If two vectors are orthogonal to a non-zero vector in $\mathbb{R}^2$ then are the two vectors scalar multiples of another? If two vectors $\bf{u}$ and $\bf{v}$ in $\mathbb{R}^2$ are orthogonal to a non-zero vector $\bf{w}$ in $\mathbb{R}^2$, then are $\bf{u}$ and $\bf{v}$ scalar multiples of one another? Prove your claim.
Attempt: From a geometric point of view it seems obvious that they must be scalar multiples of one another but I am having difficulties trying to prove it.
My approach was to use the Cauchy-Schwarz Inequality by assuming $|\bf{u}\cdot \bf{v}| < ||\bf{u}|| ||\bf{v}|| $ and somehow reaching a contradiction but I can't seem to obtain one. Maybe I need to try a different approach? It would be great (if possible) if someone can continue using my approach or show that it won't work (Assuming my answer is correct in the first place).

If either $\mathbf{u}$ or $\mathbf{v}$ are zero, then it is a scalar multiple of the other and you are done. So you may assume that $\mathbf{u}\neq\mathbf{0}$ and $\mathbf{v}\neq\mathbf{0}$.
Note that $\mathbf{u}$ and $\mathbf{w}$ are linearly independent. Hence they span $\mathbb{R}^2$, so we can write $\mathbf{v}=\alpha\mathbf{u}+\beta\mathbf{w}$. So... what is $\langle \mathbf{v},\mathbf{w}\rangle$ going to be, then, and what should it be?
