The distance from a point to a line

Cross product method

Suppose $L$ is a line in $\def\rt{\mathbb{R}^3}\def\rn{\mathbb{R}^n}\rt$. Let $A$ be a point on $L$ and let $\def\vv{\vec v}\vv$ be a direction vector along $L$.

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

Let $A$ be any point in $L$ and let $\theta$ be the angle between $AB$ and $\vv$. We have \[ d=\|\vec{AB}\|\,\sin \theta = \frac{\|\vec{AB}\|\,\|\vv\|\sin \theta}{\|\vv\|} = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}.\] So \[ \text{dist}(B,L) = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}\] where $A$ is any point in $L$.

Example

To find the distance from the point $B=(1,2,3)$ to the line \[L:\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}\] we can choose $A=(1,0,{-1})$ so that $\vec{AB}=\c024$ and taking $\vv=\c41{-5}$, we obtain \[ \vec{AB}\times \vv = \begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\0&2&4\\4&1&-5\end{vmatrix}=\c{-14}{16}{-8}=2\c{-7}{8}{-4}\] so \[ \def\dist{\text{dist}}\dist(B,L)=\frac{\|\vec{AB}\times\vv\|}{\|\vv\|}=\frac{2\sqrt{7^2+8^2+4^2}}{\sqrt{4^2+1^2+5^2}} = \frac{2\sqrt{129}}{\sqrt{42}} \approx 3.5051.\]

Dot product method

The method above relies on the cross product, so only works in $\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\def\rt{\mathbb{R}^3}\def\rn{\mathbb{R}^n}\def\vv{\vec v}\def\dist{\text{dist}}\rt$. The following alternative method works in $\rn$ for any $n$.

Observe that $\dist(B,L)$ is the length of the vector $\def\nn{\vec n}\nn=\vec{AB}-\def\pp{\vec p}\pp$ where $\pp=\def\proj{\text{proj}}\proj_{\vv}\vec{AB}$.

Example

Let's redo the previous example using this method.

We have $\vec{AB}=\c024$ and $\vv=\c41{-5}$, so \[\pp=\proj_{\vv}\vec{AB} = \left(\frac{\vec{AB}\cdot \vv}{\|\vv\|^2}\right)\vv = \frac{0(4)+2(1)+4(-5)}{4^2+1^2+5^2}\c41{-5} = -\frac{18}{42}\c41{-5}=-\frac{3}{7}\c41{-5},\] so \[ \nn=\vec{AB}-\pp=\c024-(-\frac37)\c41{-5}=\c024+\frac37\c41{-5}=\frac17\c{12}{17}{13}\] so \[ \dist(B,L)=\|\nn\|=\frac17\sqrt{12^2+17^2+13^2} = \frac17\sqrt{602} \approx 3.5051.\]

The distance between lines in $\mathbb{R}^3$

Skew lines

Suppose that $L_1$ and $L_2$ are skew lines in $\rt$: lines which are not parallel and do not cross.

Let $\vv_1$ be a direction vector along $L_1$, and let $\vv_2$ be a direction vector along $L_2$.

The shortest distance from $L_1$ and $L_2$ is measured along the direction orthogonal to both $\vv_1$ and $\vv_2$, namely the direction of $\nn=\vv_1\times\vv_2$.

Let $\Pi$ be the plane with normal vector $\nn$ which contains $L_1$.

For any point $B$ in $L_2$, we have \[\dist(L_1,L_2)=\dist(B,\Pi) = \frac{|\vec{AB}\cdot \nn|}{\|\nn\|}\] where $A$ is any point in $\Pi$; for example, we can take $A$ to be any point in $L_1$.

To summarise: for skew lines $L_1$ and $L_2$ with direction vectors $\vv_1$ and $\vv_2$, we have \[ \dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}\] where $\nn=\vv_1\times\vv_2$ and $A$ and $B$ are points with one in $L_1$ and the other in $L_2$.

Remark

What about the distance between lines which are not skew? This means that either they are non-parallel and they intersect (so that the distance between them will be zero), or they are parallel lines.

• The same method and formula work if $L_1$ and $L_2$ are non-parallel lines which intersect, and you get $\dist(L_1,L_2)=0$ in this case. The reason is that in this case $L_1$ and $L_2$ will lie in one plane, $\Pi$, and $\vec{AB}$ will also be in $\Pi$, and $\vec n$ will be orthogonal to $\Pi$. So $\frac{\vec{AB}\cdot \vec n}{\|\vec n\|}=\frac{0}{\|\vec n\|}=0=\dist(L_1,L_2)$.
• If $L_1$ and $L_2$ are parallel lines (i.e., if the vectors $\vec v_1$ and $\vec v_2$ along the lines are in the same direction), then $\vv_1\times \vv_2=0$ which isn't helpful, so this method won't work here. In this case, observe that $\dist(L_1,L_2)=\dist(A,L_2)$ where $A$ is any point in $L_1$ (because the lines are parallel), so you can use one of the formulae above for the distance from a point to a line.

Example

Consider the skew lines \begin{align*} x&=1+t_1\\L_1:\quad y&=2t_1\qquad (t_1\in \mathbb{R})\\z&=1+3t_1 \end{align*} and \[ L_2:\c xyz=\c 321+t_2\c1{-1}1,\quad t_2\in\mathbb{R}.\]

Note that we can rewrite the equation of $L_1$ in “vector form”, which is easier to digest: \[ L_1:\c xyz=\c 101+t_1\c123,\quad t_1\in\mathbb{R}\] The direction vectors are $\c123$ and $\c1{-1}1$, so we take $\nn$ to be their cross product: \[ \nn=\c123\times\c1{-1}1=\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\1&2&3\\1&-1&1\end{vmatrix}=\c52{-3}\] and if $A=(1,0,1)$ and $B=(3,2,1)$ then $A$ and $B$ are points with one in $L_1$ and the other in $L_2$, and $\vec{AB}=\c 220$. Hence \[\dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}=\frac{2(5)+2(2)+0(-3)}{\sqrt{5^2+2^2+3^2}} = \frac{14}{\sqrt{38}}.\]

Distance between lines in $\mathbb{R}^3$ in general

The formula $\dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}$ where $\nn=\vec v_1\times \vec v_2$ works for

• skew lines (not parallel, not intersecting), as we saw above,
• and actually: any non-parallel lines $L_1$, $L_2$. We can see this by noticing that if the lines above intersect, then $L_1$ lies in $\Pi$, so the formula gives $\dist(L_1,L_2)=\dist(B,\Pi)=0$ which is the correct answer.

What about parallel lines?

• The formula can't work because we'd have $\vec v_1=\vec v_2$ so $\vec n=\vec v_1\times \vec v_2=\vec 0$
• Instead: observe that when $L_1$ and $L_2$ are parallel, we have $\dist(L_1,L_2)=\dist(A,L_2)$ for any point $A$ in $L_1$
• So we can use one of of the point-to-line distance formulae we saw earlier.