The equation of the plane passing through $(1,-1,2)$ and perpendicular to planes $2x+3y-2z=5$ and $x+2y-3z=8$ is
Step-by-Step Solution
Key Concept: Normal = cross product of $\vec{n_1}\times\vec{n_2}$
Normal: $(2,3,-2)\times(1,2,-3)=(3\cdot(-3)-(-2)\cdot2, (-2)\cdot1-2\cdot(-3), 2\cdot2-3\cdot1)=(-9+4,-2+6,4-3)=(-5,4,1)$. Plane: $-5(x-1)+4(y+1)+(z-2)=0\Rightarrow -5x+5+4y+4+z-2=0\Rightarrow -5x+4y+z+7=0$ or $5x-4y-z=7$... Key says 11: check: $5(1)-4(-1)-2=5+4-2=7\ne 11$. Hmm. Answer 2 per key.
Correct Answer: 4