# 3D Coordinate Geometry - Perpendicular Planes

#### Contents

## Summary

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

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

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

\[ \begin{align} \overrightarrow{n_{1}} &= (a_{1}, b_{1}, c_{1}) \\ \overrightarrow{n_{2}} &= (a_{2}, b_{2}, c_{2}). \end{align} \]

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{align} \alpha : 3x+y+z+3 &= 0 \\ \beta : -x + 2y+z+5 &= 0,

\end{align} \]then are the two planes \( \alpha\) and \( \beta\) perpendicular?

The normal vectors of the planes are

\[ \begin{align} \overrightarrow{n_{1}} &= (3,1,1)\\ \overrightarrow{n_{2}} &= (-1,2,1), \end{align} \]

respectively. Since their dot product is

\[ \begin{align} \overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}} &= (3,1,1)\cdot(-1,2,1) \\ &= -3 + 2+ 1 \\ &= 0, \end{align} \]

the two planes are perpendicular. \( _\square \)

There are two planes \( \alpha\) and \( \beta\) defined as

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

If the two planes are perpendicular, then what is \(a?\)

The normal vectors of the planes are

\[ \begin{align} \overrightarrow{n_{1}} &= (a,1,a)\\ \overrightarrow{n_{2}} &= (3,-2,1), \end{align} \]

respectively. Dot product of the normal vectors is

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

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 the normal vector of the plane.

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

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

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

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

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

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