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 .
$

✓

Answer

$
|\vec{a} \times \vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-|\vec{a} \cdot \vec{b}|^2
$
We know that $\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \cdot \hat{n}$
$
\begin{aligned}
& \vec{a} \times \vec{b} \\
= & a b \sin \theta \hat{n} \\
\therefore \quad & |\vec{a} \times \vec{b}|
\end{aligned}=|a b \sin \theta \hat{n}|=a b \sin \theta \quad[\because|\hat{n}|=1]
$
So $\quad|\vec{a} \times \vec{b}|^2=a^2 b^2 \sin ^2 \theta\quad \quad \ldots \ldots(1)$
R.H.S. $\quad=|\vec{a}|^2|\vec{b}|^2-|\vec{a} \cdot \vec{b}|^2$
$
\begin{array}{l}
=\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\quad \quad \ldots \ldots(2)
\end{array}
$
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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