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In my textbook, they provide a theorem to calculate the angle between two vectors:

$\cos\theta = \Large\frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\|\|\vec{v}\|}$

My questions are, why does the angle have to be $0 \le \theta \le \pi$; and why do the vectors have to be in standard position?

Also, on the next page, the author writes, "the zero vector is orthogonal to every vector because $0 \cdot \vec{u} = 0$;" why is that so?

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The angle between the vectors is normally taken to be the smaller one (i.e, as you go around the angle would 'flip' from one side to the other as you cross $180^{\circ}$) and so it just naturally falls in $0\leq\theta\leq\pi$. – Robert Mastragostino Dec 22 '12 at 19:02
up vote 4 down vote accepted

2 vectors

regarding your first question:

We're always talking about the smallest angle between 2 vectors (angle $\phi$ of the picture)

regarding your second question:

When it comes to orthogonality between vectors (even in higher dimensions) it is defined as follows: vector $\vec{u}$ is orthogonal to $\vec{v}$ if their inner product equals zero. Since $\vec{0}=(0,0)$ and $\vec{u}=(u_1,u_2)$ it follows that $\vec{0}\cdot \vec{u}=0\cdot u_1+0 \cdot u_2=0$ thus every vector is orthogonal to the $\vec{0}$.

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Strictly speaking, the angle does not have to be within $0$ and $\pi$. But to find the angle between these two vectors, you would use the $\arccos$ function, and the usual $\arccos$ function has restricted domain and range: $[-1,1]$ and $[0, \pi]$. Moreover, suppose we got an angle of, say $3\pi/4$. Then we can also view the angle between the two vectors as $\pi/4$ by tilting our heads a bit.

The vectors don't have to be in standard position for measuring the angle between them to work. But if you want to have a picture in mind for the concept, then it would be that you put the tails of the two vectors together, then measure the angle between them. You can think of where the tails meet as the origin, just for convenience. But there's nothing special about the origin when it comes to the dot product.

Except this last part about orthogonality, I guess. The question isn't "why is that so," because the definition of orthogonality is that two vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if and only if $\mathbf{u}\cdot\mathbf{v}=0$. The zero vector satisfies $0\cdot\mathbf{v}=0$ for every vector $\mathbf{v}$ (just write out the dot product coordinatewise to see), so by definition it is orthogonal to every vector.

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The basic model is two lines crossing in a plane - of course in more dimensions than two that is the plane generated by the two vectors (one vector gives one dimension, two linearly independent vectors give two dimensions). If you draw two lines crossing in a plane they make two angles $\alpha$ and $\beta$, both in the given range and having $\cos\alpha =-\cos\beta$. You could choose other values corresponding to the same cosine, but you would need a context in which that would make sense (perhaps adding angles, which might put an angle out of range).

The vectors don't have to be in the same plane to use the model - bringing them to standard position, though, does identify a standard plane in which the angle makes sense.

On orthogonality, that is implicit in the definition usually adopted - but you will note that the formula for the cosine then involves dividing by zero. It would be possible to adopt the idea that orthogonal vectors were non-zero, but that would mean that every time you used the concept it would be necessary to consider special cases. It turns out that the usual definition works OK - the special cases don't cause a problem (like you might expect with dividing by zero).

In fact you will find that there are times when you will use this property of the zero vector (it is the only one which has zero dot product with every other vector) to prove other useful facts about vectors and their arithmetic - so having the word "orthogonal" to hand is convenient.

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The angle does not have to be in any range. This formula does not give the angle at all. It only gives a cosine of the angle. It so happens that if $0 \leq \theta \leq \pi$ then $\cos \theta$ could have any value from $1$ to $-1$. This sort of mapping just makes sense.

I assume by "standard position" you mean "start at the origin".
The thing is that vectors have no positions at all. They are only lengths and directions. A vector is very often associated with some point, but that is not part of the vector. When finding the angle, you only draw vectors from the origin for clarity.

The only definition of orthogonality of $\vec{x}$ and $\vec{y}$ is that $\vec{x} \cdot \vec{y} = 0$. Naturally $\vec{0}$ always satisfies this and is thus orthogonal to every other vector.

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Given two points $x$ and $y$ on the unit sphere $S^{n-1}\subset{\mathbb R}^n$ the spherical distance between them is the length of the shortest arc on $S^{n-1}$ connecting $x$ and $y$. The shortest arc obviously lies in the plane spanned by $x$ and $y$, and drawing a figure of this plane one sees that the length $\phi$ of the arc in question can be computed by means of the scalar product as $$\phi =\arccos(x\cdot y)\ \ \in[0,\pi]\ .$$ This length is then also called the angle between $x$ and $y$.

When $u$ and $v$ are arbitrary nonzero vectors in ${\mathbb R}^n$ then $u':={u\over |u|}$ and $v':={v\over |v|}$ lie on $S^{n-1}$. Geometrical intuition tells us that $\angle(u,v)=\angle(u',v')$. Therefore one defines the angle $\phi\in[0,\pi]$ between $u$ and $v$ as $$\phi:=\arccos{u\cdot v\over|u|\>|v|}\ .$$

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