JEE Main 2026 Apr 4 Shift 1, Mathematics Q15: Lines in Space

JEE Main2026Apr 4, Shift 1Mathematics

The square of the distance of the point $(- 2, - 8, 6)$ from the line $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{- 1}$ along the line $\frac{x + 5}{1} = \frac{y + 5}{- 1} = \frac{z}{2}$ is equal to:

A
$3$
B
$6$
C
$8$
D
$12$

Solution

Let $A = (- 2, - 8, 6)$ . The line through $A$ with direction $(1, - 1, 2)$ (parallel to the second given line) is

$\frac{x + 2}{1} = \frac{y + 8}{- 1} = \frac{z - 6}{2} = λ .$

Its general point is $A + λ (1, - 1, 2) = (λ - 2, - λ - 8, 2 λ + 6)$ .

This must meet the first line, parametrised as $(μ + 1, 2 μ + 1, - μ)$ . Equating coordinates:

$λ - 2 = μ + 1, - λ - 8 = 2 μ + 1, 2 λ + 6 = - μ .$

From the first, $μ = λ - 3$ . Substitute into the third: $2 λ + 6 = - (λ - 3) = 3 - λ \Rightarrow 3 λ = - 3 \Rightarrow λ = - 1$ , $μ = - 4$ (which also satisfies the second).

Intersection point $B = (- 3, - 7, 4)$ and

$A B^{2} = (- 3 + 2)^{2} + (- 7 + 8)^{2} + (4 - 6)^{2} = 1 + 1 + 4 = 6.$

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JEE Main 2026 Apr 4 Shift 1, Mathematics Q15: Lines in Space

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