JEE Main 2026 Apr 4 Shift 2, Mathematics Q13: Scalar and Vector Triple Product

JEE Main2026Apr 4, Shift 2Mathematics

The shortest distance between the lines $r = (\frac{1}{3} \hat{i} + 2 \hat{j} + \frac{8}{3} \hat{k}) + λ (2 \hat{i} - 5 \hat{j} + 6 \hat{k})$ and $r = (- \frac{2}{3} \hat{i} - \frac{1}{3} \hat{k}) + μ (\hat{j} - \hat{k})$ , $λ, μ \in R$ , is:

A
$5$
B
3
C
$23$
D
$15$

Solution

Take $a_{1} = (\frac{1}{3}, 2, \frac{8}{3})$ , $a_{2} = (- \frac{2}{3}, 0, - \frac{1}{3})$ , $b_{1} = (2, - 5, 6)$ , $b_{2} = (0, 1, - 1)$ .

$b_{1} \times b_{2} = ((- 5) (- 1) - 6 \cdot 1, 6 \cdot 0 - 2 \cdot (- 1), 2 \cdot 1 - (- 5) \cdot 0) = (- 1, 2, 2)$ , with $∣ b_{1} \times b_{2} ∣ = 3$ .

$a_{1} - a_{2} = (1, 2, 3)$ .

$(a_{1} - a_{2}) \cdot (b_{1} \times b_{2}) = (1) (- 1) + (2) (2) + (3) (2) = - 1 + 4 + 6 = 9$ .

Shortest distance $= \frac{∣9∣}{3} = 3$ .

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JEE Main 2026 Apr 4 Shift 2, Mathematics Q13: Scalar and Vector Triple Product

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