Cross product- square

Question

I recently saw the following expression somewhere-

$$\frac{1}{2} \left\| \frac{\vec{u}}{9} \times \frac{\vec{u} + \vec{v}}{9} \right\| + \frac{1}{2} \left\| \frac{\vec{u} + \vec{v}}{9}\times \frac{\vec{v}}{9} \right\| = \frac{1}{81} \left\|\vec{u}\times \vec{v}\right\|$$

How does this work? Shouldn't LHS evaluate to $\frac{ \|(\vec{u}\times\vec{v})^2\| }{162}$?

Answer 1

Note that $u \times (u + v) = (u\times u) + (u \times v) = 0 + (u \times v) = u\times v$, as the cross product is distributive over addition, and the cross product of a vector by itself is the all-zero vector ($0$ is the all-zero vector). Similarly, we obtain $(u+v) \times v = u \times v$ (this time using "distributiveness from right.") Therefore,

\begin{align} \frac{1}{2} \left\| \frac{\vec{u}}{9} \times \frac{\vec{u} + \vec{v}}{9} \right\| + \frac{1}{2} \left\| \frac{\vec{u} + \vec{v}}{9}\times \frac{\vec{v}}{9} \right\| & = \frac{1}{2 \times 9 \times 9} \|u \times (u+v)\| + \frac{1}{2 \times 9 \times 9} \|(u+v) \times v\| \\ & = \frac{\|u\times v\|}{81} \end{align}

Answer 2

You haven't understood the cross product $\times$. This is not normal multiplication, and one of the properties it obeys is $$\vec{u}\times\vec{u} = 0$$

See http://en.wikipedia.org/wiki/Cross_product for an introduction.

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Edit: Also note that taking the norm of a vector, denoted by $\lVert\vec{u}\rVert$, is not linear, so $$\lVert\vec{u}\rVert + \lVert\vec{v}\rVert \neq \lVert\vec{u} + \vec{v}\rVert$$

Answer 3

The answer you saw was correct.

First of all, the cross product is associative, i.e. $(\vec{u}+\vec{v})\times\vec{w} = (\vec{u}\times\vec{w})+(\vec{v}\times\vec{w})$. The same is true on the left-hand side too. Moreover, the cross product is linear, i.e. $(\lambda \vec{u}) \times (\mu \vec{v}) = \lambda\mu (\vec{u} \times \vec{v})$. Finally, recall that for any vector $\vec{u}$ we have $\vec{u} \times \vec{u} = \vec{0}$.

We can apply these rules to simplify the problem:

$$\frac{1}{2}\left\|\frac{\vec{u}}{9} \times \frac{\vec{u}+\vec{v}}{9}\right\| = \frac{1}{2}\left\|\frac{\vec{u} \times \vec{v}}{81}\right\|$$

$$\frac{1}{2}\left\|\frac{\vec{u}+\vec{v}}{9} \times \frac{\vec{v}}{9}\right\| = \frac{1}{2}\left\|\frac{\vec{u} \times \vec{v}}{81}\right\|$$

Putting this together, we get:

$$\frac{1}{2}\left\|\frac{\vec{u} \times \vec{v}}{81}\right\|+\frac{1}{2}\left\|\frac{\vec{u} \times \vec{v}}{81}\right\|=\left\|\frac{\vec{u} \times \vec{v}}{81}\right\|$$