JEE Main 2023 — Mathematics Vectors & 3D Geometry

2023mcqmedium

Let the line $L$ pass through the point $(0,1,2),$ intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$ . Then the distance of the point $P(1,–9,2)$ from the line $L$ is

Official previous-year question

Held on 6 Apr 2023 · Verified 6 Jul 2026.

Options

A
$\sqrt{74}$
B
$\sqrt{69}$
C
$\sqrt{54}$
D
$9$

Did you get this right?

Solution

Given,

The line $L$ pass through the point $(0,1,2),$ intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$ ,

Now plotting the diagram of above data we have,

JEE Main 2023 Mathematics, Three Dimensional Geometry — solution figure 1

Now let the direction of line which passes through $(0,1,2)$ be $(a,b,c)$

Now the line $L&{L}_{1}$ are coplanar,

So coplanar condition we get,

$|\begin{matrix}a & b & c \\ 2-1 & 3-2 & 4-3 \\ 2 & 3 & 4\end{matrix}|=0$

$\Rightarrow |\begin{matrix}a & b & c \\ 1 & 1 & 1 \\ 2 & 3 & 4\end{matrix}|=0$

$\Rightarrow a-2b+c=0.......(1)$

And the line $L$ is perpendicular to the normal vector of the plane,

So by perpendicular condition we get,

$2a+b-3c=0......(2)$

So, from equation $(1)&(2)$ we get,

$a=b=c$

So, equation of line $L$ will be,

$\Rightarrow L=\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\lambda$

So any point on $L$ can be taken as $A(\lambda ,1+\lambda ,2+\lambda )$ ,

JEE Main 2023 Mathematics, Three Dimensional Geometry — solution figure 2

Now $\vec{AP}$ will be perpendicular to the direction ratio of the line $L$ ,

So, again by perpendicular condition we get,

$\vec{AP}\cdot (\hat{i}+\hat{j}+\hat{k})=0$

$\Rightarrow \lambda -1+\lambda +10+\lambda =0$

$\Rightarrow 3\lambda +9=0$

$\Rightarrow \lambda =-3$

Hence, $A(-3,-2,-1)&P(1,-9,2)$

So, by distance formula we get, $AP=\sqrt{74}$

Your note

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Solution

Given,

The line $L$ pass through the point $(0,1,2),$ intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$ ,

Now plotting the diagram of above data we have,

JEE Main 2023 Mathematics, Three Dimensional Geometry — solution figure 1

Now let the direction of line which passes through $(0,1,2)$ be $(a,b,c)$

Now the line $L&{L}_{1}$ are coplanar,

So coplanar condition we get,

$|\begin{matrix}a & b & c \\ 2-1 & 3-2 & 4-3 \\ 2 & 3 & 4\end{matrix}|=0$

$\Rightarrow |\begin{matrix}a & b & c \\ 1 & 1 & 1 \\ 2 & 3 & 4\end{matrix}|=0$

$\Rightarrow a-2b+c=0.......(1)$

And the line $L$ is perpendicular to the normal vector of the plane,

So by perpendicular condition we get,

$2a+b-3c=0......(2)$

So, from equation $(1)&(2)$ we get,

$a=b=c$

So, equation of line $L$ will be,

$\Rightarrow L=\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\lambda$

So any point on $L$ can be taken as $A(\lambda ,1+\lambda ,2+\lambda )$ ,

JEE Main 2023 Mathematics, Three Dimensional Geometry — solution figure 2

Now $\vec{AP}$ will be perpendicular to the direction ratio of the line $L$ ,

So, again by perpendicular condition we get,

$\vec{AP}\cdot (\hat{i}+\hat{j}+\hat{k})=0$

$\Rightarrow \lambda -1+\lambda +10+\lambda =0$

$\Rightarrow 3\lambda +9=0$

$\Rightarrow \lambda =-3$

Hence, $A(-3,-2,-1)&P(1,-9,2)$

So, by distance formula we get, $AP=\sqrt{74}$

JEE Main 2023 — Mathematics Vectors & 3D Geometry

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Solution

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JEE Main 2023 — Mathematics Vectors & 3D Geometry

Options

Did you get this right?

Solution

Your note

JEE Main Vectors & 3D Geometry in other years

More JEE Main Vectors & 3D Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Probability & Statistics

Trigonometry

Keep practicing Vectors & 3D Geometry