Finding the shortest distance between two lines

I know how to find the distance between a point and a line, not between two lines.

Find the shortest distance between the lines $(-1,1,4) + t(1,1,-1)$ and $(5,3,-3) + s(-2,0,1)$

Any help would be appreciated.

Solution 1:

The distance between two lines in $ \Bbb R^3 $ is equal to the distance between parallel planes that contain these lines.

To find that distance first find the normal vector of those planes - it is the cross product of directional vectors of the given lines. For the normal vector of the form (A, B, C) equations representing the planes are:

$ Ax + By + Cz + D_1 = 0 $
$ Ax + By + Cz + D_2 = 0 $

Take coordinates of a point lying on the first line and solve for D1.
Similarly for the second line and D2.

The distance we're looking for is: $$d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$$

Solution 2:

Let $x_1$ and $y_1$ be 2 points on the line 1 and line 2 respectively. Form the difference vector $d=x_1-y_1$. Take another point $x_2$ on the line 1. Form the direction vector $x=x_1-x_2$. Project $d$ on to the direction vector $x$.

\begin{align} x_{parallel}= \frac{(d.x)}{||x||^2}x \end{align}

Now the norm of the following vector (the euclidean distance from the origin), will give you the required minimum distance.

\begin{align} x_{perp}= d-x_{parallel} \end{align}

(if they are not parallel, this will not work, instead it gives the shortest distance between the point $x_1$ and line 2.)

Solution 3:

The answer is a little tricky, first use the cross product to find $n$ by using the two direction vector. $(d_1 \times d_2)$ \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ -2 & 0 & 1 \end{bmatrix} = $i+j+2k$.

Then (with points $P$ and $S$, a point from each line) find vector $\vec{PS}. = (5,3,-3)-(-1,1,4) = (6,2,-7)$; then find the projection of $\vec{PS}$ onto $n$ and find the length of the projection. $(6,2,-7) \cdot \frac{(1,1,2)}{||1,1,2||^2}= 6^{1/2},$ or $2.44949$