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I am having problem understanding vectors. If a unit vector points in the direction of $z$ axis, then what coordinates would it have?

The paper I read says $x$ and $y$ but if it is in the direction of $z$ axis, shouldn't the coordinate be just the $z$ axis?

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    Startup Crazy: *You can't edit a new question into an already existing one*. You need to make a new question post.2012-06-23

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Just to clarify what is meant by a unit vector (a.k.a. direction vector), it is a vector which simply gives a direction (i.e. it has magnitude $1$). We can find a unit vector from some arbitrary vector, $\vec{v}\in\mathbb{R}^{n}$, using the following relation:

$\hat{v}=\frac{\vec{v}}{||\vec{v}||},$

Where $||\vec{v}||$ is the euclidean norm for $\mathbb{R}^{n}$, defined as:

$||\vec{v}||=\sqrt{\sum_{i=1}^{n}{x_{i}^{2}}}$

In answer to your question, "What is the unit vector in the direction of the $z$-axis?" It is simply $(0, 0, 1)$, i.e. the vector in the family of vectors in $\mathbb{R}^{3}$ with only the $z$-component being positive non-zero, and of unit length (i.e. of euclidean norm $1$).

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    @ZevChonoles excellent point, I just assumed he meant the positive direction because that's the one I consider to be "in the direction of", the other solution I would say is "in the opposite direction to". I guess it is somewhat a matter of interpretation.2012-06-23
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Since you are talking about $x$-, $y$-, and $z$-axes, I'll assume we're working with 3-dimensional Euclidean space, a.k.a. $\mathbb{R}^3$. The elements of this space can be thought of as vectors, which are just ordered triples of real numbers, e.g. $(1,\sqrt{2},-5)$.

The length of a vector $(a,b,c)$ is defined to be $\sqrt{a^2+b^2+c^2}$. A vector is a unit vector when it has length equal to 1.

The $z$-axis consists of those vectors that are of the form $(0,0,t)$ for some $t\in\mathbb{R}$. It is usually given the orientation where the points $(0,0,t)$ with $t>0$ are "up".

Depending on what you mean by "points in the direction of", there are either one or two unit vectors in the direction of the $z$-axis.

If you mean that you want a unit vector that has the same orientation as the $z$-axis, then you are after a vector $(0,0,t)$ where $\sqrt{0^2+0^2+t^2}=\sqrt{t^2}=1$ and $t>0$. There is only one solution, namely $t=1$, so the only unit vector that points in the direction of the $z$-axis is $(0,0,1)$.

If you instead mean that you want a unit vector that lies within the $z$-axis, then you are after a vector $(0,0,t)$ where $\sqrt{0^2+0^2+t^2}=\sqrt{t^2}=1$. There are two solutions, namely $t=1$ and $t=-1$, so the two unit vectors that point in the direction of the $z$-axis are $(0,0,1)$ and $(0,0,-1)$.