Show that $(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})$.
(N/A) We use the distributive property of the vector cross product over addition:
$(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b}) = (\vec{a}-\vec{b}) \times \vec{a} + (\vec{a}-\vec{b}) \times \vec{b}$
$= (\vec{a} \times \vec{a}) - (\vec{b} \times \vec{a}) + (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{b})$
Since the cross product of any vector with itself is the zero vector,i.e.,$\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$,and using the anti-commutative property $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:
$= \vec{0} - (-\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b}) - \vec{0}$
$= (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b})$
$= 2(\vec{a} \times \vec{b})$
Hence,the given expression is proved.
$(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b}) = (\vec{a}-\vec{b}) \times \vec{a} + (\vec{a}-\vec{b}) \times \vec{b}$
$= (\vec{a} \times \vec{a}) - (\vec{b} \times \vec{a}) + (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{b})$
Since the cross product of any vector with itself is the zero vector,i.e.,$\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$,and using the anti-commutative property $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:
$= \vec{0} - (-\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b}) - \vec{0}$
$= (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b})$
$= 2(\vec{a} \times \vec{b})$
Hence,the given expression is proved.
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