# Proof (vectors)

• Jan 6th 2008, 07:17 AM
Mtl
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
• Jan 6th 2008, 08:04 AM
galactus
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:

$(a_{2}b_{3}-a_{3}b_{2})+(a_{3}b_{1}-a_{1}b_{3})+(a_{1}b_{2}-a_{2}b_{1})$
• Jan 6th 2008, 08:27 AM
Mtl
" $
(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"
• Jan 6th 2008, 09:50 AM
Plato
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,
$\begin{array}{rcl}
\left\| {a \times b} \right\|^2 & = & \left[ {\left\| a \right\|\left\| b \right\|\sin \left( \theta \right)} \right]^2 \\
& = & \left\| a \right\|^2 \left\| b \right\|^2 \sin ^2 \left( \theta \right) \\
& = & \left\| a \right\|^2 \left\| b \right\|^2 - \left( {a \cdot b} \right)^2 \\
\end{array}$