JEE Main 2025 Jan 29 Shift 2, Mathematics Q13: Lines in Space

JEE Main2025Jan 29, Shift 2Mathematics

Let $P$ be the foot of the perpendicular from the point $(1, 2, 2)$ on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ .

Let the line $r = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ , $λ \in R$ , intersect the line $L$ at $Q$ . Then $2 (P Q)^{2}$ is equal to :

A
$27$
B
$25$
C
$29$
D
$19$

Solution

Foot of perpendicular $P$ . A general point on $L$ is $(1 + μ, - 1 - μ, 2 + 2 μ)$ . The vector from $A = (1, 2, 2)$ to this point is $(μ, - 3 - μ, 2 μ)$ , and it must be perpendicular to $d = (1, - 1, 2)$ :

$μ + (3 + μ) + 4 μ = 0 \Rightarrow 6 μ + 3 = 0 \Rightarrow μ = - \frac{1}{2}$ .

Hence $P = (\frac{1}{2}, - \frac{1}{2}, 1)$ .

Intersection $Q$ . Parameterise the second line as $(- 1 + λ, 1 - λ, - 2 + λ)$ . Equating with the parameterisation of $L$ gives the consistent solution $λ = 0$ , so $Q = (- 1, 1, - 2)$ .

Distance.

$P Q^{2} = (\frac{1}{2} - (- 1))^{2} + (- \frac{1}{2} - 1)^{2} + (1 - (- 2))^{2} = \frac{9}{4} + \frac{9}{4} + 9 = \frac{54}{4}$ .

$∴ 2 (P Q)^{2} = 2 \cdot \frac{54}{4} = 27$ .

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q13: Lines in Space

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