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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.

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

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    The projection at least looks shorter in the picture, right? So that's a good starting point. – 2017-02-14
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    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 $> \frac\pi2$. – 2017-02-14
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    If an arrow is flying right towards you, it doesn't *look* very long, does it? – 2017-02-14
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    Related: http://math.stackexchange.com/questions/1107459/is-the-norm-of-a-projection-of-a-vector-along-a-subspace-less-than-or-equal-to-t/1107485 – 2017-02-14

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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!