1. The problem statement, all variables and given/known data 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. 2. Relevant equations A*B = |A||B|cos(∠AB) A*A = |A|^2 3. 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.