Angle between U and V when given perpendicular vectors A and B

[jabber024]

Homework Statement

vector A = 3U-V
vector B = U+2V
U and V are vectors
|U| = 3|V|
Given that vector A and vector B are perpendicular vectors, find the angle between vector U and vector V.

Homework Equations

A*B = |A||B|cos(∠AB)
A*A = |A|^2

The Attempt at a Solution

Since A and B are perpendicular to each other that means that the dot product will equate zero because cos 90 deg = 0.
So substituting in the vectors I end up with something like

(3U-V)*(U+2V) = 0 = 3U*U + 5U*V - 2V*V

Given that any vector dot product itself gives you the magnitude of the vector squared and that we are trying to figure out the angle UV:
U*U = |U|^2
U*V = |U||V|cosθ
V*V = |V|^2
3U*U + 5U*V - 2V*V = 3|U|^2 + 5|U||V|cosθ - 2|V|^2

Rearrange: cosθ = (2|V|^2 - 3|U|^2)/(5|U||V|)

Substitute in |U| = 3|V| and you get -25/15. I can't get the inverse cos of a number greater than 1 and I can't figure out where I went wrong. Any help would be greatly appreciated.

 

[AlephZero]

Deleted - suggestion was wrong.

 

[jabber024]

Oh I was trying to figure out what you meant by that, but I see you changed your suggestion. It's a real head scratcher, of all the stuff I've done with vectors, this question should work out the way I did it but nothing seems to yield a realistic result.

 

[micromass]

Of course, $U$ and $V$ could also be the zero vector...