JEE Main 2025 Apr 3 Shift 1, Mathematics Q20: Lines in Space

JEE Main2025Apr 3, Shift 1Mathematics

Line $L_{1}$ passes through the point $(1, 2, 3)$ and is parallel to z-axis. Line $L_{2}$ passes through the point $(λ, 5, 6)$ and is parallel to y-axis. Let for $λ = λ_{1}, λ_{2}, λ_{2} < λ_{1}$ , the shortest distance between the two lines be $3$ . Then the square of the distance of the point $(λ_{1}, λ_{2}, 7)$ from the line $L_{1}$ is

A
40
B
32
C
25
D
37

Solution

$L_{1}$ : direction $(0, 0, 1)$ through $(1, 2, 3)$ . $L_{2}$ : direction $(0, 1, 0)$ through $(λ, 5, 6)$ .

The two direction vectors $\hat{k}$ and $\hat{j}$ are perpendicular; their cross product is $\hat{j} \times \hat{k} = \hat{i}$ (up to sign). The shortest distance between the two skew lines equals the projection of the connecting vector $(λ - 1, 3, 3)$ onto $\hat{i}$ :

$SD = ∣ λ - 1∣ = 3 \Rightarrow λ = 4 or - 2.$

With $λ_{2} < λ_{1}$ : $λ_{1} = 4, λ_{2} = - 2.$ The reference point is $P = (4, - 2, 7) .$

Distance from $P$ to $L_{1}$ (a vertical line through $(1, 2, 3)$ parallel to the $z$ -axis): only the $x$ and $y$ components contribute.

$d^{2} = (4 - 1)^{2} + (- 2 - 2)^{2} = 9 + 16 = 25.$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q20: Lines in Space

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