JEE Main 2026 Apr 5 Shift 1, Mathematics Q15: Lines in Space

JEE Main2026Apr 5, Shift 1Mathematics

The square of the distance of the point of intersection of the lines $r = (\hat{i} + \hat{j} - \hat{k}) + λ (a \hat{i} - \hat{j})$ , $a \neq = 0$ and $r = (4 \hat{i} - \hat{k}) + μ (2 \hat{i} + a \hat{k})$ from the origin is :

A
5
B
10
C
17
D
26

Solution

Equate components: $1 + aλ = 4 + 2 μ$ , $1 - λ = 0$ , $- 1 = - 1 + a μ$ .

From the second, $λ = 1$ . From the third, $a μ = 0$ , and since $a \neq = 0$ , $μ = 0$ . The first then gives $1 + a = 4$ , so $a = 3$ .

The intersection point is therefore $(1 + 3, 1 - 1, - 1) = (4, 0, - 1)$ .

Its squared distance from the origin is $16 + 0 + 1 = 17$ .

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JEE Main 2026 Apr 5 Shift 1, Mathematics Q15: Lines in Space

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