# The cosine of the angle between two vectors proof

Vector $\mathbf{v}$ and $\mathbf{v'}$ make angles $\alpha , \beta, \gamma$ and $\alpha ', \beta', \gamma$' with the coordinate axes respectively.

$\phi$ is the angle between $\mathbf{v}$ and $\mathbf{v'}$.

Why is $\cos(\phi)=\cos(\alpha)\cos(\alpha')+\cos(\beta)\cos(\beta')+\cos(\gamma)\cos(\gamma')$?

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Hint: $\mathbf{v}=\cos(\alpha)\mathbf{e}_1+\cos(\beta)\mathbf{e}_2+\cos(\gamma)\mathbf‌​{e}_3$ and similar for $\mathbf{v}'$. Then you compute $\mathbf{v}\cdot \mathbf{v}'$. – Sigur Dec 3 '12 at 18:22
I don't know why the third vector is not in boldface. Also, I supposed $\mathbf{v}$ unitary. If not, multiply by its norm. – Sigur Dec 3 '12 at 18:24
I see it now- I was being a 2D chauvinist; angles in 3 dimensions threw me. – Alyosha Dec 3 '12 at 18:29

Note that $$\cos(\alpha) = \frac{v_1}{||v||},\; \cos(\beta) = \frac{v_2}{||v||}, \;\cos(\gamma) = \frac{v_3}{||v||}$$ and $$\cos(\alpha') = \frac{v_1'}{||v'||},\; \cos(\beta') = \frac{v_2'}{||v'||}, \;\cos(\gamma') = \frac{v_3'}{||v'||}.$$ Then $$\cos(\phi) = \frac{\langle v, v' \rangle}{||v|| \cdot ||v'||} = \frac{v_1 v_1' + v_2v_2' + v_3v_3'}{||v|| \cdot ||v'||} =$$ $$\cos(\alpha) \cos(\alpha') + \cos(\beta) \cos(\beta') + \cos(\gamma) \cos(\gamma').$$
Is $\left \langle v,v' \right \rangle$ notation for the dot product? – Alyosha Dec 3 '12 at 18:34