# vector identity

• Sep 27th 2010, 03:37 PM
Random Variable
vector identity
EDIT: I'm trying to show that $\frac{| v \times w |}{|v|}= \Big |w - \frac{v \cdot w}{|v|^{2}} v \Big|$

My initial thought was to get the right side to look like $|w| |\sqrt{(1-\cos^{2} \theta}|$, but I can't seem to do it.

By the way, what happened to all my thanked posts? We're they lost when the forum was updated a few months ago? I haven't been been here in a while.
• Sep 27th 2010, 05:38 PM
Ackbeet
$\displaystyle{\left|w - \frac{v \cdot w}{|v|^{2}} v \right|=\sqrt{\left(w - \frac{v \cdot w}{|v|^{2}} v\right)\cdot\left(w - \frac{v \cdot w}{|v|^{2}} v\right)}=\sqrt{w\cdot w-2\frac{(v\cdot w)^{2}}{|v|^{2}}+(v\cdot w)^{2}}}$

$\displaystyle{=|w|\sqrt{1-2\cos^{2}(\theta)+|v|^{2}\cos^{2}(\theta)}}.$

This will equal the desired RHS if and only if $|v|=1.$ Is that stated in the problem somewhere?

[EDIT]: TheEmptySet's computations are more correct than mine. The v squared terms do cancel.
• Sep 27th 2010, 05:45 PM
TheEmptySet
Remember that for any vectors $\vec{a},\vec{b}$ we have

$|\vec{a}|^2=\vec{a}\cdot \vec{a}$ and
$(\vec{a}-\vec{b})\cdot ((\vec{a}-\vec{b}))=|\vec{a}|^2-2(\vec{a}\cdot \vec{b})+|\vec{b}|$

Using this we have

$\displaystyle \bigg| \vec{w}-\frac{\vec{v}\cdot \vec{w}}{|\vec{v}|^2}\vec{v}\bigg|^2=\left( \vec{w}-\frac{\vec{v}\cdot \vec{w}}{|\vec{v}|^2}\vec{v}\right)\cdot \left( \vec{w}-\frac{\vec{v}\cdot \vec{w}}{|\vec{v}|^2}\vec{v}\right)=$
$\displaystyle |\vec{w}|^2-\frac{2(\vec{v}\cdot \vec{w})}{|\vec{v}|^2}+\left( \frac{\vec{v}\cdot \vec{w}}{|v|^2}\right)^2(\vec{v}\cdot \vec{v})$
$\displaystyle |\vec{w}|^2-|\vec{w}|^2\cos^2(\theta)=|\vec{w}|^2(1-\cos^2(\theta))=|w|^2\sin^2(\theta)$