Find the point $$Q$$ on the line
$$(-10,-6,-14) + s(-10,4,-2)$$
that is closest to the line
$$(49,28,-26) + t(-50,-38,22)$$
I'm not sure what information I need to use to complete this question. Here is one of my attempts:
I first used both matrices that are multiplied by $s$ and $t$ to make the equation $r= P+td$ which gives me
$$r= (-10,4,-2) + t(-50,-38,22).$$
$$r= (-10-50t, 4-38t, -2+22t)$$
After that, I try to find the equation of the distance while plugging in the first matrix point.
$$d= \sqrt{(-10-50t+10)^2 + (4-38t+6)^2 + (-2+22t+14)^2}$$
Unfortunately the answer I get if I do solve this above equation is quite messy and would not allow me to find the derivative which I assume is the next step.
I can't figure out your working. Here's my suggestion:
Let $P$ be the point on the second line that is closest to the first. Then we have $$\vec{OP}=(49,28,−26)+t(−50,−38,22)\\\vec{OQ}=(−10,−6,−14)+s(−10,4,−2)\\\vec{QP}=(59,34,−12)+t(−50,−38,22)-s(−10,4,−2).$$ Since $\vec{PQ}$ is perpendicular to each of the two lines, $$\vec{QP}\cdot(−10,4,−2)=0=\vec{QP}\cdot(−50,−38,22)\\\Big((59,34,−12)+t(−50,−38,22)-s(−10,4,−2)\Big)\cdot(−10,4,−2)\\=0=\\\Big((59,34,−12)+t(−50,−38,22)-s(−10,4,−2)\Big)\cdot(−50,−38,22).$$ Solving gives $$t=25625/27434,s=-66777/54868,$$ so $$Q=(2.17048188379,-10.8681927535,-11.5659036232).$$ Extra information: $$P=(2.29700371801,-7.49427717431,-5.45068163593)\\PQ=6.9853.$$