Let P be the plane passing through the point (1,2,3) and the line of intersection of the p…

Question

Let $P$ be the plane passing through the point $(1,2,3)$ and the line of intersection of the planes $\vec r\cdot(\hat i+\hat j+4\hat k)=16$ and $\vec r\cdot(-\hat i+\hat j+\hat k)=6$. Then which of the following points does not lie on $P$?

Accepted Answer

Family: $(x+y+4z-16)+\lambda(-x+y+z-6)=0$. Through $(1,2,3)$:
$$(1+2+12-16)+\lambda(-1+2+3-6)=0\implies -1-2\lambda=0\implies\lambda=-\tfrac12.$$
Plane: $2(x+y+4z-16)-(-x+y+z-6)=0\implies 3x+y+7z-26=0$.
Testing points: $A: 9+3+14=26$, $B:18-6+14=26$, $D:-24+8+42=26$, $C: 12+2+14=28\neq26$. So $C=(4,2,2)$ does not lie on $P$.

Solutions · 1