JEE Main 2026 Jan 21 Shift 2, Mathematics Q8: Lines in Space

JEE Main2026Jan 21, Shift 2Mathematics

Let the line $L_{1}$ be parallel to the vector $- 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and pass through the point $(2, 6, 7)$ and the line $L_{2}$ be parallel to the vector $2 \hat{i} + \hat{j} + 3 \hat{k}$ and pass through the point $(4, 3, 5)$ . If the line $L_{3}$ is parallel to the vector $- 3 \hat{i} + 5 \hat{j} + 16 \hat{k}$ and intersects the lines $L_{1}$ and $L_{2}$ at the points C and D, respectively, then $C D^{2}$ is equal to :

A
171
B
290
C
312
D
89

Solution

$L_{1} : \frac{x - 2}{- 3} = \frac{y - 6}{2} = \frac{z - 7}{4}$ , so $C = (- 3 λ_{1} + 2, 2 λ_{1} + 6, 4 λ_{1} + 7)$ .

$L_{2} : \frac{x - 4}{2} = \frac{y - 3}{1} = \frac{z - 5}{3}$ , so $D = (2 λ_{2} + 4, λ_{2} + 3, 3 λ_{2} + 5)$ .

The vector $C D = D - C$ must be parallel to $(- 3, 5, 16)$ . So

$\frac{2 λ _{2} + 3 λ _{1} + 2}{- 3} = \frac{λ _{2} - 2 λ _{1} - 3}{5} = \frac{3 λ _{2} - 4 λ _{1} - 2}{16}$ .

Solving these gives $λ_{1} = - 3$ , $λ_{2} = 2$ , so $C = (11, 0, - 5)$ and $D = (8, 5, 11)$ .

$∣ C D ∣^{2} = (- 3)^{2} + 5^{2} + 1 6^{2} = 9 + 25 + 256 = 290$ .

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JEE Main 2026 Jan 21 Shift 2, Mathematics Q8: Lines in Space

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