Proving dot product and cosine

Question

Prove that $\vec p \cdot \vec q=|\vec p||\vec q|\cos a$, $a$ the angle between vector p and q.

I tried using law of cosines but I'm not supposed to do that since I need to prove law of cosines in the next exercise, also I think law of cosines is a consequence of this statement.

$\vec p•\vec q=|\vec p|comp_\vec e \vec q$ in a orthonormal system, also $\vec p•\vec p=|\vec p|^2$ — MarioM21, Oct 12 '20 at 07:22
Does this answer your question? https://math.stackexchange.com/a/846606/798113 — , Oct 12 '20 at 07:29
Yes, thanks a lot! That way it makes more sense. Also is law of cosines a consequence of that proof? — MarioM21, Oct 12 '20 at 07:42
You're welcome. Indeed, you can use this fact for to prove law of cosines. — , Oct 12 '20 at 07:48

score 1 · Accepted Answer · answered Oct 12 '20 at 07:46

Assuming you're in 2 dimensions:

$$\vec{p} = (p_x,p_y) = (|\vec{p}|cos\alpha, |\vec{p}|sin\alpha)$$

$$\vec{q} = (q_x,q_y) = (|\vec{q}|cos\beta, |\vec{q}|sin\beta)$$

$$\vec{p}\cdot\vec{q} $$ $$ = p_xq_x+p_yq_y$$ $$= |\vec{p}|cos\alpha|\vec{q}|cos\beta + |\vec{p}|sin\alpha|\vec{q}|sin\beta$$ $$= |\vec{p}||\vec{q}|(cos\alpha cos\beta + sin\alpha sin\beta)$$ $$ =|\vec{p}||\vec{q}|cos(\alpha-\beta)$$

Proving dot product and cosine

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