Prove that $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2},$ if and only if $\vec{a}$ and $\vec{b}$ are perpendicular,given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}.$

(A) Given the equation: $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$
Using the distributive property of the dot product:
$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = |\vec{a}|^{2} + |\vec{b}|^{2}$
Since $\vec{a} \cdot \vec{a} = |\vec{a}|^{2}$,$\vec{b} \cdot \vec{b} = |\vec{b}|^{2}$,and the dot product is commutative $(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a})$:
$|\vec{a}|^{2} + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2}$
Subtracting $|\vec{a}|^{2} + |\vec{b}|^{2}$ from both sides:
$2(\vec{a} \cdot \vec{b}) = 0$
$\vec{a} \cdot \vec{b} = 0$
Since $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$,the dot product of two non-zero vectors is zero if and only if they are perpendicular. Thus,$\vec{a} \perp \vec{b}$.