Good day everyone.
I have the following question, which asks to apply Lagrange multipliers technique to solve the optimization problem. So the question is following:
Suppose that L1 is the line of intersection of the planes
\begin{align} \ 2x+y+2z=15 \\\\ \ x+2y+3z=30 \end{align}
and that L2 is the line of intersection of the planes
\begin{align} \ x-y-2z=15 \\\\ \ 3x-2y-3z=20 \end{align}
Find the closest points P1 and P2 on these two skew lines. (There is a hint as well which says that I should get 10 linear equation to solve).
My attempt to solve it so far was as follows. I decided to derive equations of the lines that result from the intersection of two planes pairs. So, I used two normal vectors of two planes, and by the cross product I got the vector parallel to the intersection line under question. For the L1 line the symmetric equations are these:
\begin{align} \ -x=-\frac{\ y-15}{\ 4}=\frac{\ z}{\ 3} \end{align}
and for the second line, L2 \begin{align} \ -(x+28)=-\frac{\ y+43}{\ 3}=\ z \end{align}
Now, I got stuck on this stage. I know that I should use the squared distance formula as an optimization function but how can get to it?