Let L1 be the line passing through the point P1 = (−2, 3, 1) with direction vector →d = <−2, −1, −2>, and let L2 be the line passing through the point P2 = (4, 4, −4) with the same direction vector. Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d.
What I've solved for so far:
L1: $<-2,3,1> + t<-2,-1,2>$
L2: $<4,4-4> + u<-2,-1,2>$
Consider a point $(-2,3,1)$ and a point on L2 which is $<4-2u, 4-u, -4-2u>$
Distance between the points is:
d$^2$ = (-2u+6)$^2$ + (-u+1)$^2$ + (-2u-5)$^2$
I then differentiated both sides to get:
(4u-12) + (u-1) + (4u+10) = 0
u = $\frac13$
Shortest distance between the lines is then:
d$^2 = (-2(\frac13$)+6)$^2$ + (-($\frac13$)+1)$^2$ + (-2($\frac13$)-5)$^2$
d = $\sqrt{61}$
So I've gotten d to equal $\sqrt{61}$. I don't know where to go from here and solve for Q1 and Q2.
If the lines have the same direction vector, then they must be parallel.
So the plane perpendicular to line $P_1+td = (-2-2t, 3-t, 1-2t)$ will also be perpendicular to the line $P_2+ud = (4-2u, 4-u, -4-2u)$.
Since that plane passes through the point $$Q_1=(-2,3,1)$$ and is perpendicular to the direction $(2,1,2)$. Its equation is $2(x+2) +1(y-3) + 2(z-1) = 0$. Which simplifies to
$$2x+y+2z = -1$$
It will intersect the other line when $(4-2u)+(4-u)+2(-4-2u) = -1$, which is when $u = \dfrac 13$. This gives us the point $$Q_2 = \frac 13(10, 11, -14)$$. Then
$$d = \left \|(-2,3,1) - \frac 13(10, 11, -14) \right \| = \sqrt{61}$$