How to understand dot product is the angle's cosine?

Question

How can one see that a dot product gives the angle's cosine between two vectors. (assuming they are normalized)

Thinking about how to prove this in the most intuitive way resulted in proving a trigonometric identity: $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$.

But even after proving this successfully, the connection between and cosine and dot product does not immediately stick out and instead I rely on remembering that this is valid while taking comfort in the fact that I've seen the proof in the past.

My questions are:

1. How do you see this connection?

2. How do you extend the notion of dot product vs. angle to higher dimensions - 4 and higher?

Answer 1

Here's one way to remember it easily: assume one of the two unit vectors is $(1,0)$ (by an appropriate choice of coordinates we may assume we are working in $2$ dimensions, and then that one of the vectors is the standard basis vector). Then the dot product is just the $x$-coordinate of the other, which is by definition the cosine of the angle between them.

Answer 2

The dot product is basically a more flexible way of working with the Euclidean norm. You know that if you have the dot product $\langle x, y \rangle$, then you can define the Euclidean norm via $$\lVert x\rVert = \sqrt{\langle x, x \rangle}.$$

Conversely, it turns out that you can recover the dot product from the Euclidean norm using the polarization identity $$\langle x, y \rangle = \frac{1}{4} \left(\lVert x + y\rVert^2 - \lVert x - y\rVert^2 \right).$$

Okay, so how can you see the relationship between the dot product and cosines? The key is the law of cosines, which in vector language says that $$\lVert a - b\rVert^2 = \lVert a\rVert^2 + \lVert b\rVert^2 - 2 \lVert a\rVert \lVert b\rVert \cos \theta$$

where $\theta$ is the angle between $a$ and $b$. On the other hand, by bilinearity and symmetry we see that $$\lVert a - b\rVert^2 = \langle a - b, a - b \rangle = \lVert a\rVert^2 + \lVert b\rVert^2 - 2 \langle a, b \rangle$$

so it follows that $$\langle a, b \rangle = \lVert a\rVert \lVert b\rVert \cos \theta$$

as desired.

Any two vectors in an $n$-dimensional Euclidean space together span a Euclidean space of dimension at most $2$, so the connection between the dot product and angles in general reduces to the case of $2$ dimensions.

Answer 3

Suppose $x,y$ are unit vectors and $x\cdot y=a$. Let $w=ax$. If we can show that $w$ is the orthogonal projection of $y$ on $x$, that does it, by definition of the cosine. So is $y-w$ orthogonal to $x$? Let's find the dot product: $(y-w)\cdot x = (y\cdot x) - (w\cdot x)= a - a(x\cdot x) = a-a=0$.

Answer 4

Let $u=(a, b)$ and $v=(c, d)$ be two vectors having angles $p$, $q$ with x axis then $\cos p=a/\lVert u\rVert$ and $\sin p=b/\lVert u\rVert$ and $\cos q=c/\lVert v\rVert$, $\sin q=d/\lVert v\rVert$ then $\cos(p-q)= (a/\lVert \rVert) (c/\lVert v\rVert)+(b/\lVert u\rVert)(d/\lVert v\rVert)$ then $ac+bd=\lVert u\rVert\,\lVert v\rVert\cos(p-q$) then $\langle u, v\rangle = \lVert u\rVert\lVert v\rVert\cos(p-q)$