Consider the three planes P_1: 3x+15y+21z=9, P_2: x-3y-z=5, P_3: 2x+10y+14z=5.

Question

Consider the three planes $P_1:\ 3x+15y+21z=9$, $P_2:\ x-3y-z=5$, $P_3:\ 2x+10y+14z=5$. Then which one of the following is true?

Accepted Answer

Simplify: $P_1:\ x+5y+7z=3$; $P_3:\ x+5y+7z=2.5$. Both have normal $(1,5,7)$ --- parallel planes (and distinct, since the constants differ). $P_2$'s normal $(1,-3,-1)$ is not proportional to $(1,5,7)$.

Solutions · 1