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In trying to understand the geometric interpretation of dot product, I read that it is the length of the projection of one vector onto another.

enter image description here

My question is: how is it that the projection of u be shorter than the magnitude of u?

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    $\begingroup$ The projection at least looks shorter in the picture, right? So that's a good starting point. $\endgroup$ – littleO Feb 14 '17 at 1:27
  • $\begingroup$ It is the (signed) length of projection if and only if the other vector has length $1$. However, the length of the projection is smaller than the length of the vector because in a triangle the side opposite to the largest angle is the longest and, in a triangle, no angle can be $> \frac\pi2$. $\endgroup$ – user228113 Feb 14 '17 at 1:29
  • $\begingroup$ If an arrow is flying right towards you, it doesn't look very long, does it? $\endgroup$ – Hans Lundmark Feb 14 '17 at 7:31
  • $\begingroup$ Related: math.stackexchange.com/questions/1107459/… $\endgroup$ – Travis Feb 14 '17 at 16:49
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Using Pythagora's theorem, one has: $$\|u\|^2=\|\textrm{proj}_vu\|^2+\|u-\textrm{proj}_vu\|^2.$$ Hence, one has: $$\|u\|^2\geqslant\|\textrm{proj}_vu\|^2.$$ Which proves the claim.

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If the angle between $u$ and $v$ is $\theta$ , then the projection of $u$ onto the vector $v$ is given by $|u|cos(\theta)$,now since $|\cos(\theta)| \leq 1$,thus the projection must be in magnitude $\leq |u|$ , hence follows.

Mathematically it is due to the $cos(\theta)$ present in the projection term,hope this helps!

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