# Using cross product to find angle between two vectors

by yayscience
Tags: angle, cross, product, vectors
 P: 5 1. The problem statement, all variables and given/known data Find the angle between \begin{align*} \vec{A} = 10\hat{y} + 2\hat{z} \\ and \\ \vec{B} = -4\hat{y}+0.5\hat{z} \end{align*} using the cross product. The answer is given to be 161.5 degrees. 2. Relevant equations $$\left| \vec{A} \times \vec{B} \right| = \left| \vec{A} \right| \left| \vec{B} \right|sin(\theta)$$ 3. The attempt at a solution $$\left| \vec{A} \times \vec{B} \right| =$$ $$\left| \begin{array}{ccc} \hat{x} & \hat{y} & \hat{z} \\ 0 & 10 & 2 \\ 0 & -4 & 0.5 \end{array} \right| = \left| 13\hat{x} \right| = 13$$ The magnitude of A cross B is 13. Next we find the magnitude of vectors A and B: $$\left| \vec{A} \right| = \sqrt{10^2+2^2} = \sqrt{104} = 10.198039$$ and $$\left| \vec{B} \right| = \sqrt{(-4)^2+(\frac{1}{2})^2} = \sqrt{16.25} = 4.0311289$$ multiplying the previous two answers we get: 41.109609 So now we should have: $$\frac{13}{41.109609} = sin(\theta)$$ Solving for theta, we get: 18.434951 degrees. This is frustrating: 180-18.434951 = the correct answer. I'm not quite sure where I'm going wrong here. I must be making the same mistake repeatedly. Another problem was the same thing, but with the numbers changed, and I also got the 180-{the answer I was getting} = {the correct answer}, but when I tried the example using the SAME methodology, I got the correct answer. Can someone please share some relevant wisdom in my direction?