# proof of inner product-dot product connection

I have to prove: $$\langle a,b\rangle = \sum_{1}^{n} a_i\cdot b_i = \|a\|\cdot\|b\|\cos\alpha$$

I know how to get formula for dot product through projection, but don't know how to connect $\sum_{1}^{n} a_i\cdot b_i = \|a\|\cdot \|b\|\cos\alpha$

Any ideas?

• Hint: Draw a picture. – Sean Roberson Jul 7 '17 at 21:23
• There is nothing to prove unless the definition of $\alpha$ is given. In higher dimensions ($n\geq 3$), this is actually the definition of $\alpha$. – Jack Jul 7 '17 at 21:35

The key is to draw a triangle an apply the law of cosines. A triangle whose side lengths are $\|a\|,\|b\|,\|a - b\|$ should satisfy $$\|a - b\|^2 = \|a\|^2 + \|b\|^2 + 2\|a\|\|b\| \cos \alpha$$ However, expanding the definition of length via the inner product yields $$\|a-b\|^2 = \langle a-b,a-b\rangle = \|a\|^2 + \|b\|^2 + 2 \langle a,b \rangle$$ We must conclude that $\langle a,b \rangle = \|a\| \|b\| \cos \alpha$, as desired.