Prove true or false with counter example $\lVert\vec{u}\rVert=\lVert\vec{v}\rVert \implies (\vec{u}+\vec{v}) \cdot (\vec{u}-\vec{v})=0 $

Question

Problem

Prove true or false with counter example $$\lVert\vec{u}\rVert=\lVert\vec{v}\rVert \implies (\vec{u}+\vec{v}) \cdot (\vec{u}-\vec{v})=0 $$

when $\{\vec{u},\vec{v}\}\in \mathbb{R}^n$

Attempt to solve

Let $\{\vec{u},\vec{v}\}\in \mathbb{R}^2$ then i can display $ \vec{u},\vec{v}$ as:

$$ \vec{u}=\begin{bmatrix} a \\ b \end{bmatrix}, \vec{v}= \begin{bmatrix} b \\ a \end{bmatrix} $$

$\lVert\vec{u}\rVert=\lVert\vec{v}\rVert$ can be only true when both vectors have same elements when order doesn't matter.

where $\{a,b\}\in \mathbb{R}$

$$ \sqrt{a^2+b^2}=\sqrt{b^2+a^2} \implies \begin{bmatrix} a + b \\ b + a \end{bmatrix} \cdot \begin{bmatrix} a - b \\ b - a \end{bmatrix}=0 $$

$$ \implies (a+b)\cdot (a-b) + (b+a) \cdot (b-a) $$ $$ = a\cdot a + a \cdot (-b) + b \cdot a + b \cdot (-b)+ b \cdot b + b \cdot (-a) + a \cdot b + a \cdot (-a)$$

$$ = a^2-a^2+ab-ab+ab-ab+b^2-b^2 = 0 $$

which is true when $ \{\vec{u},\vec{v}\} \in \mathbb{R}^2 $ but not in $\mathbb{R}^n$. How do you prove this for $\mathbb{R}^n$ case ?

Answer 1

\begin{align}(\vec u+\vec v).(\vec u-\vec v)&=\vec u.\vec u-\vec u.\vec v+\vec v.\vec u-\vec v.\vec v\\&=\|\vec u\|^2-\|\vec v\|^2\\&=0\end{align}

Answer 2

\begin{align*} \|u\|^2 & =\|v\|^2\\ u \cdot u & =v \cdot v\\ (u \cdot u)+(u \cdot v)-(u \cdot v)-(v \cdot v)&=0\\ (u+v) \cdot (u-v)& =0. \end{align*}