Hey Brendan.

Are you stuck with all of the problems? I'll start off with the ones you wrote.

Prove |AXB|^2 + (A.B)^2 = |A|^2|B|^2

Now |AXB|^2 = sin^2(theta)*|A|^2*|B|^2 since we have AXB = n*sin(theta)*|A|*|B|. If we take the magnitude of this vector and square it, we get

|AXB| = |n*sin(theta)*|A|*|B|) = |n|*|sin(theta)*|A|*|B|| but n is a unit vector which means |n| = 1 so we get

|AXB| = |sin(theta)|*|A|*|B| and squaring gives us

|AXB|^2 = sin^2(theta)*|A|^2*|B|^2.

Now for the dot product we know that <A,B> = |A|*|B|*cos(theta) so squaring this gives

(<A,B>)^2 = |A|^2|B|^2*cos^2(theta).

Adding the two gives us

|A|^2|B|^2sin^2(theta) + |A|^2|B|^2cos^2(theta) but sin^2(theta) + cos^2(theta) = 1 which means the total is

|A|^2|B|^2(1) = |A|^2|B|^2.

If you weren't introduced to the sin(theta) form for AXB this would have been very confusing.

For del(F), you need to specify what F is explicitly: you have given r but r by itself doesn't mean anything. Please describe what that the r vectors mean (it gives a definition for phi in Q3 but not for r and I don't know why).