# Thread: Definition of angle between two vectors?

1. ## Definition of angle between two vectors?

Let A and B be two vectors in 3 dimensions.

Here seems to be the current treatment of the matter in Calculus textbooks:

If A = <x1, y1, z1> and B = <x2, y2, z2>, then the dot product is defined to be x1x2 + y1y2 + z1z2. Then, the distributive property and a few other useful properties are proven from the definition of the dot product. Next, a triangle consisting of sides A, B, and A-B is constructed. The law of cosines is applied to the angle between A and B. This finally yields the formula for calculating the cosine of that angle.

What I'm asking is basically, is there a rigorous definition of angles? Because, as you can see, the angle between A and B in the above discussion is just sort of taken to be a "de facto" entity.

I've imagined that one can use the differential equation y" + y = 0 to define the sine and cosine functions, prove the Pythagorean identity, and finally show that sine and cosine parametrize the unit circle. Once the unit circle is parametrized, I figure one can define angles between lines based on where they intersect the unit circle. Is there a simpler way to define angles? Or is that why all these introductory textbooks don't provide definitions?

2. First, one must understand that a vector is in some sense a hybrid object. Having both length and direction is not strictly a set. We use triples of real numbers to represent vectors and endow them with length and direction.

In axiomatic geometry we define an angle as the union of two rays which have a common end point. We can extend that idea to vectors with the above understanding. In that understanding all vectors can be seen as having the origin as a common endpoint. Thus, any two vector have an angle between them measuring from $0$ to $\pi$.

3. Thank you for that attempt, but you haven't enlightened me yet.

Now, I understand that 3D vectors are rigorously defined as ordered triples of real numbers. I am aware that they can be visualized as arrows emanating from the origin. And so you're suggesting that they can be treated as rays, by which angles are defined in axiomatic geometry.

But... I'm afraid I don't see the definition. Suppose I state a more tangible question, so that you have a better idea of what I want:

Let A and B be arbitrary non-zero vectors in 3D.
Provide a definition of "the angle between A and B" so that the following equation can be validly proven:

$\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|cos \ \theta$

where $\theta$ is the angle, as you define it. Ah, yes, and the definition shouldn't be in terms of the dot product, of course. I recognize that this may be a futile question to ask.

Could you or anyone provide me with any insight into the rigorous aspects of angles? Could anyone state a fully rigorous definition of angles and prove a simpler result than the one above, leaving the dot product proof as an exercise for me?

4. In my view, you are confusing apples and oranges.
If you go to a good calculus textbook this whole topic is carefully laid out.
Stewart in chapter 13 does an excellent job of doing just what you are asking for in section 3 on the dot product. Not just Stewart, but most others do also.

5. If $\mathbf{u}$ and $\mathbf{v}$ are two vectors in an inner product space V, then $|<\mathbf{u},\mathbf{v}>|\leq ||\mathbf{u}|| \ ||\mathbf{v}||$.

If $\mathbf{u}$ or $\mathbf{v}$ are 0, then $<\mathbf{u},\mathbf{v}>=0= ||\mathbf{u}|| \ ||\mathbf{v}||$.

If neither are 0, then let $\mathbf{p}$ be the vector projection of $\mathbf{u}$ onto $\mathbf{v}$. $\mathbf{p}$ is orthogonal to $\mathbf{p}-\mathbf{u}$.

By the Pythagorean law,

$||\mathbf{p}||^2+||\mathbf{p}- \mathbf{u}||^2=||\mathbf{u}||^2$.

$\displaystyle ||\mathbf{p}||^2=\frac{<\mathbf{u},\mathbf{v}>^2}{ ||\mathbf{v}||^2}$

$\displaystyle \frac{<\mathbf{u},\mathbf{v}>^2}{||\mathbf{v}||^2} =||\mathbf{u}||^2-||\mathbf{u}-\mathbf{p}||^2$

$\displaystyle <\mathbf{u},\mathbf{v}>^2=||\mathbf{v}||^2||\mathb f{u}||^2-||\mathbf{v}||^2||\mathbf{u}-\mathbf{p}||^2\leq ||\mathbf{v}||^2||\mathbf{u}||^2$

$\displaystyle <\mathbf{u},\mathbf{v}>\leq ||\mathbf{v}|| \ ||\mathbf{u}||$

This is the Cauchy-Schwarz Inequality, but from this, we can obtain.

$\displaystyle -1\leq\frac{<\mathbf{u},\mathbf{v}>}{||\mathbf{u}|| \ ||\mathbf{v}||}\leq 1$

What ranges from -1 to 1?

Cosine.

$\displaystyle cos(\theta)=\frac{<\mathbf{u},\mathbf{v}>}{||\math bf{u}|| \ ||\mathbf{v}||}$

6. Originally Posted by Mazerakham
Thank you for that attempt, but you haven't enlightened me yet.

Now, I understand that 3D vectors are rigorously defined as ordered triples of real numbers. I am aware that they can be visualized as arrows emanating from the origin. And so you're suggesting that they can be treated as rays, by which angles are defined in axiomatic geometry.

But... I'm afraid I don't see the definition. Suppose I state a more tangible question, so that you have a better idea of what I want:

Let A and B be arbitrary non-zero vectors in 3D.
Provide a definition of "the angle between A and B" so that the following equation can be validly proven:
In three dimensions, two intersecting lines form a plane. "The angle between A and B" is defined precisely as it is in plane geometry- the union of two rays.

$\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|cos \ \theta$

where $\theta$ is the angle, as you define it. Ah, yes, and the definition shouldn't be in terms of the dot product, of course. I recognize that this may be a futile question to ask.

Could you or anyone provide me with any insight into the rigorous aspects of angles? Could anyone state a fully rigorous definition of angles and prove a simpler result than the one above, leaving the dot product proof as an exercise for me?

7. To everyone reading this thread, I apologize if I appear stubborn, but I haven't seen an answer.

HallsofIvy: the definition of an angle as "the union of two rays" does not, in my opinion, lead to a rigorous proof of the equation $\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|cos \ \theta$ because, even if the cosine function is rigorously defined, I don't see how "the union of two rays" is a real number of which we can take the cosine. Perhaps I should have been more specific in that I'm looking for rigorous definitions in the realm of Real Analysis--not Euclidean Geometry.

dwsmith: You were closer to what I'm looking for, but there are a few holes I would like you to clear up for me:

1. Let u and v be two non-zero 3D vectors. Please define "the vector projection of u onto v"

2. Let p and u be two non-zero 3D vectors. Please define the statement "p is orthogonal to u".

I will not ask you to prove the pythagorean law. I will do that myself, after you provide those two definitions. In fact, I think with some rigorous definitions of (1) and (2), I think the Cauchy Schwartz inequality would follow, assuming the definitions are strong enough to prove the pythagorean law. Finally, the original question I didn't see answered:

Besides the fact that you haven't proven

$\displaystyle cos(\theta)=\frac{<\mathbf{u},\mathbf{v}>}{||\math bf{u}|| \ ||\mathbf{v}||}$

you also didn't define $\theta$ in that equation, which is pretty much the basis of this discussion.

If I am somehow not articulating my question clearly enough, please quote me and point to whatever needs clarification. I'm sure all you brilliant people out there will figure out what I'm asking for. In the meantime, I'm working on a definition of $\theta(u,v)$, and I'll post it if I work it out.

8. Originally Posted by Mazerakham
1. Let u and v be two non-zero 3D vectors. Please define "the vector projection of u onto v"

2. Let p and u be two non-zero 3D vectors. Please define the statement "p is orthogonal to u".
The vectors $p~\&~u$ are orthogonal if and only if $p\cdot u=0$. Another way of saying the angle between them measures $\frac{\pi}{2}$.

For #1: Given the vectors $u~\&~v$ the vector projection of u onto v is the vector $\frac{u\cdot v}{v\cdot v}v$.

That vector being a multiple of $v$ is parallel to $v$.

The vector $u-\frac{u\cdot v}{v\cdot v}v$ is orthogonal to $v$.

To see that look $\left(u-\frac{u\cdot v}{v\cdot v}v \right)\cdot v=0$.
We say that $u = u_\parallel + u_ \bot$ where $u_\parallel =\frac{u\cdot v}{v\cdot v}v ~\&~ u_ \bot =u - u_\parallel.$

We assume non-zero vectors.

9. Originally Posted by dwsmith
$\displaystyle <\mathbf{u},\mathbf{v}>^2=||\mathbf{v}||^2||\mathb f{u}||^2-||\mathbf{v}||^2||\mathbf{u}-\mathbf{p}||^2\leq ||\mathbf{v}||^2||\mathbf{u}||^2$

$\displaystyle <\mathbf{u},\mathbf{v}>\leq ||\mathbf{v}|| \ ||\mathbf{u}||$

This is the Cauchy-Schwarz Inequality, but from this, we can obtain.

$\displaystyle -1\leq\frac{<\mathbf{u},\mathbf{v}>}{||\mathbf{u}|| \ ||\mathbf{v}||}\leq 1$
Starting from here.

We have,

$\displaystyle <\mathbf{u},\mathbf{v}>^2=||\mathbf{v}||^2||\mathb f{u}||^2-||\mathbf{v}||^2||\mathbf{u}-\mathbf{p}||^2\leq ||\mathbf{v}||^2||\mathbf{u}||^2\Rightarrow <\mathbf{u},\mathbf{v}>^2\leq ||\mathbf{v}||^2||\mathbf{u}||^2$

Now we are going to divide out $||\mathbf{v}||^2||\mathbf{u}||^2$

$\displaystyle \frac{<\mathbf{u},\mathbf{v}>^2}{||\mathbf{u}||^2 \ ||\mathbf{v}||^2}\leq 1$

Taking the square root we obtain

$\displaystyle \frac{<\mathbf{u},\mathbf{v}>}{||\mathbf{u}|| \ ||\mathbf{v}||}\leq \pm 1$

Let's write this in a different format.

$\displaystyle \displaystyle -1\leq\frac{<\mathbf{u},\mathbf{v}>}{||\mathbf{u}|| \ ||\mathbf{v}||}\leq 1$

We know from trig, pre-calc, or any other course that cosine is not greater than $1$ or less than $-1$. If not, please consult the unit circle.

From this Cosine can be written like

$-1\leq cos(\theta)\leq 1$ where $\theta$ represents degrees in the unit circle in either radians or degrees.

Now, let's make a direct substitution (which I am not going to make since it is or should be obvious what the substitution will be).

What does $Cosine(\theta)=\mbox{???}$

10. Originally Posted by Mazerakham
Let A and B be two vectors in 3 dimensions.

Here seems to be the current treatment of the matter in Calculus textbooks:

If A = <x1, y1, z1> and B = <x2, y2, z2>, then the dot product is defined to be x1x2 + y1y2 + z1z2. Then, the distributive property and a few other useful properties are proven from the definition of the dot product. Next, a triangle consisting of sides A, B, and A-B is constructed. The law of cosines is applied to the angle between A and B. This finally yields the formula for calculating the cosine of that angle.

What I'm asking is basically, is there a rigorous definition of angles? Because, as you can see, the angle between A and B in the above discussion is just sort of taken to be a "de facto" entity.

I've imagined that one can use the differential equation y" + y = 0 to define the sine and cosine functions, prove the Pythagorean identity, and finally show that sine and cosine parametrize the unit circle. Once the unit circle is parametrized, I figure one can define angles between lines based on where they intersect the unit circle. Is there a simpler way to define angles? Or is that why all these introductory textbooks don't provide definitions?
What you are having trouble with, it seems, is the idea of definition versus equality. You have an intuitive notion of angle in $\mathbb{R}^k,\text{ }k=2,3$ and want to understand how one proves that this intuitive notion is equivalent to the definition via the inner product. But, first you must define what you mean by the intuitive notion! Moreover, your intuitive notion is useless in higher dimensions and so we take $\displaystyle \cos(\theta)=\frac{\left\langle \vec{u},\vec{v}\right\rangle}{\|\vec{u}\|\|\vec{v} \|}$ as a definition! It's isn't something to be proven, it just 'is'. If you don't like that definition, then create your own. But you can't ask for a proof, the best you can do is ask for the intuition as to why the angle was defined that way. This is completely analogous to asking to prove that $\displaystyle \sum_{n\in\mathbb{N}}\frac{(-1)^n x^{2n+1}}{(2n+1)!}=\sin(x)$ where you have an intuitive definition of sine already stored away! the problem is you can't prove the above in most cases since...in most cases this is true by definition