Proof (vectors)

**Mtl** · January 6th 2008, 07:17 AM

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

**galactus** · January 6th 2008, 08:04 AM

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

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

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

Expand out and confirm the equality.

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

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

**Mtl** · January 6th 2008, 08:27 AM

" $ (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"

**Plato** · January 6th 2008, 09:50 AM

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.
$\left\| a \right\|\left\| b \right\|\sin \left( \theta \right) = \left\| {a \times b} \right\|\,\& \,\left\| a \right\|\left\| b \right\|\cos \left( \theta \right) = a \cdot b$ .

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

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