Shortest distance between 2 lines with the same direction vector.

Question

Let $L_1$ be the line passing through the point $P_1=(2, 1, −2)$ with direction vector $v=[−3, 2, 3]$, and let $L_2$ be the line passing through the point $P_2=(1, −4, −1)$ with the same direction vector. Find the shortest distance $d$ between these two lines, and find a point $Q_1$ on $L_1$ and a point $Q_2$ on $L_2$ so that $d(Q_1,Q_2) = d$.

I've tried using the cross product on the direction vectors and using them to find points on $L_1$ and $L_2$. But that didn't help one bit, and I don't really know how else to go about this. Any help?

Answer 1

Since $L_1$ and $L_2$ have the same direction vector of $v=[-3,2,3]$ then they are parallel. If $\theta$ is the angle between $P_1 - P_2$ and $v$, then the distance between them is just

$ d = \| P_1 - P_2 \| \sin \theta = \dfrac{\| (P_1 - P_2) \times v \| }{\| v \|}$

This evaluates to

$ d = \dfrac{\| (1,5,-1) \times (-3, 2, 3) \|}{\sqrt{22} } = \dfrac{ \| (17, 0, 17) \| }{\sqrt{22} } = \dfrac{ 17 }{\sqrt{11} } $

To check this result, let $P$ slide on the second line $L_2$, then

$ P = P_2 + t v = ( 1, -4, -1 ) + t (-3, 2, 3) = (1 - 3 t, -4 + 2 t , -1 + 3 t) $

The squared distance between $P_1$ and $P$ is

$d^2 = \| P - P_1 \|^2 = \| (-1 - 3 t, -5 + 2 t, 1 + 3 t ) \|^2 = 22 t^2 - 8 t + 27 $

The minimum of $d^2$ is at $t = - \dfrac{-8}{44} = \dfrac{2}{11} $

Substitute this $t$, you get

$ d^2 = 22 \bigg(\dfrac{2}{11}\bigg)^2 - 8 \bigg(\dfrac{2}{11}\bigg) + 27 = \dfrac{289}{11} $

Therefore, the distance is

$ d= \sqrt{\dfrac{289}{11} } = \dfrac{17}{\sqrt{11}} $

Answer 2

You may find some success by reading up on a similar question here: Closest points between two lines