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The shortest distance between the lines whose vector equations are r⃗=(i^+2j^+3k^)+λ(2i^+3j^+4k^)\vec r=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+4\hat k) and r⃗=(2i^+4j^+5k^)+μ(4i^+6j^+8k^)\vec r=(2\hat i+4\hat j+5\hat k)+\mu(4\hat i+6\hat j+8\hat k) is:

Solution

✅ Correct Option: 1

The direction vectors are proportional, so the lines are parallel and the formula ∣(a⃗2−a⃗1)×b⃗∣∣b⃗∣\frac{|(\vec a_2-\vec a_1)\times\vec b|}{|\vec b|} applies.

Here a⃗2−a⃗1=i^+2j^+2k^\vec a_2-\vec a_1=\hat i+2\hat j+2\hat k and b⃗=2i^+3j^+4k^\vec b=2\hat i+3\hat j+4\hat k, whose cross product is 2i^−k^2\hat i-\hat k with magnitude 5\sqrt5.

Since ∣b⃗∣=29|\vec b|=\sqrt{29}, the distance is 529\frac{\sqrt5}{\sqrt{29}}.

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