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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?

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

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