$\|x\|=\|y\|$ then $x+y$ and $x-y$ are orthogonal

Question

Let $\vec{x},\vec{y}$ be members of $R^n$. If $\|x\|=\|y\|$, then $\vec x+\vec y$ and $\vec x-\vec y$ are orthogonal, prove this.

Not sure how to approach this, except for the fact that I know $\|\vec x\|^2=\vec x \cdot \vec x$.

Answer 1

By bilinearity of inner products, $\langle x+y,x-y\rangle = \langle x,x\rangle - \langle x,y\rangle + \langle y,x\rangle -\langle y,y\rangle = \|x\|^2-\|y\|^2=0.$

Answer 2

Hint:

To prove to vectors orthogonal we must show the dot product is zero,

$$(\vec x- \vec y) \cdot (\vec x+ \vec y)=0$$

$$\vec x \cdot (\vec x+\vec y)-\vec y \cdot (\vec x+\vec y)=0$$

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It's worth noting that the dot product follows the commutative and distributive rules similar to regular multiplication.