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The shortest distance between lines whose vector equations are: r⃗=(1−t)i^+(t−2)j^+(3−2t)k^\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k} and r⃗=(s+1)i^+(2s−1)j^−(2s+1)k^\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}, is:

Solution

✅ Correct Option: 3

Lines: a1⃗=(1,−2,3)\vec{a_1}=(1,-2,3), b1⃗=(−1,1,−2)\vec{b_1}=(-1,1,-2); a2⃗=(1,−1,−1)\vec{a_2}=(1,-1,-1), b2⃗=(1,2,−2)\vec{b_2}=(1,2,-2).

b1⃗×b2⃗=(2,−4,−3)\vec{b_1}\times\vec{b_2}=(2,-4,-3) with magnitude 29\sqrt{29}, and a2⃗−a1⃗=(0,1,−4)\vec{a_2}-\vec{a_1}=(0,1,-4).

Shortest distance =∣0−4+12∣29=829=\frac{|0-4+12|}{\sqrt{29}}=\frac{8}{\sqrt{29}}.

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