Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

up vote 1 down vote accepted

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 $\vec{a_{\perp}} = \left( 3,2,1 \right) - \left( 2,2,2 \right) = \left( 1,0,-1 \right) = \frac{\vec{e}}{||\vec{e}||}_2 \times \left( \vec{a} \times \frac{\vec{e}}{||\vec{e}||}_2 \right)$

share|improve this answer

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 $\lvert \vec{e}\rvert^2$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.