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.

  • $\begingroup$ Can you use differential calculus? $\endgroup$
    – DonAntonio
    Oct 11 '12 at 3:20
  • $\begingroup$ No, I cannot, I think I may have figured it out by doing the do product of the vectors t and s with the cross product of s and t, then subtracting them. Incidentally the shortest distance is 0, I believe. $\endgroup$
    – Unknown
    Oct 11 '12 at 3:25

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.)

  • 1
    $\begingroup$ This method does not give the correct answer because $x_{perp}$ is not guaranteed to be orthogonal to line 2. In fact, there are infinitely many lines that pass through $y_1$ and give a different distance. Counterexample: Consider lines $f: x=(0,0,0)+(1,0,0)t$ and $g: x=(0,0,1)+(0,1,0)t$. Choose $x_1=(2,0,0)$, $x_2=(1,0,0)$, $y_1=(0,1,1)$. Then $d=(2,-1,-1)$, $x=(1,0,0)$, $x_{para}=(2,0,0)$, $x_{perp}=(0,-1,-1)$, which has norm $\sqrt{2}$, but the correct answer is clearly $1$. $\endgroup$
    – durianice
    Jan 20 at 19:11

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: $$d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$$


the answer is a little tricky, first use cross product to find n by using the two direction vector.$(d1*d2)$, |i,j,k 1,1,-1 -2,0,1| = $i+j+2k$. then let point $p$ and $s$ be on the two line respectively, find vector $ps. = (5,-3,-3)-(-1,1,4) = (6,2,-7)$ then find the projection of $ps$ 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 $2.44949.$


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