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• Equation of a plane $\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{smallmatrix}\right|}\def\nn{\vec n}\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\def\uu{\vec u}\def\vv{\vec v}\def\ww{\vec w}\def\bR{\mathbb R}\Pi$ 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:

1. $\Pi_1$ and $\Pi_2$ are orthogonal planes;
2. $\nn_1\cdot\nn_2=0$;
3. $\nn_1$ is a vector in $\Pi_2$;
4. $\nn_2$ 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 $2x+13y-17z=5$.

The distance to a plane

Distance from $A$ to $\Pi$

• $A$: any point in $\def\rt{\mathbb R^3}\rt$. $\Pi$: a plane with normal vector $\nn$.
• $\nn$ is direction of shortest path from $A$ to $\Pi$
• Let $B$ be any point in the plane $\Pi$.

• (shortest) distance from $A$ to $\Pi$ is $\text{dist}(A,\Pi)=\|\def\pp{\vec p}\pp\|$
• where $\pp=\text{proj}_{\nn}{\vec{AB}}$.
• Do some algebra: we get $\text{dist}(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}$.

Example

Find the distance from $A=(1,-4,3)$ to the plane $\Pi:2x-3y+6z=1$.

• choose any point $B$ in $\Pi$, e.g. $B=(2,1,0)$
• $\nn=\c2{-3}6$ and $\vec{AB}=\c15{-3}$
• So $\def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7$.

The distance from the origin to a plane

• We write $0=(0,0,0)$ for the origin in $\rt$
• Distance from $0$ to a plane $\Pi:ax+by+cz=d$ ?
• Take $B=(d/a,0,0)$ (assuming that $a\ne 0$)
• We get $\dist(0,\Pi)=\frac{|d|}{\|\nn\|}$ where $\nn$ is the normal vector $\nn=\c abc$.
• In particular, if $\nn$ is a unit vector, then $\dist(0,\Pi)=|d|$.
• As $d$ varies (with $\nn$ fixed), we obtain parallel planes at different distances to the origin $0$
• The larger $d$ is, the further the plane is from $0$.

The distance between planes

• Let $\Pi_1$ and $\Pi_2$ be two planes. What is the (shortest) distance between them?
• If they're not parallel, then they intersect! So $\dist(\Pi_1,\Pi_2)=0$.
• If they're parallel, then $\dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2)$ for any point $A$ in $\Pi_1$.
  • Why?
  • Since the planes are parallel, $\dist(A,\Pi_2)$ doesn't change if $A$ changes in $\Pi_1$
  • So this is also the shortest distance, for any choice of $A$ in $\Pi_1$

Example

The distance between the planes $3x+4y-2z=5$ and $3x+4y-3z=1$ is $0$, since the normal vectors $\c34{-2}$ and $\c34{-3}$ are not scalar multiples of one another, so they are in different directions, so the planes are not parallel.

Example

The planes $\Pi_1:3x+4y-2z=5$ and $\Pi_2:3x+4y-2z=1$ have the same normal vector $\c34{-2}$, so they are parallel. Their distance is given by $\dist(A,\Pi_2)$ where $A$ is any point in $\Pi_1$, and to find this we also need a point $B$ in $\Pi_2$.

We can choose $A=(1,0,-1)\in \Pi_1$ and $B=(1,0,1)\in \Pi_2$. (Of course, there are lots of different possible choices here, but they should all give the same answer!) Then $\vec {AB}=\c002$ and \[ \dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2) = \frac{|\nn\cdot \vec{AB}|}{\|n\|}=\frac{|0+0+(-2)2|}{\sqrt{3^2+4^2+(-2)^2}} = \frac4{\sqrt{29}}.\]

Exercise: a formula for the distance between parallel planes

Show that the distance between the parallel planes $\Pi_1:ax+by+cz=d_1$ and $\Pi_2:ax+by+cz=d_2$ is \[\dist(\Pi_1,\Pi_2)=\frac{|d_2-d_1|}{\|\nn\|},\] where $\nn=\c abc$.

Example

To find the distance between $x+3y-5z=4$ and $2x+6y-10z=11$ we can rewrite the second equation as $x+3y-5z=11/2$ to see that this is a parallel plane to the first, with common normal vector $\nn=\c13{-5}$. By the formula in the exercise the distance between these planes is \[ \frac{|\tfrac{11}2-4|}{\|\nn\|} = \frac{|\tfrac 32|}{\sqrt{1^2+3^2+(-5)^2}} = \frac3{2\sqrt{35}}.\]