Question

Find the perpendicular distance from the origin to the plane passing through the points $(1,-2,5),(0,-5,-1)$ and $(-3,5,0)$.

Answer 1

We can get two vectors in the plane by subtracting pairs of points in the

plane:

$\left[\begin{array}{c}1\\ -2\\ 5\end{array}\right]-\left[\begin{array}{c}0\\ -5\\ -1\end{array}\right]=\left[\begin{array}{c}1\\ 3\\ 6\end{array}\right]$

$\left[\begin{array}{c}-3\\ 5\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ -5\\ -1\end{array}\right]=\left[\begin{array}{c}-3\\ 10\\ 1\end{array}\right]$

The cross product of these two vectors will be in the unique direction orthogonal to both, and hence in the direction of the normal vector to the plane.

$\left[\begin{array}{c}1\\ 3\\ 6\end{array}\right]$x$\left[\begin{array}{c}-3\\ 10\\ 1\end{array}\right]=\left[\begin{array}{c}-57\\ -19\\ 19\end{array}\right]$

The Equation of Plane $ax+by+cz+d=0$ where $\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$ is the normal vector to the plane

Hence, $\left[\begin{array}{c}-57\\ -19\\ 19\end{array}\right]$ is normal vector to the plane

$-57x-19y+19z+d=0$

Putting one of three points in this Equation we can get $d$

$-57\ast 0-5\ast (-19)+19\ast -1+d=0\Rightarrow d=76$

So, Equation of plane $=-57x-19y+19z+76=0\Rightarrow -3x-y+z+4=0$

Distance of Origin from this plane $=\frac{-3\left(0\right)-\left(0\right)+\left(0\right)+4}{\sqrt{3^2+1+1}}=\frac{4}{\sqrt{11}}$