Angle between $\left(\vec{u} + \vec{v}\right)$ and $\left(\vec{u} - \vec{v}\right)$ Given that the vectors $\vec{u}$ and $\vec{v}$ are not null, can we state that the angle between $\left(\vec{u} + \vec{v}\right)$ and $\left(\vec{u} - \vec{v}\right)$ is $\pi/2$?
If so, how can one prove it?
Thank you.
 A: No.
Given any vectors $\vec a,\vec b$, the angle $\theta$ between them can be defined as:
$$\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$$
So in this context, 
$$0=\cos\frac{\pi}2=\frac{(\vec u+\vec v)\cdot(\vec u-\vec v)}{|\vec u+\vec v||\vec u-\vec v|}$$
$$\iff(\vec u+\vec v)\cdot(\vec u-\vec v)=0$$
$$\iff\vec u^2-\vec v^2=0$$
which is not always true.
A: No. This is true if and only if $\;\lVert \vec u+\vec v\rVert=\lVert \vec u-\vec v\rVert$.
A: Let $\vec a$ and $\vec b$ be any two vectors that are neither equal
nor exactly opposite one another.
Let $\vec u = \frac12(\vec a + \vec b)$ and let
$\vec v = \frac12(\vec a - \vec b)$.
Then neither $\vec u$ nor $\vec v$ is null,
while at the same time
$\vec u + \vec v = \vec a$ and $\vec u - \vec v =\vec b$.
In other words, unless you can give us some additional restrictions
on $\vec u$ and $\vec v$ other than "not null", the vectors
$\vec u + \vec v$ and $\vec u - \vec v$ could be
just about any two vectors in the vector space,
with only the restrictions that they can not be exactly equal
nor can one simply be the negation of the other.
The angle between them could be $0$, $\pi$, or anything in between.
