# 3D Coordinate Geometry - Perpendicular Planes

#### Contents

## Summary

Let two planes $\alpha$ and $\beta$ be defined as follows:

$\begin{aligned} \alpha : a_{1}x+b_{1}y+c_{z}+d_{1} &= 0 \\ \beta : a_{2}x+b_{2}y+c_{z}+d_{2} &= 0, \end{aligned}$

then the normal vectors of the planes $\overrightarrow{n_{1}}$ and $\overrightarrow{n_{2}},$ respectively, are

$\begin{aligned} \overrightarrow{n_{1}} &= (a_{1}, b_{1}, c_{1}) \\ \overrightarrow{n_{2}} &= (a_{2}, b_{2}, c_{2}). \end{aligned}$

When the two planes $\alpha$ and $\beta$ are perpendicular, their normal vectors are also perpendicular and their dot product is 0. That is,

$\overrightarrow{n_{1}} \perp \overrightarrow{n_{2}} \implies \overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}} = 0.$

Hence, the condition for the two planes to be perpendicular to each other is

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0 .$

## Example Problem

If two planes $\alpha$ and $\beta$ are

$\begin{aligned} \alpha : 3x+y+z+3 &= 0 \\ \beta : -x + 2y+z+5 &= 0, \end{aligned}$

then are the two planes $\alpha$ and $\beta$ perpendicular?

The normal vectors of the planes are

$\begin{aligned} \overrightarrow{n_{1}} &= (3,1,1)\\ \overrightarrow{n_{2}} &= (-1,2,1), \end{aligned}$

respectively. Since their dot product is

$\begin{aligned} \overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}} &= (3,1,1)\cdot(-1,2,1) \\ &= -3 + 2+ 1 \\ &= 0, \end{aligned}$

the two planes are perpendicular. $_\square$

There are two planes $\alpha$ and $\beta$ defined as

$\begin{aligned} \alpha : ax+y+az-4 &= 0 \\ \beta : 3x -2y+z+7 &= 0. \end{aligned}$

If the two planes are perpendicular, then what is $a?$

The normal vectors of the planes are

$\begin{aligned} \overrightarrow{n_{1}} &= (a,1,a)\\ \overrightarrow{n_{2}} &= (3,-2,1), \end{aligned}$

respectively. Dot product of the normal vectors is

$\begin{aligned} \overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}} &= (a,1,a) \cdot (3,-2,1) \\ &= 3a-2+a \\ &= 4a-2. \end{aligned}$

When two planes are perpendicular, the dot product of their normal vectors is 0. Hence,

$4a-2=0 \implies a = \frac{1}{2}. \ _ \square$

What is the equation of the plane which passes through point $A=(2,1,3)$ and is perpendicular to line segment $\overline {BC} ,$ where $B=(3, -2, 3)$ and $C=(0,1,3)?$

The direction vector which passes through points $B=(3,-2,3)$ and $C=(0,1,3)$ is

$\overrightarrow{BC} = (-3, 3, 0),$

which is the same as the normal vector of the plane.

Thus, the equation of the plane which passes through point $A=(2,1,3)$ is

$\begin{aligned} -3(x-2) +3(y-1) +0(z-3) &=0 \\ \Rightarrow -x+y + 1 &= 0 . \ _\square \end{aligned}$

What is the equation of the plane which is perpendicular to line segment $\overline{AB}$ and passes through point $A,$ where $A=(2,0,3)$ and $B=(3,2,-1)?$

The direction vector which passes through points $A=(2,0,3)$ and $B=(3,2,-1)$ is

$\overrightarrow{AB} = (1, 2, -4),$

which is the same as the normal vector of the plane.

Since the plane passes through point $A=(2,0,3) ,$ the equation of the plane is

$\begin{aligned} 1(x-2)+2(y-0) -4(z-3) &= 0 \\ \Rightarrow x+2y-4z+10 &= 0. \ _\square \end{aligned}$

**Cite as:**3D Coordinate Geometry - Perpendicular Planes.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/3d-coordinate-geometry-perpendicular-planes/