Proving vectors are orthogonal

Question

if $||\vec{a}+\vec{b}||^2=||\vec{a}||^2+||\vec{b}||^2$ for $\vec{a},\vec{b} \in \Bbb{R^n}$ how would you show that $\vec{a},\vec{b}$ are orthogonal?

I cant seem to find any connection or algebraic manipulation to show that $\vec{a} \cdot \vec{b}=0$

Answer 1

Note that $\|a+b\|^2=\langle a+b,a+b\rangle=\langle a,a \rangle+2 \langle a,b\rangle+\langle b,b\rangle=\|a\|^2+\|b\|^2+2\langle a,b \rangle$.

Subtract the LHS from the RHS and deduce the result.

The third equality follows from bilinearity of inner (dot) product.

Answer 2

$\|a+b\|^2 = (a+b)\cdot(a+b) = a\cdot a + a\cdot b + b\cdot a + b\cdot b = \|a\|^2 + \|b\|^2 + 2a\cdot b$

if $\|a+b\|^2 = \|a\|^2 + \|b\|^2$ then $2a\cdot b = 0$

This is the expression of familiar theorems in the language of linear algebra.

$\|a+b\|^2 = \|a\|^2 + \|b\|^2$ is the Pythagorean theorem.

$\|a+b\|^2 = \|a\|^2 + \|b\|^2 + 2a\cdot b$ is the law of cosines.

Answer 3

\begin{align*} ||\vec a + \vec b||^2 &= (\vec a+\vec b)\cdot (\vec a+\vec b)\\ &= \vec a \cdot \vec a + 2\; \vec a \cdot \vec b + \vec b \cdot \vec b\\ &= ||\vec a||^2+2\; \vec a \cdot \vec b+ ||\vec b||^2\\ \end{align*} On the other hand \begin{align*} ||\vec a + \vec b||^2 &= ||\vec a||^2 + ||\vec b||^2 \end{align*}

Hence, $\vec a \cdot \vec b = 0$, implying $\vec a$ and $\vec b$ are perpendicular.