Let a plane P pass through the point (3,7,-7) and contain the line x-2-3=y-32=z+21.

Question

Let a plane $P$ pass through the point $(3,7,-7)$ and contain the line $\dfrac{x-2}{-3}=\dfrac{y-3}{2}=\dfrac{z+2}{1}$. If the distance of the plane $P$ from the origin is $d$, then $d^2$ is equal to \_\_\_\_.

Answer: 3

Accepted Answer

The line passes through $A(2,3,-2)$ with direction $\vec d=(-3,2,1)$. Since $P$ also passes through $(3,7,-7)$, a second vector in the plane is $\overrightarrow{A(3,7,-7)}=(1,4,-5)$.
$$\vec n=\vec d\times(1,4,-5)=\hat i(2(-5)-1(4))-\hat j((-3)(-5)-1(1))+\hat k((-3)(4)-2(1))=(-14,-14,-14)\parallel(1,1,1).$$
Plane: $1(x-2)+1(y-3)+1(z+2)=0\Rightarrow x+y+z-3=0$.
$$d=\frac{|0+0+0-3|}{\sqrt3}=\sqrt3 \implies d^2=3.$$

Solutions · 1