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Question:
Let $P_1$ and $P_2$ be two planes given by $ P_1: 10 x+15 y+12 z-60=0,$ $ P_2:-2 x+5 y+4 z-20=0 $ Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$ ?
Let $P_1$ and $P_2$ be two planes given by $ P_1: 10 x+15 y+12 z-60=0,$ $ P_2:-2 x+5 y+4 z-20=0 $ Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$ ?
Updated On: Mar 16, 2024
- $\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}$
- $\frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3}$
- $\frac{x}{-2}=\frac{y-4}{5}=\frac{z}{4}$
- $\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}$
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The Correct Option is D
Solution and Explanation
D
$\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}$
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