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Table of Contents
4. Find the equation of the plane containing the points $A=(1,2,0)$, $B=(3,0,1)$ and $C=(4,3,-2)$.
Solution: $\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\vec{AB}=\c2{-2}1$ and $\vec{AC}=\c31{-2}$ are both vectors in this plane. We want to find a normal vector $\nn$ which is be orthogonal to both of these. The cross product of two vectors is orthogonal to both, so we can take the cross product of $\vec{AB}$ and $\vec{AC}$: \[ \nn=\vec{AB}\times\vec{AC}=\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1\\#4\end{vmatrix}}\cp2{-2}131{-2}=\c378\] so the equation of the plane is $3x+7y+8z=d$, and we find $d$ by subbing in a point in the plane, say $A=(1,2,0)$, which gives $d=3(1)+7(2)+8(0)=17$. So the equation is \[ 3x+7y+8z=17.\]
Orthogonal planes and parallel planes
Let $\Pi_1$ be a plane with normal vector $\nn_1$, and let $\Pi_2$ be a plane with normal vector $\nn_2$.
- $\Pi_1$ and $\Pi_2$ are orthogonal or perpendicular planes if they meet at right angles. The following conditions are equivalent:
- $\Pi_1$ and $\Pi_2$ are orthogonal planes;
- $\nn_1\cdot\nn_2=0$;
- $\nn_1$ is a vector in $\Pi_2$;
- $\nn_2$ is a vector in $\Pi_1$.
- $\Pi_1$ and $\Pi_2$ are parallel planes if they have the same normal vectors. In other words, if $\Pi_1$ has equation $ax+by+cz=d_1$ then any parallel plane $\Pi_2$ has an equation with the same left hand side: $ax+by+cz=d_2$.
Examples
1. Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$.
Solution: The plane $x-y+z=5$ has normal vector $\c1{-1}1$, so this is a vector in $\Pi$. Moreover, $\vec{AB}=\c3{-5}4$ is also a vector in $\Pi$, so it has normal vector \[ \nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}.\] So the equation of $\Pi$ is $x-y-2z=d$ and subbing in $A=(1,3,-3)$ gives $d=1-3-2(-3)=4$, so the equation of $\Pi$ is \[ x-y-2z=4.\]
2. The plane parallel to $2x-4y+5z=8$ passing through $(1,2,3)$ is $2x-4y+5z=2(1)-4(2)+5(3) = 10$, or $2x-4y+5z=10$.
3. Find the equation of the plane $\Pi$ which contains the line of intersection of the planes \[ \Pi_1: x-y+2z=1\quad\text{and}\quad \Pi_2: 3x+2y-z=4,\] and is perpendicular to the plane $\Pi_3:2x+y+z=3$.
Solution: To find the line of intersection of $\Pi_1$ and $\Pi_2$, we must solve the system of linear equations \begin{gather}x-y+2z=1\\3x+2y-z=4.\end{gather} We can solve this linear system in the usual way, by applying EROs to the matrix $\begin{bmatrix}1&-1&2&1\\3&2&-1&4\end{bmatrix}$: \begin{align*} \def\go#1#2{\begin{bmatrix}#1\\#2\end{bmatrix}} \def\ar#1{\\[6pt]\xrightarrow{#1}&} &\go{1&-1&2&1}{3&2&-1&4} \ar{R2\to R2-3R1} \go{1&-1&2&1}{0&5&-7&1} \ar{R1\to 5R1+R2} \go{5&0&3&6}{0&5&-7&1} \ar{R1\to\tfrac15R1,\,R2\to\tfrac15R2} \go{1&0&3/5&6/5}{0&1&-7/5&1/5} \end{align*} So the line $L$ of intersection is given by \[ L: \c xyz=\c{\tfrac65}{\tfrac15}0+t\c{-\tfrac35}{\tfrac75}1,\quad t\in\mathbb{R}.\] So $\c{-\tfrac35}{\tfrac75}1$ is a direction vector along $L$, and also $5\c{-\tfrac35}{\tfrac75}1=\c{-3}75$ is a vector along $L$. So $\c{-3}75$ is a vector in the plane $\Pi$. Moreover, taking $t=2$ gives the point $(0,3,2)$ in the line $L$, so this is a point in $\Pi$.
Since $\Pi$ is perpendicular to $\Pi_3$, which has normal vector $\nn_3=\c211$, the vector $\c211$ is in $\Pi$.
So a normal vector for $\Pi$ is \[ \nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}\] hence $\Pi$ has equation $-2x-13y+17z=d$, and subbing in the point $(0,3,2)$ gives $d=0-13(3)+17(2)=-39+34=-5$, so $\Pi$ has equation $-2x-13y+17z=-5$, or \[ 2x+13y-17z=5.\]
The distance to a plane
The distance from a point to a plane
Let $\Pi$ be a plane in $\def\rt{\mathbb{R}^3}\rt$ with equation $ax+by+cz=d$, so that $\def\nn{\vec n}\nn=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c abc$ is a normal vector to $\Pi$. Also let $A$ be any point in $\rt$.
The shortest path from $A$ to a point in $\Pi$ goes in the same direction as $\nn$. Let $B$ be any point in the plane $\Pi$.
From the diagram, we see that the shortest distance from $A$ to $\Pi$ is given by \[ \text{dist}(A,\Pi)=\|\def\pp{\vec p}\pp\|\] where \[ \pp=\text{proj}_{\nn}{\vec{AB}}.\] Using the formula for $\text{proj}_{\vec w}\vec v$ and the fact that $\|c\vec v\|=|c|\,\|\vec v\|$ where $c$ is a scalar and $\vec v$ is a vector, we obtain the formula \[ \text{dist}(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}.\]
Example
To find the distance from $A=(1,-4,3)$ to the plane $\Pi:2x-3y+6z=1$, choose any point $B$ in $\Pi$; for example, let $B=(2,1,0)$. Then $\nn=\c2{-3}6$ and $\vec{AB}=\c15{-3}$, so \[ \def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7.\]