Find the two points where the shortest distance occurs on two lines

Question

Find the point P on $\vec{AB}$ and point Q on $\vec{CD}$ such that $\vec{PQ}$ is the shortest distance between the lines AB and CD, given $\vec{AB} = \begin{pmatrix} 1\\ 0\\ 2\\ \end{pmatrix} + u\begin{pmatrix} -2\\ 2\\ 1\\ \end{pmatrix} ,\vec{CD} = \begin{pmatrix} 0\\ 1\\ 1\\ \end{pmatrix} + v\begin{pmatrix} 2\\ -1\\ -2\\ \end{pmatrix} $ the normal vector $n=\begin{pmatrix} -3\\ -2\\ -2\\ \end{pmatrix} $

and the shortest distance is $\frac{3}{\sqrt{17}}$. I figured all this out from 4 given points but don't know how to find points P and Q. Please help, I'm stuck...

Answer 1

Once you know that the normal vector is $n$, the vector equation $$ \begin{pmatrix} 1\\ 0\\ 2\\ \end{pmatrix} + u\begin{pmatrix} -2\\ 2\\ 1\\ \end{pmatrix} +w\begin{pmatrix} -3\\ -2\\ -2\\ \end{pmatrix} = \begin{pmatrix} 0\\ 1\\ 1\\ \end{pmatrix} + v\begin{pmatrix} 2\\ -1\\ -2\\ \end{pmatrix} $$ is equivalent to a system of three linear equations in three unknowns, which indeed has the unique solution $$ u=\frac{19}{17},v=-\frac{15}{17},w=\frac{3}{17}. $$ That tells you that the points on $\vec{AB}$ and $\vec{CD}$ are $$ \begin{pmatrix} 1\\ 0\\ 2\\ \end{pmatrix} + \frac{19}{17}\begin{pmatrix} -2\\ 2\\ 1\\ \end{pmatrix} \quad\text{and}\quad\begin{pmatrix} 0\\ 1\\ 1\\ \end{pmatrix} -\frac{15}{17}\begin{pmatrix} 2\\ -1\\ -2\\ \end{pmatrix}, $$ respectively. (It also tells you that the distance between the two lines is the norm of $\frac3{17}(-3\ {-2}\ {-2})$, or $\frac3{\sqrt{17}}$ as you indicated.)

Answer 2

If the normal vector is not provided in the question, you can still solve the question.

Any point $P$ on line $\vec{AB}$ given $u$ can be represented as $\begin{pmatrix} 1-2u \\ 2u\\ 2+u \end{pmatrix}$ , whilst point any point $Q$ on line $\vec{CD}$ given $v$ can be represented as $\begin{pmatrix} 2v \\ 1-v\\ 1-2v \end{pmatrix}$, $\vec{PQ}=$ $\begin{pmatrix} -1+2u+2v \\ 1-2u-v\\ -1-u-2v \end{pmatrix}$

You can interpret the shortest distance between $\vec{AB}$ and $\vec{CD}$ at points $P$ and $Q$ as a separate line that is perpendicular to both $\vec{AB},\vec{CD}$ at the same time, similar to how the shortest distance from a point to a line functions: $$ \vec{PQ} \perp \vec{AB} \\ \vec{PQ} \perp \vec{CD} $$ The dot product between two perpendicular vectors is 0, which gives: $$ \vec{PQ} \cdot \vec{AB} = \begin{pmatrix} -1+2u+2v \\ 1-2u-v\\ -1-u-2v \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 2\\ 1 \end{pmatrix} = 0 \\ \vec{PQ} \cdot \vec{CD} = \begin{pmatrix} -1+2u+2v \\ 1-2u-v\\ -1-u-2v \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1\\ -2 \end{pmatrix} = 0 $$ Which gives the following system of equations $$ \begin{aligned} 3-9u-8v = 0\\ -1+8u+9v=0 \end{aligned} $$ Solving for $u,v$ gives $u=\frac{19}{17},v=-\frac{15}{17}$, which can be substituted back to calculate the coordinates of $P$ and $Q$, giving $P=(\frac{38}{17}, \frac{53}{17}, -\frac{21}{17}), Q=(-\frac{30}{17}, \frac{32}{17}, \frac{47}{17})$