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Consider the the following problem.

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Define $u'$ and $v'$ to be the projections of $u$ and $v$ onto the plane perpendicular to the axis of $\xi$. If $\omega \in \mathbb{R}^{3 \times 1}$ is a unit vector in the direction of the axis of $\xi$, then

$u' = u -\omega \omega ^{T} u$ and $v' = v -\omega \omega ^{T} v$

Is something wrong with these equations?

I thought that $\omega \omega ^{T}u = u$ (dot product) so it becomes $u' = u - u = 0$

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1 Answer 1

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Note that $\omega$ is a column unit vector and hence the inner product (or dot product) is $\omega^T \omega = 1$.

The outer product $\omega \omega^T$ is a $3 \times 3$ matrix (which is called the projection matrix) and hence $\omega \omega^T u \neq u$ unless $u$ and $\omega$ are in the same direction.

The outer product $\omega \omega^T$ takes any vector and projects it onto the space spanned by the vector $\omega$.

$\omega^T u$ is the scalar projection of the vector $u$ onto the $\xi$ axis.

$\omega (\omega^T u)$ is the vector projection of the vector $u$ onto the $\xi$ axis.

So you subtract of the vector projection of $u$ onto the $\xi$ axis to get $u'$, the vector projection of $u$ onto the plane perpendicular to the $\xi$ axis.

The same thing holds for $v$

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