JEE Main 2026 Apr 6 Shift 2, Mathematics Q14: Lines in Space

JEE Main2026Apr 6, Shift 2Mathematics

The shortest distance between the lines $\frac{x - 4}{1} = \frac{y - 3}{2} = \frac{z - 2}{- 3}$ and $\frac{x + 2}{2} = \frac{y - 6}{4} = \frac{z - 5}{- 5}$ is:

A
$\frac{5 6}{6}$
B
$25$
C
$35$
D
$45$

Solution

Points: $A = (4, 3, 2)$ , $B = (- 2, 6, 5)$ ; direction vectors $d_{1} = ⟨ 1, 2, - 3 ⟩$ , $d_{2} = ⟨ 2, 4, - 5 ⟩$ .

$d_{1} \times d_{2} = ⟨ 2 \cdot (- 5) - (- 3) \cdot 4, (- 3) \cdot 2 - 1 \cdot (- 5), 1 \cdot 4 - 2 \cdot 2 ⟩ = ⟨ 2, - 1, 0 ⟩$ , magnitude $5$ .

$A B = ⟨ - 6, 3, 3 ⟩$ . Scalar triple product:

$A B \cdot (d_{1} \times d_{2}) = (- 6) (2) + (3) (- 1) + (3) (0) = - 15$ .

SD $= \frac{∣ - 15∣}{5} = \frac{15}{5} = 35$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q14: Lines in Space

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