$(a\vec{x}+b\vec{v})$ and $(b\vec{x}+a\vec{v})$ are orthogonal. How would you show that $\vec{x}=\vec{0}=\vec{v}$

Question

Suppose that $\vec{x}$, $\vec{v}$ ∈ $\Bbb{R^2}$ are vectors such that for every choice of scalars $a, b ∈ \Bbb{R}$, the vectors $(a\vec{x}+b\vec{v})$ and $(b\vec{x}+a\vec{v})$ are orthogonal. How would you show that $\vec{x}=\vec{0}=\vec{v}$

I did the dot product of both vectors, and expanded but i have reached a point where im stuck

Answer 1

Take $a = b = 1 \rightarrow (\vec{x} + \vec{v})$ orthogonal to $(\vec{x} + \vec{v})$ hence $(\vec{x} + \vec{v}) = \vec{0} \rightarrow \vec{x} = -\vec{v}$. Then take $a = 0, b = 1$ thus $\vec{x}$ orthogonal to $\vec{v}$ and since $\vec{x} = -\vec{v}$ it's possible only if $\vec{x} = \vec{v} = \vec{0}$.

Answer 2

Let $a=1, b=0$. Then $x\cdot v=0$.

Now let $a=b=1$. Then$(x+v)\cdot (x+v)=0$. So ${\lvert x \rvert}^2+{\lvert v\rvert} ^2=0 \implies {\lvert x\rvert}^2={\lvert v\rvert}^2 =0 \implies x=v=0$...

Answer 3

Let $f(a,b)=(a\vec x+b\vec v)\cdot(b\vec x+a\vec v)=0$

We have forall $a,b$ that $\quad f(a,b)-f(a,-b)=2ab\left(\lVert\vec x\lVert^2+\lVert\vec v\lVert^2\right)=0\iff \vec x=\vec v=\vec 0$