Prove that `(veca+ vecb)*( veca+ vecb)=|veca|^2+| vecb|^2`, if and only if `veca , vecb`are perpendicular, given ` veca!= vec0, vecb!= vec0`

Explanation

To 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, we will follow these steps:

### Step 1: Expand the left-hand side
We start with the expression \((\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})\).

Using the distributive property of the dot product, we can expand this as follows:
\[
(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}
...