Let L1 be the line passing through the point $P_1=(−11, 10, −1)$ with direction vector $\vec d_1=\begin{bmatrix}−2\\ 3\\ −1\end{bmatrix}t$, and let $L_2$ be the line passing through the point $P_2=(−8, 9, 10)$ with direction vector $\vec d_2=\begin{bmatrix}−1\\ 3\\1\end{bmatrix}t$. 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 dont really understand how to properly approach this question, how do i start?


Hints: any point on $L_1$ can be written as $(-11-2t,10+3t,-1-t)$ where t is a parameter. $Q_1$ is one such point. Likewise for $L_2$ with parameter, say s.

Note that the line connecting $Q_1$ and $Q_2$ is perpendicular to both $d_1$ and $d_2$ so, must be along their cross product $d_3$. Parametrise that direction by u. So,

$$Q_1 + u d_3=Q_2$$

The three components of this give 3 equations for s, t, u which are easily solved and you get everything you need.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.