
7

Solution:
Let, the given point be P≡-i^+2j^+6k^ and A≡2, 3, -4, hence the position vector of the point A is 2i^+3j^-4k^.

We know that the vector joining two points x1, y1, z1 and x2, y2, z2 is x2-x1i^+y2-y1j^+z2-z1k^,

Hence, the vector AP→=-1-2i^+2-3j^+6+4k^

⇒AP→=-3i^-j^+10k^

And, AD is the projection of AP→ on the given line i.e. 6i^+3j^-4k^

We know that the projection of a vector a→ on a vector b→ is a→·b→b→

Hence, AD=AP→⋅n→n→

⇒AD=-3i^-j^+10k^·6i^+3j^-4k^62+32+-42

⇒AD=-18-3-4036+9+16

⇒AD=61 units

The distance between the points x1, y1, z1 and x2, y2, z2 is x1-x22+y1-y22+z1-z22

Thus, AP=-1-22+2-32+6+42=110 units

Now, in triangle ADP, by Pythagoras theorem, we have AP2=PD2+AD2

⇒PD=AP2-AD2=110-61=7 units.

Topic: Mathematics, Vector Algebra, JEE Main
The distance of the point having position vector -i^+2j^+6k^ from the straight line passing through the point 2, 3, -4 and parallel to the vector, 6i^+3j^-4k^ is

Options:
A 43
B 6
C 213
D 7

The correct answer is:
7

Solution:
Let, the given point be P≡-i^+2j^+6k^ and A≡2, 3, -4, hence the position vector of the point A is 2i^+3j^-4k^.

We know that the vector joining two points x1, y1, z1 and x2, y2, z2 is x2-x1i^+y2-y1j^+z2-z1k^,

Hence, the vector AP→=-1-2i^+2-3j^+6+4k^

⇒AP→=-3i^-j^+10k^

And, AD is the projection of AP→ on the given line i.e. 6i^+3j^-4k^

We know that the projection of a vector a→ on a vector b→ is a→·b→b→

Hence, AD=AP→⋅n→n→

⇒AD=-3i^-j^+10k^·6i^+3j^-4k^62+32+-42

⇒AD=-18-3-4036+9+16

⇒AD=61 units

The distance between the points x1, y1, z1 and x2, y2, z2 is x1-x22+y1-y22+z1-z22

Thus, AP=-1-22+2-32+6+42=110 units

Now, in triangle ADP, by Pythagoras theorem, we have AP2=PD2+AD2

⇒PD=AP2-AD2=110-61=7 units.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The distance of the point $-\hat{i}+2 \hat{j}+6 \hat{k}$ from the straight line that passes through the point $2 \hat{i}+3 \hat{j}-4 \hat{k}$ and is parallel to the vector $6 \hat{i}+3 \hat{j}-4 \hat{k}$ is

Options:
A
9

B
8

C
7

D
10

The correct answer is:
7

Solution:
Point is $(-1,2,6)$
Line passes through the point $(2,3,-4)$ parallel to vector whose direction ratios is $6,3,-4$.
Equation is $\frac{x-2}{6}=\frac{y-3}{3}=\frac{z+4}{-4}=\lambda$
Any point on this line is given by $x=6$ $\lambda+2, y=3 \lambda+3, z=-4 \lambda-4$
Now, d. Rs of line passing through $(-1,2,6)$ and $\perp$ to this line is $\{(x+1),(y-2),(z-6)\}$
So, $6(x+1)+3(y-2)-4(z-6)=0$ $\Rightarrow 6 x+3 y-4 z+24=0$
Now, $6(6 \lambda+2)+3(3 \lambda+3)+4(4 \lambda+4)$ $+24=0$
$\Rightarrow 61 \lambda+61=0 \Rightarrow \lambda=-1$
So, $x=-4, y=0, z=0$
Now, distance between $(-1,2,6)$ and $(-4,0,0)$ is
$\sqrt{9+4+36}=\sqrt{49}=7$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (26 May Online)
The distance of the point of intersection of the lines $2 x-3 y+5=0$ and $3 x+4 y=0$ from the line $5 x-2 y=0$ is

Options:
A $\frac{130}{17 \sqrt{29}}$
B $\frac{13}{7 \sqrt{29}}$
C $\frac{130}{17}$
D None of these

The correct answer is:
$\frac{130}{17 \sqrt{29}}$

Solution:
Points of intersection are $y=\frac{15}{17}, x=\frac{-4 \times 15}{3 \times 17}=\frac{-20}{17}$
Therefore
$D=\left|\frac{5\left(\frac{-20}{17}\right)-2\left(\frac{15}{17}\right)}{\sqrt{29}}\right|=\frac{130}{17 . \sqrt{29}}$

Topic: Mathematics, Straight Lines, JEE Main
The distance of the point P3,4,4 from the point of intersection of the line joining the points Q3,-4,-5 and R2,-3,1 and the plane 2x+y+z=7, is equal to _____.

Options:

The correct answer is:
7

Solution:
We have, point P3,4,4

Equation of line joining the points Q3,-4,-5 and R2,-3,1 is x-3-1=y+41=z+56=r

Any point on above line is -r+3,r-4,6r-5

Now, satisfying it in the given plane 2x+y+z=7, we get

2-r+3+r-4+6r-5=7

⇒r=2

So, required point of intersection is T1,-2,7.

Hence, PT=3-12+4+22+4-72=7.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The distance of the point P4,6,-2 from the line passing through the point -3,2,3 and parallel to a line with direction ratios 3,3,-1 is equal to:

Options:
A 3
B 6
C 23
D 14

The correct answer is:
14

Solution:
The equation of line passing through the point -3,2,3 and parallel to a line with direction ratios 3,3,-1 will be,

x+33=y-23=z-3-1=λ

Now let any point on the line will be, M3λ-3,3λ+2,3-λ

We know the direction ratios of the line joining the points x1,y1,z1 and x2,y2,z2 is given by

x2-x1,y2-y1,z2-z1

Therefore, the direction ratios of line PM,

D.R of PM3λ-7,3λ-4,5-λ

Since, PM is perpendicular to the line

⇒ 33λ-7+33λ-4-15-λ=0

⇒ λ=2

⇒ M3,8,1

Now distance of point P from point M,

⇒PM=3-42+8-62+1+22

⇒PM=14

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (25 Jan Shift 1)