datavorous/entrance-exam-dataset huggingface dataset:
Let P be a plane containing the line x-13=y+64=z+52 and parallel to the line x-34=y-2-3=z+57. If the point (1,-1,α) lies on the plane P, then the value of 5α is equal to ___ .

Options:

The correct answer is:
38

Solution:
Equation of plane is x-1y+6z+53424-37=0

Now (1,-1,α) lies on it so

05α+53424-37=0⇒5α+38=0⇒5α=38

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let P be a plane lx+my+nz=0 containing the line, 1-x1=y+42=z+23. If plane P divides the line segment AB joining points A-3,-6,1 and B2,4,-3 in ratio k:1 then the value of k is equal to :

Options:
A 1.5
B 3
C 2
D 4

The correct answer is:
2

Solution:
Point C is

2k-3k+1,4k-6k+1,-3k+1k+1

x-1-1=y+42=z+23

Plane lx+my+nz=0

l-1+m2+n3=0

-l+2m+3n=0 ...(1)

It also satisfy point 1,-4,-2

l-4m-2n=0 ...(2)

Solving (1) and (2)

2m+3n=4m+2n

n=2m

l-4m-4m=0

l=8 m

l8=m1=n2

l:m:n=8:1:2

Plane is 8x+y+2z=0

It will satisfy point C

82k-3k+1+4k-6k+1+2-3k+1k+1=0

16k-24+4k-6-6k+2=0

14k=28 ∴k=2

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P be a plane lx+my+nz=0 containing the line, 1-x1=y+42=z+23. If plane P divides the line segment AB joining points A-3,-6,1 and B2,4,-3 in ratio k:1 then the value of k is equal to :

Options:
A 1.5
B 3
C 2
D 4

The correct answer is:
2

Solution:
Point C is

2k-3k+1,4k-6k+1,-3k+1k+1

x-1-1=y+42=z+23

Plane lx+my+nz=0

l-1+m2+n3=0

-l+2m+3n=0 ...(1)

It also satisfy point 1,-4,-2

l-4m-2n=0 ...(2)

Solving (1) and (2)

2m+3n=4m+2n

n=2m

l-4m-4m=0

l=8 m

l8=m1=n2

l:m:n=8:1:2

Plane is 8x+y+2z=0

It will satisfy point C

82k-3k+1+4k-6k+1+2-3k+1k+1=0

16k-24+4k-6-6k+2=0

14k=28 ∴k=2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Let P be a plane passing through the points (1,0,1),(1,-2,1) and (0,1,-2). Let a vector a→=αi^+βj^+γk^ be such that a→ is parallel to the plane P, perpendicular to i^+2j^+3k^ and a→·(i^+j^+2k^)=2, then (α-β+γ)2 equals______.

Options:

The correct answer is:
81

Solution:
Equation of plane is

x-1y-0z-11-121-11-00-11+2=0

⇒x-1yz-10201-13=0

⇒x-16-y0+z-1-2=0

⇒3x-1-z+1=0

⇒3x-z-2=0

Since, a→=αi^+βj^+γk^ is parallel to 3x-z-2=0, therefore normal vector of plane is perpendicular to a→, hence

⇒3α-γ=0 ...1

Now,

a→⊥i˙+2j^+3k^

⇒α+2β+3γ=0 ...2

And,

a→·(i^+j^+2k^)=0

⇒α+β+2γ=2 ...3

On solving 1, 2 & 3, we get

α=1, β=-5, γ=3

So, α-β+γ2=81

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P be a plane passing through the points (1,0,1),(1,-2,1) and (0,1,-2). Let a vector a→=αi^+βj^+γk^ be such that a→ is parallel to the plane P, perpendicular to i^+2j^+3k^ and a→·(i^+j^+2k^)=2, then (α-β+γ)2 equals______.

Options:

The correct answer is:
81

Solution:
Equation of plane is

x-1y-0z-11-121-11-00-11+2=0

⇒x-1yz-10201-13=0

⇒x-16-y0+z-1-2=0

⇒3x-1-z+1=0

⇒3x-z-2=0

Since, a→=αi^+βj^+γk^ is parallel to 3x-z-2=0, therefore normal vector of plane is perpendicular to a→, hence

⇒3α-γ=0 ...1

Now,

a→⊥i˙+2j^+3k^

⇒α+2β+3γ=0 ...2

And,

a→·(i^+j^+2k^)=0

⇒α+β+2γ=2 ...3

On solving 1, 2 & 3, we get

α=1, β=-5, γ=3

So, α-β+γ2=81

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let P be a plane passing through the points 2,1,0,4,1,1 and 5,0,1 and R be any point 2,1,6 .Then the image of R in the plane P is

Options:
A 6,5,2
B 6,5,-2
C 4,3,2
D 3,4,-2

The correct answer is:
6,5,-2

Solution:
Solving for Equation of Plane, we get, x+y-2z=3

Now applying relation to find Image of point about a Plane, we get,

x-21=y-11=z-6-2=-22+1-12-36⇒x,y,z=6,5,-2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Let $\mathrm{P}$ be a point inside a triangle $\mathrm{ABC}$ with $\angle \mathrm{ABC}=90^{\circ}$. Let $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ be the images of $\mathrm{P}$ under reflection in $\mathrm{AB}$ and $\mathrm{BC}$ respectively. The distance between the circumcenters of triangles $\mathrm{ABC}$ and $\mathrm{P}_{1} \mathrm{P} \mathrm{P}_{2}$ is

Options:
A $\frac{\mathrm{AB}}{2}$
B $\frac{\mathrm{AP}+\mathrm{BP}+\mathrm{CP}}{3}$
C $\frac{\mathrm{AC}}{2}$
D $\frac{\mathrm{AB}+\mathrm{BC}+\mathrm{AC}}{2}$

The correct answer is:
$\frac{\mathrm{AC}}{2}$

Solution:
$\mathrm{M}$ is circumcentre of $\triangle \mathrm{ABC}$
$\Rightarrow \mathrm{M}\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)$
$\& \mathrm{~N}$ is circumcentre of $\triangle \mathrm{ABC}$ $\mathrm{N}=(0,0)=\mathrm{B}\left(\right.$ Mid-point of $\left.\mathrm{P}_{1}, \& \mathrm{P}_{2}\right)$
So $\mathrm{MN}=\frac{\mathrm{AC}}{2}$

Topic: Mathematics, Straight Lines, JEE Main
Let \( P \) be a point on a variable hyperbola \( x^{2}-y^{2}=a^{2} \). Locus of \( P \) such that \( P \) is nearest to the line \( y=2 x \) is, a is a parameter

Options:
A \( x+2 y=2 \)
B \( x=4 y \)
C \( x-2 y=0 \)
D \( 2 x+y=2 \)

The correct answer is:
\( x-2 y=0 \)

Solution:
Let Ph, k=(a secâ¡θ, a tanâ¡θ) on x2-y2=a2
2x-2ydydx=0 ⇒ dydx=xy, dxdy at P=cosec θ
For nearest point cosec θ=2, θ= π6
∴ h=asecâ¡π6, k=a tanâ¡π6
h=a23, k=a 13, h=2k ⇒ x-2y=0

Topic: Mathematics, Hyperbola, JEE Main
Let $P$ be a point on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line through $P$ parallel to the $Y$ -axis mects the circle $x^{2}+y^{2}=9$ at $Q$ where $P, Q$ are on the same side of the $X$ -axis. If $R$ is a point on $P Q$ such that $\frac{P R}{R Q}=\frac{1}{2},$ then the locus of $R$ is

Options:
A $\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1$
B $\frac{x^{2}}{49}+\frac{y^{2}}{9}=1$
C $\frac{x^{2}}{9}+\frac{y^{2}}{49}=1$
D $\frac{9 x^{2}}{49}+\frac{y^{2}}{9}=1$

The correct answer is:
$\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1$

Solution:
Since, point $P$ on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
$\therefore$
$P(3 \cos \theta, 2 \sin \theta)$
Now, equation of line parallel of $Y$ -axis is
$$
x=3 \cos \theta
$$
and above line meets circle at $Q$
$\therefore$
$Q(3 \cos \theta, 3 \sin \theta)$
Given, $\frac{P R}{R Q}=\frac{1}{2}$
$\therefore h=\frac{3 \cos \theta+6 \cos \theta}{3}, k=\frac{3 \sin \theta+4 \sin \theta}{3}$
$\Rightarrow \quad h=3 \cos \theta, \quad k=\frac{7}{3} \sin \theta$
$\Rightarrow \quad \cos \theta=h / 3, \sin \theta=\frac{3 k}{7}$
Now, $\cos ^{2} \theta+\sin ^{2} \theta=h^{2} / 9+\frac{9 k^{2}}{49}=1$
Hence, locus of a point is $\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1$

Topic: Mathematics, Ellipse, JEE Main
Let P be a point on the hyperbola H:x29-y24=1, in the first quadrant such that the area of triangle formed by P and the two foci of H is 213. Then, the square of the distance of P from the origin is

Options:
A 18
B 26
C 22
D 20

The correct answer is:
22

Solution:
Plotting the diagram of given data we get,

Given equation of hyperbola is x29-y24=1.

⇒a2=9, b2=4

We know that, b2=a2e2-1

⇒e2=1+b2a2

⇒e2=1+49

⇒e2=139

⇒e=133

Now, S1S2=2ae

⇒S1S2=2×3×133

⇒S1S2=213

It is given that, area of ΔPS1S2 is 213.

⇒12×β×S1S2=213

⇒12×β×213=213

⇒β=2

Since, α,β lies on the hyperbola,

⇒α29-β24=1

⇒α29-1=1

⇒α2=18

⇒α=32

Distance of P from origin is given by,

OP=α2+β2

⇒OP=18+4

⇒OP=22

⇒OP2=22

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Let P be a point on the parabola x2=4y. If the distance of P from the center of the circle x2+y2+6x+8=0 is minimum, then the equation of the tangent to the parabola at P is

Options:
A x+y+1=0
B x+4y-2=0
C x+2y=0
D x-y+3=0

The correct answer is:
x+y+1=0

Solution:
Let P2t,t2.

Equation normal at P to x2=4y be

y-t2=-1tx-2t.

It passes a normal through -3,0.

⇒0-t2=-1t-3-2t

⇒t3+2t+3=0

⇒t+1t2-t+3=0

⇒t=-1

So, point P is -2,1.

Equation of a tangent to x2=4y at -2,1.

⇒x-2=2y+1

∴x+y+1=0.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2018 (16 Apr Online)
Let P(α,β) be a point on the parabola y2=4x. If P also lies on the chord of the parabola x2=8y whose mid point is 1,54, then (α-28)(β-8) is equal to _______.

Options:

The correct answer is:
192

Solution:
Given,

Parabola x2=8y

We know that,

Chord with mid point x1,y1 is T=S1

xx1-8y+y12=x1-28y1

⇒xx1-4y+y1=x1-28y1

Now, putting the point x1,y1=1,54 we get,

⇒x-4y+54=1-8×54=-9

∴x-4y+4=0 …i

(α,β) lies on (i) and also on y2=4x

∴α-4β+4=0 …ii

And β2=4α …iii

Solving (ii) and (iii)

β2=4(4β-4)

⇒β2-16β+16=0

⇒β=8±43

And α=4β-4=28±163

∴α, β=28+163,8+43 and (28-163,8-43)

Now, solving

(α-28)(β-8)=(163)(43)

⇒(α-28)(β-8)=192

Topic: Mathematics, Parabola, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let P be a point on the parabola, y2=12x and N be the foot of the perpendicular drawn from P , on the axis of the parabola. A line is now drawn through the mid-point M of PN, parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is 43, then :

Options:
A PN=4
B MQ=13
C MQ=14
D PN=3

The correct answer is:
MQ=14

Solution:
Comparing the equation of the parabola, with standard equation y2=4ax, we get a=3.

Figure is drawn on the basis of the given information.

Parametric form of parabola is x=at2, y=2at

∴ P=3t2,6t, N=3t2,0, M=3t2,3t

⇒Q=3t24,3t

Equation of NQ: y-0=3t-03t24-3t2x-3t2

i.e., y=-43t·x-3t2⇒y=-43t·x+4t

⇒4t=43⇒t=13

PN=6t=2

MQ=3t2-3t24=9t24=14

Topic: Mathematics, Parabola, JEE Main
Let P be a point on the parabola, y2=12x and N be the foot of the perpendicular drawn from P , on the axis of the parabola. A line is now drawn through the mid-point M of PN, parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is 43, then :

Options:
A PN=4
B MQ=13
C MQ=14
D PN=3

The correct answer is:
MQ=14

Solution:
Comparing the equation of the parabola, with standard equation y2=4ax, we get a=3.

Figure is drawn on the basis of the given information.

Parametric form of parabola is x=at2, y=2at

∴ P=3t2,6t, N=3t2,0, M=3t2,3t

⇒Q=3t24,3t

Equation of NQ: y-0=3t-03t24-3t2x-3t2

i.e., y=-43t·x-3t2⇒y=-43t·x+4t

⇒4t=43⇒t=13

PN=6t=2

MQ=3t2-3t24=9t24=14

Topic: Mathematics, Parabola, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Let \( P \) be a variable point on the parabola \( y=4 x^{2}+1 \). Then, the locus of the mid-point of the point \( P \) and the foot of the perpendicular drawn from the point \( P \) to the line \( y=x \) is:

Options:
A \( (3 x-y)^{2}+(x-3 y)+2=0 \)
B \( 2(3 x-y)^{2}+(x-3 y)+2=0 \)
C \( (3 x-y)^{2}+2(x-3 y)+2=0 \)
D \( 2(x-3 y)^{2}+(3 x-y)+2=0 \)

The correct answer is:
\( 2(3 x-y)^{2}+(x-3 y)+2=0 \)

Solution:
We have,

y=4x2+1

L:y=x

Let the foot of perpendicular from P to line y=x is Q.

Let P≡x,y, Q≡c,c and R≡h,k where, R is the mid-point of PQ.

Clearly,

PQ⊥L

⇒k-ch-c=-1

⇒c=h+k2

And,

R≡x+c2,y+c2

⇒R≡x2+h4+k4,y2+h4+k4

Hence,

h=x2+h4+k4⇒x=3h2-k2

k=y2+h4+k4⇒y=3k2-h2

Now,

Y=4x2+1

⇒3k-h2=43h-k22+1

⇒3k-h=23h-k2+2

Required locus is

23x-y2+x-3y+2=0

Topic: Mathematics, Straight Lines, JEE Main
Let $\mathrm{P}$ be an arbitrary point on the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}-1, \mathrm{a}>\mathrm{b}>0 .$ Suppose $\mathrm{F}_{1}$ and $\mathrm{F}_{2}$ are the foci of the ellipse. The locus of the centroid of the triangle $\mathrm{PF}_{1} \mathrm{~F}_{2}$ as $\mathrm{P}$ moves on the ellipse is-

Options:
A a circle
B a parabola
C an ellipse
D a hyperbola

The correct answer is:
an ellipse

Solution:
$$
\begin{array}{l}
\mathrm{P} \rightarrow \mathrm{a} \cos \theta, \mathrm{b} \sin \theta \\
\mathrm{G} \rightarrow\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{3}, \frac{\Sigma \mathrm{y}_{\mathrm{i}}}{3}\right) \\
\mathrm{F}_{1} \rightarrow(\mathrm{ae}, 0) \quad \mathrm{F}_{2} \rightarrow(-\mathrm{ae}, 0) \\
\mathrm{h}=\frac{\operatorname{acos} \theta+a \mathrm{e}-\mathrm{ae}}{3} \quad ; \quad \cos \theta=\frac{3 \mathrm{~h}}{\mathrm{a}}
\end{array}
$$
$$
\begin{array}{l}
k=\frac{b \sin \theta}{3} ; \quad \sin \theta=\frac{3 k}{b} \\
\cos ^{2} \theta+\sin ^{2} \theta=1 \\
\frac{9 h^{2}}{a^{2}}+\frac{9 k^{2}}{b^{2}}=1 \Rightarrow \frac{x^{2}}{\left(a^{2} / 9\right)}+\frac{y^{2}}{\left(b^{2} / 9\right)}=1 \\
\text { (Ellipse) }
\end{array}
$$

Topic: Mathematics, Ellipse, JEE Main
Let $P$ be an $m \times m$ matrix such that $P^{2}=P$. Then $(1+P)^{n}$ equals -

Options:
A $I+P$
B $I+n P$
C $I+2^{n} P$
D $I+\left(2^{n}-1\right) P$

The correct answer is:
$I+\left(2^{n}-1\right) P$

Solution:
$\begin{array}{l}\mathrm{P}^{2}=\mathrm{P} \\ \mathrm{P}^{-1} \mathrm{P}^{2}=\mathrm{P}^{-1} \mathrm{P} \\ \mathrm{P}=\mathrm{I} \\ (\mathrm{I}+\mathrm{P})^{\mathrm{n}}=(2 \mathrm{P})^{\mathrm{n}}=2^{\mathrm{n}} \mathrm{P}^{\mathrm{n}}\end{array}$
$=2^{n} \mathrm{P}$
$=\mathrm{P}+\left(2^{\mathrm{n}}-1\right) \mathrm{P}$
$=\mathrm{I}+\left(2^{\mathrm{n}}-1\right) \mathrm{P}$

Topic: Mathematics, Matrices, JEE Main
Let P be the image of the point 3,1,7 with respect to the plane x-y+z=3. The equation of the plane passing through P and parallel to x-2y+3z=7 is

Options:
A x-2y+3z=2
B 2x-4y+6z=7
C x+2y-6z+2=0
D x-2y+3z+2=0

The correct answer is:
x-2y+3z+2=0

Solution:
Image of 3,1,7 w.r.t. plane x-y+z-3=0 is
x-31=y-1-1=z-71=-263=-4
x=-1,y=5,z=3

So, image is -1,5,3
Equation of the plane through P and parallel to x-2y+3z=7 is
1x+1-2y-5+3z-3=0
x+1-2y+10+3z-9=0
x-2y+3z+2=0

Topic: Mathematics, Straight Lines, JEE Main
Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(1,6,4)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$. Then $2 \alpha+\beta+\gamma$ is equal to_______

Options:

The correct answer is:
11

Solution:
$\begin{aligned} & \mathrm{A}(\mathrm{t}, 2 \mathrm{t}+1,3 \mathrm{t}+2) \\ & \overrightarrow{\mathrm{QA}}=(\mathrm{t}-1) \hat{\mathrm{i}}+(2 \mathrm{t}-5) \hat{\mathrm{j}}+(3 \mathrm{t}-2) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{QA}} \cdot \overrightarrow{\mathrm{b}}=0 \\ & (\mathrm{t}-1)+2(2 \mathrm{t}-5)+3(3 \mathrm{t}-2)=0 \\ & 14 \mathrm{t}=17 \\ & \alpha=\frac{20}{14} \quad \beta=\frac{12}{14} \quad \gamma=\frac{102}{14} \\ & 2 \alpha+\beta+\gamma=\frac{154}{14}=11\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(3,-3,1)$ in the line $\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$ and $\mathrm{R}$ be the point $(2,5,-1)$. If the area of the triangle $P Q R$ is $\lambda$ and $\lambda^2=14 K$, then $K$ is equal to :

Options:
A 36
B 81
C 72
D 18

The correct answer is:
81

Solution:
$\begin{aligned} & \mathrm{RQ}=\sqrt{1+64+4}=\sqrt{69} \\ & \overrightarrow{\mathrm{RQ}}=\hat{\ell}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{RS}}=\hat{\ell}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \cos \theta=\frac{\overrightarrow{\mathrm{RQ}} \cdot \overrightarrow{\mathrm{RS}}}{|\overrightarrow{\mathrm{RQ}}||\overrightarrow{\mathrm{RS}}|}=\left|\frac{1-8-2}{\sqrt{69} \sqrt{3}}\right|=\frac{9}{3 \sqrt{23}} \\ & \cos \theta=\frac{3}{\sqrt{23}}=\frac{\mathrm{RS}}{\mathrm{RQ}}=\frac{\mathrm{RS}}{\sqrt{69}} \\ & \mathrm{RS}=3 \sqrt{3} \\ & \sin \theta=\frac{\sqrt{14}}{\sqrt{23}}=\frac{\mathrm{QS}}{\sqrt{69}}\end{aligned}$
$\begin{aligned} & \mathrm{QS}=\sqrt{42} \\ & \text { area }=\frac{1}{2} \cdot 2 \mathrm{QS} \cdot \mathrm{RS}=\sqrt{42} \cdot 3 \sqrt{3} \\ & \lambda=9 \sqrt{14} \\ & \lambda^2=81.14=14 \mathrm{k} \\ & \mathrm{k}=81\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Let $P$ be the mid-point of a chord joining the vertex of the parabola $y^{2}=8 x$ to another point on it. Then, the locus of $P$ is

Options:
A $y^{2}=2 x$
B $y^{2}=4 x$
C $\frac{x^{2}}{4}+y^{2}=1$
D $x^{2}+\frac{y^{2}}{4}=1$

The correct answer is:
$y^{2}=4 x$

Solution:
Let the chord be MN. Let other end of chord joining from vertex to it lying on $y^{2}=8 x$ be $N\left(2 t^{2}, 4 t\right)$

$\therefore$ Mid-point of $M N=\left(t^{2}, 2 t\right)$ $\therefore \quad x=t^{2}, y=2 t \Rightarrow x=\left(\frac{y}{2}\right)^{2}$
$\Rightarrow \quad y^{2}=4 x$
is the required locus of $P$

Topic: Mathematics, Parabola, JEE Main
Let P be the plane containing the straight line x-39=y+4-1=z-7-5 and perpendicular to the plane containing the straight lines x2=y3=z5 and x3=y7=z8. If d is the distance of P from the point 2,-5,11, then d2 is equal to

Options:
A 1472
B 96
C 323
D 54

The correct answer is:
323

Solution:
Let <a,b,c> be direction ratios of plane containing the line x2=y3=z5 and x3=y7=z8

Since plane normal will be perpendicular to given lines then by condition of perpendicular lines,

2a+3b+5c=0 ....1 and 3a+7b+8c=0.....2

On solving 1 & 2 we get,

a24-35=b15-16=c14-9

So, DR of plane are<11,1,-5>

Now let DR of plane P be <a1,b1,c1>

Then, 11a1+b1-5c1=0 ....3

And 9a1-b1-5c1=0......4 (as plane normal will be perpendicular to given line x-39=y+4-1=z-7-5)

Now solving equation 3 & 4 we get, direction ratio 1,-1,2

So, equation of plane will be x-y+2z=21

Now finding distance of point 2,-5,11 from plane x-y+2z=21

d=2+5+22-2112+12+22

⇒d=86

⇒d2=323

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Let P be the plane passing through the intersection of the planes r→·i^+3j^-k^=5 and r→·2i^-j^+k^=3, and the point 2,1,-2. Let the position vectors of the points X and Y be i^-2j^+4k^ and 5i^-j^+2k^ respectively. Then the points

Options:
A X and X+Y are on the same side of P
B Y and Y-X are on the opposite sides of P
C X and Y are on the opposite sides of P
D X+Y and X-Y are on the same side of P

The correct answer is:
X and Y are on the opposite sides of P

Solution:
We know that family of planes equation is given by, P1+λP2=0

⇒x+3y-z-5+λ2x-y+z-3=0

2,1,-2 lies on this plane

∴λ=1⇒ So the equation of required plane is 3x+2y-8=0

Now let P=3x+2y-8=0

So, P1=3×1+2×-2-8=-9

P2=3×5+2-1-8=5

Since P1×P2<0, So X and Y are on the opposite sides of plane P

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Let P be the plane passing through the line x-11=y-2-3=z+57 and the point (2,4,-3). If the image of the point (-1,3,4) in the plane P is (α,β,γ), then α+β+γ is equal to

Options:
A 10
B 12
C 9
D 11

The correct answer is:
10

Solution:
Given,

P be the plane passing through the line x-11=y-2-3=z+57 and the point (2,4,-3),

So, equation of plane is given by,

x-2y-4z+31-372-14-2-3--5=0

⇒x-2y-4z+31-37122=0

⇒x-2-20-y-4-5+z+35=0

⇒4 x-y-z-7=0

Now finding the imageα, β, γ of the point -1,3,4 with respect to plane 4 x-y-z-7=0 we get,

x+14=y-3-1=z-4-1=-2-4-3-4-716+1+1=2

⇒x=7=α, y=1=β, z=2=γ

Hence, α+β+γ=10

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let P be the plane, passing through the point 1,-1,-5 and perpendicular to the line joining the points 4,1,-3 and 2,4,3. Then the distance of P from the point 3,-2,2 is

Options:
A 6
B 4
C 5
D 7

The correct answer is:
5

Solution:
Given,

P be the plane, passing through the point 1,-1,-5 and perpendicular to the line joining the points 4,1,-3 and 2,4,3,

So, direction ratio of Plane Normal will be, 4-2,1-4,-3-3≡2,-3,-6

Now, equation of Plane will be,

2x-1-3y+1-6z+5=0

⇒2x-3y-6z-35=0

Hence, perpendicular distance of plane from 3,-2,2 will be,

Distance =2×3-3×-2-6×2-3522+-32+-62=6+6-12-3522+-32+-62=5.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let P be the plane passing through the point (1,2,3) and the line of intersection of the planes r→·(i^+j^+4k^)=16 & r→·(-i^+j^+k^)=6. Then which of the following points does NOT lie on P?

Options:
A 4,2,2
B 6,-6,2
C -8,8,6
D 3,3,2

The correct answer is:
4,2,2

Solution:
x+y+4z-16+λ-x+y+z-6=0

Passes through 1,2,3

-1+λ-2=0λ=-12

2x+y+4z-16--x+y+z-6=03x+y+7z-26=0

Now put points given in option and then we get 4,2,2 as required answer.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Let P be the plane passing through the points 5,3,0,13,3,-2 and 1,6,2. For α∈ℕ, if the distance of the points A3,4,α and B2,α,a from the plane P are 2 and 3 respectively, then the positive value of a is

Options:
A 6
B 3
C 5
D 4

The correct answer is:
4

Solution:
Given,

P be the plane passing through the points 5,3,0,13,3,-2 and 1,6,2

Now, AB→=8i^-2k^, AC→=-4i^+3j^+2k^ ,

So, normal vector to the plane will be,

AB→×AC→=i^j^k^80-2-432=6i^-8j^+24k^

Now equation of plane will be: 6x-8y+24z=d

And given plane passes through 5,3,0

So, 6×5-8×3+24×0=d⇒d=6

Hence, equation of plane will be,
6x-8y+24z=6⇒3x-4y+12z=3

Now using the formula distance of point 3, 4, α from the plane 3x-4y+12z=3, we get,

9-16+12α-39+16+144=2⇒α=3

Now again using the distance of point 2, α, a≡2,3,a from the plane formula we get,

3×2-4×3+12×a-313=3

⇒12a-9=39

⇒12a=48

⇒a=4

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let P be the plane, which contains the line of intersection of the planes, x+y+z-6=0 and 2x+3y+z+5=0 and it is perpendicular to the xy-plane. Then the distance of the point 0, 0, 256 from P is equal to:

Options:
A 2055 units
B 175 units
C 115 units
D 635 units

The correct answer is:
115 units

Solution:
Using family of planes, equation to plane passing through the line of intersection of the given planes, is

λx+y+z-6+2x+3y+z+5=0

⇒λ+2x+λ+3y+λ+1z+5-6λ=0

If a plane is perpendicular to xy plane, then it doesn't contain the term of z.

Hence, λ+1=0⇒λ=-1.

Thus, the plane is x+2y+11=0

The length of perpendicular from a point x1, y1, z1 to the plane ax+by+cz+d=0 is ax1+by1+cz1+da2+b2+c2

Hence, required perpendicular distance of point 0, 0, 256 from plane is P=0+0+111+4 =115 units.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Let $P$ be the point $(1,0)$ and $Q$ a point on the locus $y^2=8 x$. The locus of mid point of $P Q$ is

Options:
A
$y^2-4 x+2=0$

B
$y^2+4 x+2=0$

C
$x^2+4 y+2=0$

D
$x^2-4 y+2=0$

The correct answer is:
$y^2-4 x+2=0$

Solution:
$\mathrm{P}=(1,0)$
$Q=(h, k)$ such that $k^2=8 h$
Let $(\alpha, \beta)$ be the midpoint of PQ
$$
\begin{aligned}
& \alpha=\frac{\mathrm{h}+1}{2}, \quad \beta=\frac{\mathrm{k}+0}{2} \\
& 2 \alpha-1=\mathrm{h}, \quad 2 \beta=\mathrm{k} . \\
& (2 \beta)^2=8(2 \alpha-1) \Rightarrow \beta^2=4 \alpha-2 \\
& \Rightarrow \mathrm{y}^2-4 \mathrm{x}+2=0 .
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2005
Let \( P \) be the point \( (1,0) \) and \( Q \) be the point on \( y^{2}=8 x \). The locus of mid-point of \( P Q \) is

Options:
A \( x^{2}-4 y+2=0 \)
B \( x^{2}+4 y+2=0 \)
C \( y^{2}+4 x+2=0 \)
D \( y^{2}-4 x+2=0 \)

The correct answer is:
\( y^{2}-4 x+2=0 \)

Solution:
The co-ordinates of P are 1,0.

We know that, any point on the parabola y2=4ax is at2,2at.

Hence, for the parabola y2=8x the point Q is 2t2,4t.

Let mid-point of PQ be h,k, then by using mid-point formula, we get h=2t2+12

⇒2h=2t2+1 …1

and k=4t+02 ⇒t=k2 …2

On putting the value of t from Equation 2 in Equation 1, we get

2h=2k22+1, ⇒2h=2k24+1,

⇒4h=k2+2

The locus is obtained by replacing h,k by x,y.

Hence, the locus is y2-4x+2=0.

Topic: Mathematics, Straight Lines, JEE Main
Let P be the point of intersection of the common tangents to the parabola y2=12x and the hyperbola 8x2-y2=8. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio:

Options:
A 5:4
B 2:1
C 13:11
D 14:13

The correct answer is:
5:4

Solution:
Let m is slope of common tangent and an equation of tangent to hyperbola x21-y28=1 is

y=mx±1⋅m2-8 ...1

equation of tangent to parabola y2=12x is

y=mx+3m ...2

∵ 1 & 2 is same

∴ ±m2-8=3m

⇒m4-8m2-9=0⇒m2=-1 (Reject)

⇒m4-8m2-9=0⇒m2=9⇒m=±3

∵ equation of common tangents is y=3x+1 & y=-3x-1

∴ point of intersection of common tangents is R-13,0 & Foci of hyperbolas are 3,0 & -3,0

⇒-3λ+3λ+1=-13⇒λ=54

Hence, the required ratio is 5 : 4.

Topic: Mathematics, Hyperbola, JEE Main