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The shortest distance between the lines x+17=−y−16=z+11\dfrac{x+1}{7}=\dfrac{-y-1}{6}=\dfrac{z+1}{1} and x−31=−y+52=z−71\dfrac{x-3}{1}=\dfrac{-y+5}{2}=\dfrac{z-7}{1} is

Solution

✅ Correct Option: 4

Rewriting, the lines pass through (−1,−1,−1)(-1,-1,-1) and (3,5,7)(3,5,7) with directions (7,−6,1)(7,-6,1) and (1,−2,1)(1,-2,1). Their cross product is (−4,−6,−8)(-4,-6,-8), of magnitude 116=229\sqrt{116}=2\sqrt{29}. With the joining vector (4,6,8)(4,6,8), the scalar triple product is −116-116, so the distance is 116229=229\frac{116}{2\sqrt{29}}=2\sqrt{29}.

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