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
\[ \begin{align} \overrightarrow{n_{1}} &= (a_{1}, b_{1}, c_{1}) \\ \overrightarrow{n_{2}} &= (a_{2}, b_{2}, c_{2}). \end{align} \]
ParallelAngleBisector
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 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
\[ \begin{align} 1(x-2)+2(y-0) -4(z-3) &= 0 \\ \Rightarrow x+2y-4z+10 &= 0. \ _\square \end{align} \]