Question

Let $P_1$ and $P_2$ be two planes given by
$P_1:10x+15y+12z-60=0,$
$P_2:-2x+5y+4z-20=0.$
Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$ ?

• A. $\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}$
• B. $\frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3}$
• C. $\frac{x}{-2}=\frac{y-4}{5}=\frac{z}{4}$
• D. $\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}$

Answer 1

Answer: (A), (B), (D)

line of intersection is $\frac{x}{0}=\frac{y-4}{-4}=\frac{z}{5}$
(1) Any skew line with the line of intersection of given planes can be edge of tetrahedron.
(2) any intersecting line with line of intersection of given planes must lie either in plane $P_1$ or $P_2$ can be edge of tetrahedron.