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I'm having a difficulty to understand unit vectors other than those of the axes ($i, j, k$).

Say I have two vectors $U$ and $V$, somewhere in $\Bbb R^3$ in terms of $i, j, k$, and I want to know the projection of vector $U$ over vector $V$, if their coordinates are all non-zero.

As I see it, I should make:

$$\operatorname{proj}_vu = |u| \cos(\alpha)\;,$$ where $v$ is a unit vector. But how can I find an unit vector $V$ if its coordinates are non-zero? Its length won't ever be $1$.

I apologize for my lack of TEX knowledge, I'm sure it would have made this clearer. Tried to code it somewhere else and paste an url here, but it didn't work.

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Please check that I added $\LaTeX$ correctly. (You can get some help using $\LaTeX$ here.) –  Brian M. Scott Jan 26 '13 at 23:29
    
If $v$ is any non-zero vector, then the scalar multiple $\frac1{\|v\|}v$ is a unit vector in the same direction as $v$. –  Brian M. Scott Jan 26 '13 at 23:31
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1 Answer

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If you have a vector with non-zero coordinates, then it has non-zero length. Hence, you may always divide out by the length, to obtain a vector of length 1, that points in the direction of your original vector. We call this the unit vector

$$ \hat{u} = \frac {\vec{v} } { \left\| u \right \| }$$

For example, the vector $ ( 1, 2, 2) $ has length $\sqrt{ 1^2 + 2^2 + 2^2} = \sqrt{9} = 3$, thus the unit vector will be $ ( \frac {1}{3}, \frac {2}{3}, \frac {2}{3} ) $.

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