JEE Main 2025 Jan 29 Shift 2, Mathematics Q6: Lines in Space

JEE Main2025Jan 29, Shift 2Mathematics

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 :

A
$2$
B
$10$
C
$3$
D
$23$

Solution

The direction of $L$ is perpendicular to both given lines, so it is parallel to the cross product of their direction vectors $d_{1} = (2, 1, - 2)$ and $d_{2} = (1, 3, 4)$ :

$d_{1} \times d_{2} = (1 \cdot 4 - (- 2) \cdot 3, (- 2) \cdot 1 - 2 \cdot 4, 2 \cdot 3 - 1 \cdot 1) = (10, - 10, 5) = 5 (2, - 2, 1)$ .

So $L$ : $\frac{x - 2}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{1} = λ$ .

For the yz-plane, $x = 0 \Rightarrow 2 + 2 λ = 0 \Rightarrow λ = - 1$ , giving $Q = (0, 1, 2)$ .

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

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q6: Lines in Space

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