Question

Find the foot of perpendicular from $(0,2,-2)$ to the plane $2x-3y+4z-44=0$. Find the equation of this perpendicular and the perpendicular distance between the point and the plane.

Answer 1

The given point $A(0,2,-2)$

And the direction of normal to the plane $2x-3y+4z-44=0$

is $(2,-3,4)$

Hence, any point on the line through P and along the given diameter is given by

$P(x,y,z)=(0+2k,2-3k,-2+4k)$ where $k$ is a parameter.

Now, $⟹\frac{x}{2}=\frac{y-2}{-3}=\frac{z+2}{4}=k$

This is the equation of the line.

Now for foot of the perpendicular-

$P should lie on the plane.$

$2\left(2k\right)-3(2-3k)+4(-2+4k)-44=0$

$⟹k=2$

Foot of perpendicular is $P(4,-4,6)$

Distance $AP=\sqrt{(4-0{)}^2+(2+4{)}^2+(6+2{)}^2}$

$⟹AP=\sqrt{116}$