For vectors a and b prove that : |a b|^2=|a|^2|b|^2-|a b|^2 .

Question

For vectors $\vec{a}$ and $\vec{b}$ prove that : $|\vec{a} \times \vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-|\vec{a} \cdot \vec{b}|^2 .$

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Answer

$|\vec{a} \times \vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-|\vec{a} \cdot \vec{b}|^2$
We know that $\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \cdot \hat{n}$
$\vec{a} \times \vec{b}$
$= a b \sin \theta\ \hat{n}$
$\therefore |\vec{a} \times \vec{b}|
=|a b \sin \theta\ \hat{n}|=a b \sin \theta \quad[\because| \ \hat{ n}|=1]$
So $\quad|\vec{a} \times \vec{b}|^2=a^2 b^2 \sin ^2 \theta \ldots \ldots(1)$
$\text{R.H.S}. \quad=|\vec{a}|^2|\vec{b}|^2-|\vec{a} \cdot \vec{b}|^2$
$=\vec{a} \cdot \vec{a} \vec{b} \cdot \vec{b}-|a b \cos \theta|^2$
$=a^2 b^2-a^2 b^2 \cos ^2 \theta=a^2 b^2\left(1-\cos ^2 \theta\right)$
$=a^2 b^2 \sin ^2 \theta \ldots \ldots(2)$
Hence from equations $(1)$ and $(2), $ it is clear that
$|\vec{a} \times \vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-|\vec{a} \cdot \vec{b}|^2 \quad$

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