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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?

2 Answers 2

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$$|u\times v|^2=|u|^2|v|^2\sin^2\theta$$ $$=|u|^2|v|^2(1-\cos^2\theta)$$ $$=|u|^2|v|^2-(u\cdot v)^2$$

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?

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    sorry my bad, now it's correct2012-07-12
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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.