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How can I find the unit vector of a three dimensional vector? For example, I have a problem that I am working on that tells me that I have a vector $\hat{r}$ that is a unit vector, and I am told to prove this fact:

$\hat{r} = \frac{2}{3}\hat{i} - \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k}$

I know that with a two-dimension unit vector that you can split it up into components, treat it as a right-triangle, and find the hypotenuse. Following that idea, I tried something like this, where I found the magnitude of the vectors $\hat{i}$ and $\hat{j}$, then using that vector, found the magnitude between ${\hat{v}}_{ij}$ and $\hat{k}$:

$\left|\hat{r}\right| = \sqrt{\sqrt{{\left(\frac{2}{3}\right)}^{2} + {\left(\frac{-1}{3}\right)}^{2}} + {\left(\frac{-2}{3}\right)}^{2}}$

However, this does not prove that I was working with a unit vector, as the answer did not evaluate to one. How can I find the unit vector of a three-dimensional vector?

Thank you for your time.

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    $\begingroup$ The pythagorean theorem works for any number of dimensions, not just two. $\endgroup$ Jan 28 '12 at 20:15
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Since this is homework, we are not supposed to give you the answer. But one mistake you made is in your formula for the magnitude of $r$ - the inner square root needed to be squared. So the length of $r$ is simply the square root of the sum of the squares of the $i$, $j$ and $k$ lengths.

Good luck...

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  • $\begingroup$ I see... However, whenever I evaluate this, it doesn't show that it is a unit vector: wolframalpha.com/input/… Is it possible that I am doing something wrong? $\endgroup$ Jan 28 '12 at 20:21
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    $\begingroup$ You should end up with $\sqrt((2^2+1^2+2^2)/ 3^2)$ - isn't that $1$? $\endgroup$
    – FrankH
    Jan 28 '12 at 20:41
  • $\begingroup$ Rats, I made a typo in WolframAlpha, I ended up cubing the $\hat{i}$ direction in my calculation. Thank you, Frank! $\endgroup$ Jan 28 '12 at 20:44
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Although you already have an answer, I want to show you a visualization. The dark black vector is $\hat{r}$ and in green is the projection on the XY plane (ignoring the z-axis). In blue is only the z axis component vector. These form an orthogonal triangle and if you want to know the length of the hypotenuse ($\hat{r}$) you will need the length of the other two vectors. Now the length of the green vector you said you know how to get, and the length of the blue vector is trivial. If you work it out, you will arrive at the 3D formula for vector lengths.

VectorSketch

PS. Sketches were done in GeoGebra 5.0 beta (which has some 3D capabilities now).

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