Question

find the equation of the plane if the foot of the perpendicular from origin to the plane (1,3,-5).​

Answer 1

Let (x, y, z) be a point on the plane.

The foot of the perpendicular from origin to the plane is at (1, 3, -5).​ So the vector $\left<1,\ 3,\ -5\right>$ is perpendicular to the plane.

Also the vector $\left<x-1,\ y-3,\ z+5\right>$ lies along the plane, since both the points (x, y, z) and (1, 3, -5) lie on the plane.

The two vectors are perpendicular to each other and so their dot product is equal to zero, i.e.,

$\longrightarrow\left<x-1,\ y-3,\ z+5\right>\cdot\left<1,\ 3,\ -5\right>=0$

$\longrightarrow x-1+3(y-3)-5(z+5)=0$

$\longrightarrow\underline{\underline{x+3y-5z-35=0}}$

This is the equation of the required plane.