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

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Three Dimensional Geometry

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 :

Answer: (C)

The direction of

L

must be perpendicular to both given lines, so it is along

\hat{i} 21 \hat{j} 13 \hat{k} - 2 4 = \hat{i} (4 + 6) - \hat{j} (8 + 2) + \hat{k} (6 - 1) = (10, - 10, 5)

i.e. along

(2, - 2, 1)

.

Line

L

through

P (2, - 1, 3)

(2 + 2 λ, - 1 - 2 λ, 3 + λ)

.

Setting the

x

-coordinate to

0

(yz-plane) gives

2 + 2 λ = 0 \Rightarrow λ = - 1

, so

Q = (0, 1, 2)

P Q = (2 - 0)^{2} + (- 1 - 1)^{2} + (3 - 2)^{2} = 4 + 4 + 1 = 3

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option