# Cross product: problem with breakdown of vector to parallel + orthogonal components

Vector $\vec{a}$ can be broken down into its components $\vec{a}_\parallel$ and $\vec{a}_\perp$ relative to $\vec{e}$.

• $\vec{a}_\parallel = (\vec{a}\vec{e})\vec{e}$

and

• $\vec{a}_\perp = \vec{e} \times (\vec{a} \times \vec{e})$ (f1)

The orthogonal part can be found via application of the triple product:

• $\vec{a}_\perp = \vec{a} - \vec{a}_\parallel = \vec{a}(\vec{e}\vec{e}) - \vec{e}(\vec{e}\vec{a}) = \vec{e} \times (\vec{a} \times \vec{e})$ (f2)

This one causes me problems. I tried to use some values for the formulas and disaster strikes:

$\vec{a} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$ and $\vec{e} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ makes $\vec{a}_\parallel = \begin{pmatrix} 6 \\ 6 \\ 6 \end{pmatrix}$.

I thought to follow the values "all the way" through the last formula f2. So I calculate for $\vec{a}_{\perp subtraction} = \vec{a} - \vec{a}_\parallel = \begin{pmatrix} -3 \\ -4 \\ -5 \end{pmatrix}$ but I find for $\vec{a}_{\perp cross} = \vec{e} \times (\vec{a} \times \vec{e}) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} =\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix}$.

Can someone point out my error? Must be some mis-calculation, as $\vec{a}_{\perp cross} + \vec{a}_\parallel \neq \vec{a}$ and I do not really find the two $\perp$-vectors parallel.

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I think you are not normalizing correctly, because the vector you get is just a scalar multiple of $\overline{a}$. Are you sure there is no assumption on the lengths of the vectors, e.g., that $\overline{e}$ is a unit vector? – Arturo Magidin Mar 3 '11 at 16:33
Ok, that seems to be the problem. I'll re-read the passages to find where I missed this instruction. – Leonidas Mar 3 '11 at 17:08
Next time I'll look up the definition of all used components of a formula. Normalizing a vector ... so easy ... argh :) – Leonidas Mar 3 '11 at 17:25

To get the projection along $\vec{e}$ i.e. $\vec{a_{||}}$, you need to project your vector $a$ along the unit vector in the direction of $\vec{e}$. Similarly, if you want the component of $\vec{a}$ perpendicular to $\vec{e}$, $\vec{a_{\perp}} = \frac{\vec{e}}{||\vec{e}||}_2 \times \left( \vec{a} \times \frac{\vec{e}}{||\vec{e}||}_2 \right)$
Hence, $\vec{a_{||}} = \left( \vec{a} \cdot \frac{\vec{e}}{||\vec{e}||}_2 \right) \frac{\vec{e}}{||\vec{e}||}_2$ and $\vec{a_{\perp}} = \vec{a} - \vec{a_{||}} = \frac{\vec{e}}{||\vec{e}||}_2 \times \left( \vec{a} \times \frac{\vec{e}}{||\vec{e}||}_2 \right)$.
In your case, $\vec{a_{||}} = \frac{6}{\sqrt{3}} \frac1{\sqrt{3}} \left(1,1,1\right)^T = \left( 2,2,2 \right)$ and hence $\vec{a_{\perp}} = \left( 3,2,1 \right) - \left( 2,2,2 \right) = \left( 1,0,-1 \right) = \frac{\vec{e}}{||\vec{e}||}_2 \times \left( \vec{a} \times \frac{\vec{e}}{||\vec{e}||}_2 \right)$
Your formulas are valid only if $\vec{e}$ is a unit vector, but the $\vec{e}$ you're using isn't. To use an arbitrary non-zero vector $\vec{e}$, you'd need to divide the right-hand side of both formulas by $\lvert \vec{e}\rvert^2$.