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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 $\vec{AC}$: \[ \nn=\vec{AB}\times\vec{AC}=\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\cp2{-2}131{-2}=\c378\] 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 $\nn_2$.

1. $\Pi_1$ 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 $\Pi_1$.
2. $\Pi_1$ 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 \[ \def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7.\]

Remark: the distance from the origin to a plane

If we write $0=(0,0,0)$ for the origin in $\rt$ and apply the formula above to the plane $\Pi:ax+by+cz=d$ with $B=(d/a,0,0)$ (assuming that $a\ne 0$) then we obtain \[ \dist(0,\Pi)=\frac{|d|}{\|\nn\|}\] where $\nn$ is the normal vector $\nn=\c abc$.

So as $d$ varies (with the normal vector $\nn$ 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 parallel planes

If $\Pi_1$ and $\Pi_2$ are parallel planes, then the shortest distance between them is given by \[ \dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2)\] for any point $A$ is $\Pi_1$. The reason is that for parallel planes, changing $A$ to a different point in $\Pi_1$ does not change $\dist(A,\Pi_2)$.

Of course, if the planes $\Pi_1$ and $\Pi_2$ are not parallel, then they intersect (in many points: in a whole line). So for non-parallel planes we always have $\dist(Pi_1,Pi_2)=0$.

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 \[ \frac{|\tfrac{11}2-4|}{\|\nn\|} = \frac{|\tfrac 32|}{\sqrt{1^2+3^2+(-5)^2}} = \frac3{2\sqrt{35}}.\]

Lines in $\mathbb{R}^3$

A line $L$ in $\rt$ has an equation of the form \[ L: \c xyz=\c abc+t\c def, \quad t\in \mathbb R\] where $a,b,c,d,e,f$ are fixed numbers. This is called a parametric equation, since the variable $t$ is not fixed: it is a free parameter, allowed to take any real value.

Setting $t=0$ we see that $A=(a,b,c)$ is a point in the line $L$. Also, setting $t=1$ shows that $B=(a+d,b+e,c+f)$ is another point in $L$, so the vector $\vec {AB}=\c def$ is a direction along the line $L$.

Example

To find the parametric equation of the line in $\rt$ which passes through $A=(2,1,-3)$ and $B=(4,-1,5)$, note that $\vec{AB}=\c 2{-2}{8}$, so the equation is \[ L: \c xyz=\c 21{-3}+t\c 2{-2}8,\quad t\in \mathbb{R}.\]