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I have a problem in 3D where I have three vectors: $\boldsymbol{\omega}$, $\boldsymbol{m}$ and $\boldsymbol{T}$.

The first $\boldsymbol{\omega}$ is the variable I solve for, $\boldsymbol{m}$ is fixed and the last is related to the others by

$$ \boldsymbol{T} \sim (\boldsymbol{m}^{T} \boldsymbol{m} - I) \boldsymbol{\omega} $$

Hopefully I have this correct, the $(\boldsymbol{m}^{T} \boldsymbol{m} )$ term creates a $3 \times 3$ array and $I$ is the identity matrix.

I think this should be some kind of orthogonal projection of $\boldsymbol{\omega}$ onto a plane orthogonal to $\boldsymbol{m}$ but I can't seem to picture it.

So does anyone have an idea or references for what kind of transformation this is?

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up vote 1 down vote accepted

Given that you're multiplying $\omega$ from the left to get another vector, I can only presume that $T$, $m$, and $\omega$ are column vectors, in which case you should have $mm^T$ instead of $m^T m$, which would just be a scalar.

Now, if $m$ is a unit vector, then $m m^T$ is precisely the orthogonal projection onto the line $\mathbb{R}m$ spanned by $m$ (otherwise, you need to take $\tfrac{1}{\|m\|^2}mm^T$). Hence, $I - mm^T$ (or more generally, $I - \tfrac{1}{\|m\|^2}mm^T$ if $m$ isn't a unit vector) is, for completely general reasons, the orthogonal projection onto the orthogonal complement $(\mathbb{R}m)^\perp$ of $\mathbb{R}m$, that is, onto the plane orthogonal to $m$. Thus, if $m$ is a unit vector, then your relation implies that $T$ is proportional to the orthogonal projection of $\omega$ onto the plane orthogonal to $m$, as you suspected.

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Fantastic thanks for that – Greg Ashton Jul 22 '13 at 15:00

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