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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?

  • 0
    The Law of Cosines certainly drives the result in general. Evidently, $u\cdot u = |u|^2$ in any dimension; that's just the Distance Formula. Writing $w$ for $u-v$, we have $|w|^2 = (u-v)\cdot (u-v) = u \cdot u + v \cdot v - 2 u\cdot v = |u|^2 + |v|^2 - 2 u\cdot v$. The Law of Cosines tells us that this should be $|w|^2 = |u|^2 + |v|^2 - 2 |u||v|\cos\psi$ for angle $\psi$ between $u$ and $v$; consequently, we must have that $u\cdot v = |u||v|\cos\psi$. Nevertheless, I like that I can "see" the angle-difference formula in the 2d dot product.2012-03-03

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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.

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    makes sense, however, unfortunately this proof just relies on another fact I have been brought up to believe by faith which is the law of cosines.2017-07-16
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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.

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    @shuhalo [for example here](https://math.stackexchange.com/questions/947867/how-to-prove-invariance-of-dot-product-to-rotation-of-coordinate-system)2018-07-05
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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: $(y-w)\cdot x = (y\cdot x) - (w\cdot x)= a - a(x\cdot x) = a-a=0$.

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    How would you show that a zero dot product means 90 degree angle?2012-03-07
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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) : http://www.math.ubc.ca/~feldman/m100/trigId.pdf

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    Thanks for contributing to math.SE! One small suggestion, you could use [MathJax](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) to format your formula2017-05-23
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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 $\langle u, v\rangle = \lVert u\rVert\lVert v\rVert\cos(p-q)$

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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.