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- Equation of a plane $\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1\\#4\end{smallmatrix}\right|}\def\nn{\vec n}\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\def\uu{\vec u}\def\vv{\vec v}\def\ww{\vec w}\def\bR{\mathbb R}\Pi$ in $\mathbb{R}^3$ is $\vec n\cdot \c xyz=d$
- $\vec n$ is a fixed vector, called the normal vector to the plane
- if $\vec n=\c abc$, then the equation of $\Pi$ is $ax+by+cz=d$
- $d$ is a fixed number
Example 4
Find the equation of the plane $\Pi$ containing $A=(1,2,0)$, $B=(3,0,1)$ and $C=(4,3,-2)$.
Solution
- $\vec{AB}=\c2{-2}1$ and $\vec{AC}=\c31{-2}$ are both vectors in $\Pi$
- Need $\nn$, orthogonal to both. Use cross product!
- $\nn=\vec{AB}\times\vec{AC}=\cp2{-2}131{-2}=\c378$
- Equation is $3x+7y+8z=d$; find $d=17$ by subbing in $A=(1,2,0)$
- Answer: $ 3x+7y+8z=17$.
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 parallel planes if $\nn_1$ and $\nn_2$ are are in the same direction
- If $\Pi_1$ has equation $ax+by+cz=d_1$ then any parallel plane $\Pi_2$ has an equation which may be written with the same left hand side: $ax+by+cz=d_2$.
- i.e., we can assume that $\nn_1=\nn_2=\c abc$.
Example
The plane parallel to $2x-4y+5z=8$ passing through $(1,2,3)$ is
- $2x-4y+5z=2(1)-4(2)+5(3) = 10$
- i.e., $2x-4y+5z=10$.
Orthogonal 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 $\Pi_1$.
Example 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 $x-y+z=5$.
- $x-y+z=5$ has normal $\c1{-1}1$; this is in $\Pi$.
- $\vec{AB}=\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\c3{-5}4$ is also a vector in $\Pi$
- Normal for $\Pi$: $\def\nn{\vec n}\nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}$.
- Sub in $A$ (or $B$): get $x-y-2z=4$.
Example 2
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 $\Pi_3:2x+y+z=3$.
- First find the line of intersection of $\Pi_1$ and $\Pi_2$
- Solve $x-y+2z=1$, $3x+2y-z=4$
- $\left[\begin{smallmatrix}1&-1&2&1\\3&2&-1&4\end{smallmatrix}\right]\to_{EROs}\left[\begin{smallmatrix}1&0&3/5&6/5\\0&1&-7/5&1/5\end{smallmatrix}\right]$
- Line $L$ of intersection is $\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$.
- $\Pi$ contains $L:\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$, orthogonal to $\Pi_3: 2x+y+z=3$
- $5\c{-3/5}{7/5}1=\c{-3}75$ is a vector along $L$, so in $\Pi$
- $\Pi_3$ has normal vector $\nn_3=\c211$, which is in $\Pi$.
- Normal vector for $\Pi$: $\nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}$
- $\Pi$ has equation $-2x-13y+17z=d$
- $\Pi$ has equation $-2x-13y+17z=d$ and contains $L:\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$
- Take $t=2$: $(0,3,2)$ in $L$, so in $\Pi$
- Sub in: $d=0-13(3)+17(2)=-39+34=-5$
- Answer: $-2x-13y+17z=-5$, or $2x+13y-17z=5$.