JEE Main 2026 Apr 5 Shift 1, Mathematics Q13: Lines in Space

JEE Main2026Apr 5, Shift 1Mathematics

The square of the distance of the point $P (5, 6, 7)$ from the line $\frac{x - 2}{2} = \frac{y - 5}{3} = \frac{z - 2}{4}$ is equal to :

Solution

A general point on the line is $Q = (2 + 2 λ, 5 + 3 λ, 2 + 4 λ)$ with direction $d = (2, 3, 4)$ .

The foot of perpendicular from $P (5, 6, 7)$ satisfies $P Q \cdot d = 0$ :

$2 (2 λ - 3) + 3 (3 λ - 1) + 4 (4 λ - 5) = 0 \Rightarrow 29 λ - 29 = 0 \Rightarrow λ = 1$ .

So $Q = (4, 8, 6)$ and $P Q = (- 1, 2, - 1)$ . Therefore $∣ P Q ∣^{2} = 1 + 4 + 1 = 6$ .

Concept behind this question

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