I've read that taking a dot product is just projecting one vector on the other, so a perpendicular vector will have no components in the other vectors direction. But shouldn't this leave the length unchanged so it has its original magnitude like multiplying it by 1?
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$\begingroup$To clarify, the projection of $\vec u$ on $\vec v$ is the vector $$\left(\frac{\vec u\cdot \vec v}{\|\vec v\|^2}\right)\vec v = \left(\frac{\vec u\cdot\vec v}{\|\vec v\|}\right)\frac{\vec v}{\|\vec v\|}.$$The dot product is a scalar quantity. But the length of the projection is always strictly less than the original length unless $\vec u$ is a scalar multiple of $\vec v$.
$\endgroup$ $\begingroup$$$u.v=|u||v|\cos \theta $$
If $ \theta =\pi/2$ we have $\cos \theta =0$
Thus perpendicular vectors have zero dot product.
$\endgroup$ $\begingroup$The dot product is that way by definition, this particular definition gives the expected Euclidean Norm. A consistent dot product can be and is defined differently, for example in physics & differential geometry the metric tensor is solved for and ascribes a different inner product at every space-time coordinate, which is the means for modeling curved spaces.
$\endgroup$ $\begingroup$Visually, nonzero $u, v \in \mathbb{R}^n$ are perpendicular iff the smaller (in $[0, \pi]$) angle between them is $\pi/2$.
With $u \cdot v$ defined as $\sum_{i = 1}^n u_i v_i$, the the law of cosines implies that$$u \cdot v= \| u \| \|v\| \cos \theta $$If $ \theta =\pi/2$, then $\cos \theta = 0$, whence $u \cdot v = 0$; perpendicular vectors have zero dot product.
Conversely, if $u \cdot v = 0$, then multiplying by $\frac 1 {\| u \| \| v \|}$ gives $\cos \theta = 0$, so the angle between $u$ and $v$ must be $\theta = \arccos 0 = \pi/2$.
In conclusion, two nonzero vectors are perpendicular iff their dot product is zero. If we assert that the zero vector is perpendicular to everything, then this equivalence applies to all vectors, so the geometric statement of "perpendicular" coincides with the algebraic statement of $u \cdot v = 0$. This motivates the definition that two vectors in any inner product space are "orthogonal" iff $\langle u, v \rangle = 0$; in the case of Euclidean $\mathbb{R}^n$, orthogonal vectors are perpendicular vectors.
(This was originally an edit to Mohammad Riazi-Kermani's answer, but the edit was taken down.)
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