Proof with cross product

**a.a** · May 25th 2008, 02:41 PM

Proove that the magnitude of vector a cross b is equal to :
sqrt. ((a dot b) (b dot b) - (a dot b)^2)
I have no idea how to go about proveing this...:S

**galactus** · May 25th 2008, 03:42 PM

It would appear you mean LaGrange's identity.

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

$||a\times{b}||^{2}=(a_{2}b_{3}-a_{3}b_{2})^{2}+(a_{3}b_{1}-a_{1}b_{3})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}$

$||a||^{2}||b||^{2}-(a\cdot{b})^{2}=(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_ {1}^{2}+b_{2}^{2}+b_{3}^{2})-(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^{2}$

Now, just multiply out the right sides and verify the equality.

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