Finding points in 3D plane that are closest possible

Question

Find points P,Q which are closest possible with P lying on the line

$x$ $=$ $-1-8t$, $y$ $=$ $-9+5t$, $z$ $=$ $1+3t$

and Q lying on the line

$x$ $=$ $4+6s$, $y$ $=$ $-253+12s$, $z$ $=$ $705+7s$.

So this is what I did:

I computed the cross product like this:

$det\begin{pmatrix}i&j&k\\ -8&5&3\\ 6&12&7\end{pmatrix}$

and got $-i+72j-102k$

And, therefore, the normal vector is:

$\begin{pmatrix}-1\\ 72\\ -102\end{pmatrix}$

But I am not sure how to proceed after this. I believe I need to calculate the projection but I am not sure how with the variables I have been given. And also, what exactly would I need to do after calculating this projection.

Any help would be highly appreciated!

Answer 1

Hint:

$\overrightarrow{PQ}$ should be parallel to the normal vector.

Express $\overrightarrow{PQ}$ in terms of $s$ and $t$. Then find $s$ and $t$ using proportion.

Answer 2

$\vec{PQ}$ is $(5+6s-8t,-244+12s-5t,704+7s-3t)$. Now note that this vector must be perpendicular to $(-8,5,3)$ as well as $(6,12,7)$ (these are the two vectors that you have put into your determinant).

These conditions give you two equations in the variables $s$ and $t$, which you can easily solve.