10.4#8

Hello Dr. Taylor!

I got the answer correct by finding the cross product of
the two vectors, then finding the ((magnitude of the cross
product)/(distance between Q and R)). Could you explain why this works
though? I would've thought it would have something to do with projection.

Thanks!

*******************
Oh. Yeah, I never thought about that, but you can do that.  Here's a picture of the points Q,R,P:

There is a dotted line passing through the two points Q, R, and a displacement vector from Q to P.  Notice that I've drawn a right triangle, one corner of which is at Q, the other corner at P, and the right angle on the line passing through Q and R.  I suppose that when you say "the two vectors" you mean the displacement vectors u=R-Q and v=P-Q.  Note that the hypotenuse of the triangle is ||v||. Then the part of v along the direction of  u  is (u.v)/(u.u) u, this is the displacement vector that goes from Q to the place where the blue line intersects the line QR, and has magnitude ||v|| Cos(θ) where θ is the angle RQP.  Of course a little trigonometry tells you that the opposite leg of the right triangle must have magnitude ||v||Sin(θ) = ||uxv||/||u||