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 $\vec{b}.$

To show that two vectors are perpendicular,their dot product must be equal to $0$.
Let $\vec{u} = |\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ and $\vec{v} = |\vec{a}| \vec{b}-|\vec{b}| \vec{a}$.
Calculate the dot product $\vec{u} \cdot \vec{v}$:
$\vec{u} \cdot \vec{v} = (|\vec{a}| \vec{b}+|\vec{b}| \vec{a}) \cdot (|\vec{a}| \vec{b}-|\vec{b}| \vec{a})$
Using the distributive property of the dot product:
$= |\vec{a}|^2 (\vec{b} \cdot \vec{b}) - |\vec{a}||\vec{b}| (\vec{b} \cdot \vec{a}) + |\vec{b}||\vec{a}| (\vec{a} \cdot \vec{b}) - |\vec{b}|^2 (\vec{a} \cdot \vec{a})$
Since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ and $\vec{x} \cdot \vec{x} = |\vec{x}|^2$:
$= |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}||\vec{b}| (\vec{a} \cdot \vec{b}) + |\vec{a}||\vec{b}| (\vec{a} \cdot \vec{b}) - |\vec{b}|^2 |\vec{a}|^2$
$= |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2$
$= 0$
Since the dot product is $0$,the vectors $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ and $|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$ are perpendicular to each other.