Let $L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$ ?

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The common perpendicular has direction $\vec d_1	imes\vec d_2=(-1,2,-1)$. The plane through $L_1$ containing it is $11x+2y-7z+6=0$, and the plane through $L_2$ containing it is $7x+y-5z+7=0$. Their intersection is the line of shortest distance, and $\left(\frac{14}3,-3,\frac{22}3ight)$ satisfies both. Answer (1).

Q. Let $L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$ ?

Explanation

Reviewed by Dr. Abubacker

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