Vectors & 3D GeometryeasyPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Shortest Distance Between Two Skew Lines = 3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The shortest distance between the lines r⃗=(13i^+2j^+83k^)+λ(2i^−5j^+6k^)\vec r=\left(\dfrac13\hat i+2\hat j+\dfrac83\hat k\right)+\lambda\left(2\hat i-5\hat j+6\hat k\right) and r⃗=(−23i^−13k^)+μ(j^−k^)\vec r=\left(-\dfrac23\hat i-\dfrac13\hat k\right)+\mu\left(\hat j-\hat k\right), λ,μ∈R\lambda,\mu\in\mathbb{R}, is
A5\sqrt5
B33correct
C232\sqrt3
D15\sqrt{15}
Solution
Step 1: Direction vectors b⃗1=2i^−5j^+6k^\vec b_1=2\hat i-5\hat j+6\hat k, b⃗2=j^−k^\vec b_2=\hat j-\hat k. Cross product:
b⃗1×b⃗2=∣i^j^k^2−5601−1∣=−i^+2j^+2k^,∣b⃗1×b⃗2∣=1+4+4=3.\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-5&6\\0&1&-1\end{vmatrix}=-\hat i+2\hat j+2\hat k,\qquad |\vec b_1\times\vec b_2|=\sqrt{1+4+4}=3.
 Step 2: a⃗1−a⃗2=(13+23)i^+2j^+(83+13)k^=i^+2j^+3k^\vec a_1-\vec a_2=\left(\dfrac13+\dfrac23\right)\hat i+2\hat j+\left(\dfrac83+\dfrac13\right)\hat k=\hat i+2\hat j+3\hat k. Step 3: Shortest distance:
S.D.=∣(i^+2j^+3k^)⋅(−i^+2j^+2k^)∣3=∣−1+4+6∣3=93=3.\text{S.D.}=\frac{\left|(\hat i+2\hat j+3\hat k)\cdot(-\hat i+2\hat j+2\hat k)\right|}{3}=\frac{|-1+4+6|}{3}=\frac{9}{3}=3.
 Correct answer: (2)
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