JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Three Dimensional Geometry

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) $35$ ✓ Correct
(D) $45$

Answer: (C)

Solution

With

b_{1} = (1, 2, - 3)

b_{2} = (2, 4, - 5)

and

a_{2} - a_{1} = (- 6, 3, 3)

, the shortest distance is

\frac{( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} )}{∣ b _{1} \times b _{2} ∣}

. Here

b_{1} \times b_{2} = (2, - 1, 0)

with magnitude

5

, and

(a_{2} - a_{1}) \cdot (2, - 1, 0) = - 12 - 3 = - 15

. Hence the distance is

\frac{15}{5} = 35

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option

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