Question:
Show that $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$, for any two nonzero vectors $\vec{a}$ and $\vec{b}$
Solution:
$(|\vec{a}| \vec{b}+|\vec{b}| \vec{a}) \cdot(|\vec{a}| \vec{b}-|\vec{b}| \vec{a})$
$=|\vec{a}|^{2} \vec{b} \cdot \vec{b}-|\vec{a}||\vec{b}| \vec{b} \cdot \vec{a}+|\vec{b}||\vec{a}| \vec{a} \cdot \vec{b}-|\vec{b}|^{2} \vec{a} \cdot \vec{a}$
$=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{b}|^{2}|\vec{a}|^{2}$
$=0$
Hence, $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ and $|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$ are perpendicular to each other.
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