 # A linear map f:RR^3 to RR^3 is defined as vec(u) xx (vec(r) xx vec(u)) where vec(u) is a unit vector. What would its geometry look like? Javion Henry 2022-07-21 Answered
A linear map $f:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$ is defined as $\stackrel{\to }{u}×\left(\stackrel{\to }{r}×\stackrel{\to }{u}\right)$ where $\stackrel{\to }{u}$ is a unit vector. What would its geometry look like?
I know that I can rewrite this map as $\left(\stackrel{\to }{u}\cdot \stackrel{\to }{u}\right)\stackrel{\to }{r}-\left(\stackrel{\to }{u}\cdot \stackrel{\to }{r}\right)\stackrel{\to }{u}=\stackrel{\to }{r}-\left(\stackrel{\to }{u}\cdot \stackrel{\to }{r}\right)\stackrel{\to }{u}$
However I am not sure what to do from here.
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Clearly $\left(\stackrel{\to }{u}\cdot \stackrel{\to }{r}\right)\stackrel{\to }{u}$ is the component of $\stackrel{\to }{r}$ along the unit vector $\stackrel{\to }{u}$. So $\stackrel{\to }{u}×\left(\stackrel{\to }{r}×\stackrel{\to }{u}\right)=\stackrel{\to }{r}-\left(\stackrel{\to }{u}\cdot \stackrel{\to }{r}\right)\stackrel{\to }{u}$ is the component of $\stackrel{\to }{r}$ which is perpendicular to the unit vector $\stackrel{\to }{u}$ .
The map is neither injective nor surjective. It is not injective because for $\stackrel{\to }{a}$ and for $\stackrel{\to }{a}+\lambda \stackrel{\to }{u}$ we get the same value for any real $\lambda$. It is not surjective because the range of the function doesn't contain any vector in ${\mathbb{R}}^{3}$ that has non zero component in the direction of $\stackrel{\to }{u}$ such as $\stackrel{\to }{u}$ itself.

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