See this statement: $ |u×v|^2=|u|^2\cdot|v|^2-(u\cdot v)^2 $
I need to prove this is right.
I only found that: $ u×v=|u|\cdot|v|\cdot\sin\theta $ and $ u.v=|u|\cdot|v|\cdot\cos\theta $
Does this helps?
See this statement: $ |u×v|^2=|u|^2\cdot|v|^2-(u\cdot v)^2 $
I need to prove this is right.
I only found that: $ u×v=|u|\cdot|v|\cdot\sin\theta $ and $ u.v=|u|\cdot|v|\cdot\cos\theta $
Does this helps?
$|u\times v|^2=|u|^2|v|^2\sin^2\theta$ $=|u|^2|v|^2(1-\cos^2\theta)$ $=|u|^2|v|^2-(u\cdot v)^2$
Your equation isn't right. Take $(1,0,0)\times(0,1,0)=(0,0,1)$. The dot product is $0$, and the lengths of the starting vectors are equal. The formula above gives $1$ (the correct length), while yours would give $0$. Where did you find that formula?
According to what you wrote: $|u\times v|^2=|u|^2|v|^2\sin^2\theta$ $|u|^2-|v|^2-(u\cdot v)^2=|u|^2-|v|^2-|u|^2|v|^2\cos^2\theta$ So if the left sides are equal also the right ones are: $|u|^2|v|^2\sin^2\theta=|u|^2-|v|^2-|u|^2||v|^2\cos^2\theta\Longleftrightarrow$ $\Longleftrightarrow|u|^2|v|^2=|u|^2-|v|^2$which of course is far from being true in general. Check your equalities.