Why would length of vector projection be different from length of a vector

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.

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

• The projection at least looks shorter in the picture, right? So that's a good starting point. – littleO Feb 14 '17 at 1:27
• 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 $> \frac\pi2$. – user228113 Feb 14 '17 at 1:29
• If an arrow is flying right towards you, it doesn't look very long, does it? – Hans Lundmark Feb 14 '17 at 7:31
• – Travis Feb 14 '17 at 16:49

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.
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!