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  4. The shortest distance between the lines (x-2/3)=(y+1/2)=(z-6/2) and (x-6/3)=(1-y/2)=(z+8/0) is equal to

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Q. The shortest distance between the lines $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to

JEE MainJEE Main 2023Three Dimensional Geometry

Solution:

Shortest distance between the lines
$=\frac{\begin{vmatrix}4 & 2 & -14 \\ 3 & 2 & 2 \\ 3 & -2 & 0\end{vmatrix}}{\begin{vmatrix}\hat{i} & \hat{ j } & \hat{ k } \\ 3 & 2 & 2 \\ 3 & -2 & 0\end{vmatrix}}$
$=\frac{16+12+168}{|-4 \hat{i}+6 \hat{j}-12 \hat{k}|}=\frac{196}{14}=14$