# Vector Analysis Question

equation 1:

$-i\omega\underline \xi - 2i(\underline k \cdot \underline \Omega )\underline u = 0$

with:

$\underline \xi = i \underline k \times \underline u$

$\underline k \cdot \underline u = 0$

taking:

$i\underline k \times \text{(1)}$

derive:

$\omega = \pm 2\frac{(\underline k \cdot \underline \Omega )}{|\underline k|}$

I can get pretty close and can post my working if required - but I don't do anything complicated - any help = many thanks

-
Can you write a couple of sentences describing what you want to do or to know? – Mariano Suárez-Alvarez Feb 7 '11 at 20:31
Derive the bottom equation by taking ik cross product with equation 1 – user6757 Feb 7 '11 at 20:51

You have $$\underline 0=i\underline k\times(-i\omega\underline \xi - 2i(\underline k \cdot \underline \Omega )\underline u))=-i^2\omega(\underline k\times \underline \xi)-2i^2(\underline k \cdot \underline \Omega)(\underline k\times\underline u)$$

But by the triple cross product expansion formula

$$\underline k\times \underline \xi=\underline k\times (i\underline k\times \underline u)=i|k|^2\underline u$$ since $\underline u\cdot\underline k=0$.

Thus the first equation can be restated as

$$-i^3\omega |k|^2\underline u=2i^2(\underline k \cdot \underline \Omega)(\underline k\times\underline u)$$

Taking norms, you will get what you want.

Remember that $|\underline k\times\underline u|=|\underline k||\underline u|$ because $\underline u\cdot\underline k=0$ (see this).

-