JEE Main 2020 (Online) 4th September Morning ShiftQuestion 1
If the equation of a plane $P$, passing through the intersection of the planes, $x+4y-z+7=0$ and $3x+y+5z=8$ is $ax+by+6z=15$ for some $a,b\in\mathbf{R}$, then the distance of the point $(3,2,-1)$ from the plane $P$ is ________.
3D Geometry
Answer$3$
Solution
Plane: $(x+4y-z+7)+k(3x+y+5z-8)=0\Rightarrow(1+3k)x+(4+k)y+(5k-1)z=8k-7$. Matching $ax+by+6z=15$: $\dfrac{5k-1}{6}=\dfrac{8k-7}{15}\Rightarrow 75k-15=48k-42\Rightarrow k=-1$. Plane: $-2x+3y-6z+15=0$, i.e. $2x-3y+6z=15$ ($a=2,\ b=-3$). Distance of $(3,2,-1)$: $\dfrac{|-6+6+6+15|}{\sqrt{4+9+36}}=\dfrac{21}{7}=3$.