The distance to a plane

Distance from $A$ to $\Pi$

• $\def\dist{\text{dist}}\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{smallmatrix}\right|}\def\nn{\vec n}\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\def\uu{\vec u}\def\vv{\vec v}\def\ww{\vec w}\def\bR{\mathbb R}\def\rt{\bR^3}A$: 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, 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

What is the distance between the planes $3x+4y-2z=5$ and $3x+4y-3z=1$?

• The normal vectors are $\c34{-2}$ and $\c34{-3}$
  • They aren't scalar multiples of one another
  • So they're they are in different directions
• So the planes are not parallel.
• So they intersect, and the distance is $0$.

Example

Find the distance between the planes $\Pi_1:3x+4y-2z=5$ and $\Pi_2:3x+4y-2z=1$.

• Same normal vector $\nn=\c34{-2}$, so parallel planes.
• Distance is $\dist(A,\Pi_2)$ where $A$ is any point in $\Pi_1$
  • To find this we also need a point $B$ in $\Pi_2$.
• Choose $A=(1,0,-1)$, $B=(1,0,1)$.
• $\vec {AB}=\c002$ so $\dist(A,\Pi_2) = \frac{|\nn\cdot \vec{AB}|}{\|n\|}=\frac{|0+0+(-2)2|}{\sqrt{3^2+4^2+(-2)^2}} = \frac4{\sqrt{29}}.$

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 $\nn=\c abc$.

• Important: need equations with the same left hand side.

Example

What is the distance between $x+3y-5z=4$ and $2x+6y-10z=11$?

• Rewrite the second equation as $x+3y-5z=11/2$
• So the planes are parallel, with common normal vector $\nn=\c13{-5}$.
• By the formula, the distance 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.

• Called a parametric equation
  • because of the variable $t$: a free parameter

• 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}$).

The distance from a point to a line

How can we find $d=\text{dist}(B,L)$, the distance from the point $B$ to a line $L$?

• Let $A$ be a point on $L$
• Let $\def\vv{\vec v}\vv$ be a direction vector along $L$.
• 
• $\text{dist}(B,L) =\|\vec{AB}\|\,\sin \theta = \frac{\|\vec{AB}\|\,\|\vv\|\sin \theta}{\|\vv\|} = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}$.

Example

Find the distance from the point $B=(1,2,3)$ to the line \[L:\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}.\]

• Choose $A=\c10{-1}$, in $L$, and $\vv=\c41{-5}$, along $L$
• $\vec{AB}=\c024$, so $ \vec{AB}\times \vv = \cp02441{-5}=\c{-14}{16}{-8}=2\c{-7}{8}{-4}$
• So $\def\dist{\text{dist}}\dist(B,L)=\frac{\|\vec{AB}\times\vv\|}{\|\vv\|}=\frac{2\sqrt{7^2+8^2+4^2}}{\sqrt{4^2+1^2+5^2}} = \frac{2\sqrt{129}}{\sqrt{42}} \approx 3.5051.$

Alternative formula

• $\dist(B,L)=\tfrac{\|\vec{AB}\times \vv\|}{\|\vv\|}$ only works in $\rt$
  • because the cross product is only defined in $\rt$
• 
• $\def\pp{\vec p}\dist(B,L)=\|\nn\|=\|\vec{AB}-\pp\|$
  • where $\pp=\def\proj{\text{proj}}\proj_{\vv}\vec{AB}$.
• This works in $\bR^n$ for any $n$

Example, again

Find the $\dist(B,L)$ for $B=(1,2,3)$ and $L:\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}$.

• $\vec{AB}=\c024$ and $\vv=\c41{-5}$
• $\pp=\proj_{\vv}\vec{AB} = \left(\frac{\vec{AB}\cdot \vv}{\|\vv\|^2}\right)\vv = \frac{-18}{42}\c41{-5}= -\frac{3}{7}\c41{-5}$
• $\nn=\vec{AB}-\pp=\c024-(-\frac37)\c41{-5}=\frac17\c{12}{17}{13}$
• $\dist(B,L)=\|\nn\|=\frac17\sqrt{12^2+17^2+13^2} \approx 3.5051$.