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 $\def\rt{\mathbb{R}^3}\def\dist{\text{dist}}\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}0=(0,0,0)$ 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.

The variable $t$ is called a “free parameter”: it's “free” because it can take any value, and it's a “parameter” because this is just another name for a variable.

The equation above is called a parametric equation for the line $L$, because of the free parameter $t$.

What do $a,b,c,d,e,f$ mean?

• Set $t=0$: $A=(a,b,c)$ is a point in $L$
• Set $t=1$: $B=(a+d,b+e,c+f)$ is another
• So $\vec {AB}=\c def$ is a direction along $L$.

Example

Find the parametric equation of the line $L$ in $\rt$ which passes through $A=(2,1,-3)$ and $B=(4,-1,5)$.

• Note that $\vec{AB}=\c 2{-2}{8}$
• So this is a direction vector along the line $L$
• So the equation is $ L: \c xyz=\c 21{-3}+t\c 2{-2}8,\quad t\in \mathbb{R}$.
• (Same as $ L: \c xyz=\c 4{-1}5+t\c 2{-2}8,\quad t\in \mathbb{R}$).