Let a straight line L pass through the point P(2,-1,3) and be perpendicular to the lines frac x -1 2 =frac y +1 1 =frac z -3 -2 and frac x -3 1 =frac y -2 3 =frac z +2 4 . If the line L intersects the yz-plane at the point Q , then the distance between the points P and Q is

Mathematics · 3D Geometry

JEE Main 2025 — 29 January, Evening Shift — Question 47

Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{1}=\frac{\mathrm{z}-3}{-2} \quad$ and $\quad \frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}+2}{4}$ . If the line $L$ intersects the yz-plane at the point $Q$ , then the distance between the points P and Q is

Option A:
2
Option B:
$\sqrt{10}$
Option C:
3
Correct
Option D:
$2 \sqrt{3}$

Answer: C

Step-by-step solution

Vector parallel to ‘L’ $=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4\end{array}\right|=10 \hat{i}-10 \hat{j}+5 \hat{k}$

$=5(2 \hat{i}-2 \hat{j}+\hat{k})$

Equation of 'L'\ $\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-3}{1}=\lambda$ (say)

Let $\mathrm{Q}(2 \lambda+2,-2 \lambda-1, \lambda+3)$

$\Rightarrow 2 \lambda+2=0 \Rightarrow \lambda=-1$

$\Rightarrow \mathrm{Q}(0,1,2)$ \ $d(P, Q)=3$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Straight Lines in 3D Geometry

Let P be the foot of the perpendicular from the point (1,2,2) on the line L: x-1/1=y+1/-1=z-2/2 . Let the line overrightarrow r=(-hat i+hat…

Answer: C

Step-by-step solution

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