1. ## Proof (vectors)

OK, so I have a proof which I have am having a hard time prooving. Well here it is |a cross b| = ((a dot a) (b dot b) - (a dot b)sqrd) to the power of 1/2

(Also, it seems i have posted in the wrong section, if moving the thread to the Euclidean Geometry section is at all possible please let me know

2. This looks very similar to Lagrange's Identity. Which states:

$\displaystyle \Vert{a\times{b}}\Vert^{2}=\Vert{a}\Vert^{2}\Vert{ b}\Vert^{2}-(a\cdot{b})^{2}$

For 3-space, you could use $\displaystyle \langle{a_{1},a_{2},a_{3}}\rangle$

Expand out and confirm the equality.

$\displaystyle \Vert{a\times{b}}\Vert$ may be written as:

$\displaystyle (a_{2}b_{3}-a_{3}b_{2})+(a_{3}b_{1}-a_{1}b_{3})+(a_{1}b_{2}-a_{2}b_{1})$

3. "$\displaystyle (a_{2}b_{3}-a_{3}b_{2})+(a_{3}b_{1}-a_{1}b_{3})+(a_{1}b_{2}-a_{2}b_{1})$"

Should not each term be squared and the term as a whole be "squarerooted"

4. There are many ways to do this, no one is correct (except the one your instructor wants). Here is one way using basic definitions and equivalents.
$\displaystyle \left\| a \right\|\left\| b \right\|\sin \left( \theta \right) = \left\| {a \times b} \right\|\,\& \,\left\| a \right\|\left\| b \right\|\cos \left( \theta \right) = a \cdot b$.

Therefore,
$\displaystyle \begin{array}{rcl} \left\| {a \times b} \right\|^2 & = & \left[ {\left\| a \right\|\left\| b \right\|\sin \left( \theta \right)} \right]^2 \\ & = & \left\| a \right\|^2 \left\| b \right\|^2 \sin ^2 \left( \theta \right) \\ & = & \left\| a \right\|^2 \left\| b \right\|^2 - \left( {a \cdot b} \right)^2 \\ \end{array}$