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The shortest distance between the lines x+31=y−22=z+43\frac{x+3}{1} = \frac{y-2}{2} = \frac{z+4}{3} and x+3−3=y+72=z−64\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4} is :

Solution

✅ Correct Option: 4

d1⃗×d2⃗=(2,−13,8)\vec{d_1} \times \vec{d_2} = (2, -13, 8), magnitude =237= \sqrt{237}.

P2P1⃗=(0,−9,10)\vec{P_2 P_1} = (0, -9, 10).

d=∣(0)(2)+(−9)(−13)+(10)(8)∣237=197237d = \frac{|(0)(2) + (-9)(-13) + (10)(8)|}{\sqrt{237}} = \frac{197}{\sqrt{237}}.

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