Testing for Orthogonality

Question

When testing for orthogonality, why is it that when $\vec{v} \cdot \vec{w}=0$ do we say that these vectors are perpendicular? What does the dot product do for us that we end up with that conclusion?

Answer 1

It is a well known fact that for two vectors we have $a\cdot b = |a||b|\cos \theta $ where $|a|$ and $|b|$ are the magnitudes of the vector and $\theta$ is the angle between the two vectors (in the plane spanned by them). If $a \cdot b = 0$, and $a\neq 0$,$b \neq 0$, we must have $\cos \theta = 0$. But this means $\theta =\pm \pi/2$, or that they are orthogonal geometrically.

By analogy, with any inner product space, we say that $f$ and $g$ are orthogonal if $\langle f, g \rangle = 0$.

Answer 2

This is how we define perpendicularity/orthogonality. In fact, we can define the angle between nonzero vectors via the dot product: As $|v\cdot w|\le |v|\,|w|$, the number $\frac{v\cdot w}{|v|\,|w|}$ is between $-1$ and $1$, inclusive, and hence there exists exactly one $\alpha\in[0,\pi]$ such that $\cos\alpha=\frac{v\cdot w}{|v|\,|w|}$.

Answer 3

It may be the inspiration for such a name comes from what happens in the plane $\,\Bbb R^2\,$, where two vectors (let's anchor them say at the origin) are perpendicular iff the product of the slopes of the lines on which they lay is minus one , so

$$(a,b)\perp(x,y)\iff \frac ba\frac yx=-1\iff ax+by=0$$

and since the usual, euclidean dot product in the real plane is $\,\;(a,b)\cdot (x,y):=ax+by\,\;$ there you have a relation between the two things.