How to understand dot product is the angle's cosine?
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:
How do you see this connection?
How do you extend the notion of dot product vs. angle to higher dimensions - 4 and higher?
7 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ 6 $\begingroup$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: $(y-w)\cdot x = (y\cdot x) - (w\cdot x)= a - a(x\cdot x) = a-a=0$.
$\endgroup$ 4 $\begingroup$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 $\langle u, v\rangle = \lVert u\rVert\lVert v\rVert\cos(p-q)$
$\endgroup$ 0 $\begingroup$v1=[x1,y1]=[|v1|*cos(a),|v1|*sin(a)]
v2=[x2,y2]=[|v2|*cos(b),|v1|*sin(b)]
v1 dot v2 = x1*x2 + y1*y2 = |v1||v2|( cos(a)*cos(b)+sin(a)sin(b) ) = |v1||v2|cos(a-b) = |v1||v2|*cos(theta)
Besides, the proof of cos(a-b)==cos(a)*cos(b)+sin(a)*sin(b) :
$\endgroup$ 1 $\begingroup$A 3-D rotation should be a transformation $T:\mathbb{R}^{3}\to\mathbb{R}^3$ that preserves the Euclidean norm. For any two vectors $\mathbf{x}, \mathbf{y}$, $T(\mathbf{x}):=\mathbf{a}$ and $T(\mathbf{y}):=\mathbf{b}$. As $T$ also preserves the angle between two vectors, the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$ is congruent to the one formed by $\mathbf{x}$ and $\mathbf{y}$, and thus the diagonal $$T(\mathbf{x}+\mathbf{y})= \mathbf{a}+\mathbf{b}$$ $$\lVert \mathbf{x}+\mathbf{y}\rVert = \lVert \mathbf{a}+\mathbf{b}\rVert $$ So $\langle \mathbf{a}+\mathbf{b}, \mathbf{a} +\mathbf{b}\rangle = \langle \mathbf{x}+\mathbf{y}, \mathbf{x} +\mathbf{y}\rangle$. By definition, $\lVert \mathbf{a}\rVert = \lVert \mathbf{x}\rVert$ and $\lVert \mathbf{b}\rVert=\lVert \mathbf{y}\rVert$, and by the commutativity of of the dot product and its distributivity over addition, $$\langle \mathbf{a}+\mathbf{b}, \mathbf{a} +\mathbf{b}\rangle = \langle \mathbf{a}, \mathbf{a} \rangle + 2\langle \mathbf{a}, \mathbf{b} \rangle + \langle \mathbf{b}, \mathbf{b} \rangle$$ $$\langle \mathbf{x}+\mathbf{y}, \mathbf{x} +\mathbf{y}\rangle = \langle \mathbf{x}, \mathbf{x} \rangle + 2\langle \mathbf{x}, \mathbf{y} \rangle + \langle \mathbf{y}, \mathbf{y} \rangle$$ So this shows that $\langle T(\mathbf{x}), T(\mathbf{y})\rangle = \langle \mathbf{x}, \mathbf{y} \rangle $ i.e. the dot product is preserved under any rotation. Which then leads rather directly to Stephen's answer.
$\endgroup$ $\begingroup$This is what helped me understand it most:
from here
primarily the picture on the right side.
Let the pink left side of the rectangle be a vector v you're projecting onto.
Let the black line labeled with a '1' be a vector u that you're projecting onto v.
The bit of pink on the top left is the orthogonal "line of projection" from the tip of u to the span of v.
The angle $\alpha - \beta$ in the pink triangle is the angle between v and u.
Imagine you're tilting the diagram a bit clockwise so that the long side ($cos\beta$) of the orange triangle is horizontal.
Now the $cos\beta$ side of the orange triangle is the x component of u in an imagined cartesian coordinate system, and the $sin\beta$ side of the orange triangle is the y component.
The blue side on the right is equal in length to the pink side on the left since this is a rectangle. On the blue side we can see the length is made up of $cos\beta \cdot cos\alpha + sin\beta \cdot sin\alpha$, which, if u and v are unit length, is the dot product.
Here's another diagram of the same scenario with a "normal" axis, and then also a rotated grid corresponding to v, the orange vector being projected onto
$cos\beta$ is the length of the x component of u, a unit vector $\beta$ counter-clockwise from the x axis.
$cos\alpha$ is the same for v. It's also the factor by which a vector pointing along the x axis (such as the x part of u) is scaled when it's projected onto v. This is because: say you have a right triangle with hypotenuse of length 1 lying on the x axis, with another side aligned with v. That side will have angle $\alpha$ with the hypotenuse (because v is $\alpha$ counter-clockwise from the x axis) and so its length will be $cos\alpha$.
So the x component of u is of length $cos\beta$, then when being "scaled" by projecting onto $v$, it becomes $cos\beta \cdot cos\alpha$.
The case with y components and $sin$ is analogous.
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