The shortest distance between the lines x−44=y+25=z+33 and x−13=y−34=z−42 is

l→1×l→2=i^j^k^453342=−2i^+j^+k^d=(a→−b→)⋅l→1×l→2l→1×l→2=(3i^−5j^−7k^)⋅(−2i^+j^+k^)6=−6−5−76=36

The shortest distance between the lines x−44=y+25=z+33 and x−13=y−34=z−42 is

l→1×l→2=i^j^k^453342=−2i^+j^+k^d=(a→−b→)⋅l→1×l→2l→1×l→2=(3i^−5j^−7k^)⋅(−2i^+j^+k^)6=−6−5−76=36

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$The shortest distance between the lines \frac{x - 4}{4} = \frac{y + 2}{5} = \frac{z + 3}{3} and \frac{x - 1}{3} = \frac{y - 3}{4} = \frac{z - 4}{2}$ is

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$2 \sqrt{6}$

$3 \sqrt{6}$

$6 \sqrt{2}$

$6 \sqrt{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} {\vec{l}}_{1} \times {\vec{l}}_{2} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{array}| \\ = - 2 \hat{i} + \hat{j} + \hat{k} \\ d = |\frac{(\vec{a} - \vec{b}) \cdot ({\vec{l}}_{1} \times {\vec{l}}_{2})}{|{\vec{l}}_{1} \times {\vec{l}}_{2}|}| \\ = |\frac{(3 \hat{i} - 5 \hat{j} - 7 \hat{k}) \cdot (- 2 \hat{i} + \hat{j} + \hat{k})}{\sqrt{6}}| \\ = |\frac{- 6 - 5 - 7}{\sqrt{6}}| \\ = 3 \sqrt{6} \end{array}$