Minimization of the distance between 2 vectors

Question

Find the value of $t$ for which the vector

${v} = \begin{pmatrix} 2 \\ -3 \\ -3 \end{pmatrix} + \begin{pmatrix} 7 \\ 5 \\ -1 \end{pmatrix} t$

is closest to

${a} = \begin{pmatrix} 4 \\ 4 \\ 5 \end{pmatrix}.$

My original way to attempt this was to find an equation relating the distance and $t$, and then finding the absolute minimum if said equation, but the instructions say to use only linear algebraic techniques(cross product, dot product, components, projections, etc.)

Is there a better way to solve this problem?

Thanks in advance!

Answer 1

You can write also $$(-2+7t,-7+5t,-8-t)(7,5,-1)=0$$

Answer 2

You can find the distance using the area of a triangle as follows.

1) Pick two points $p$ and $q$ on the line.

2) Find the area of the triangle $apq$ by dividing the norm of the cross product of $ap$ and $aq$ by $2$

3) Find the distance between $p$ and $q$

4) Divide twice the area of the triangle $apq$ by the distance $pq$ and you get the required distance.