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).
 A: 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?
A: Without using dimensionality arguments:
Suppose $(a,b)$ and $(c,d)$ are both orthogonal to $(e,f)\ne (0,0)$.
Then, from the definition of orthogonality: $$ ae+bf  =0 $$ and $$ ce+df =0 .$$
If $e=0$, we must have $f\ne 0$, which implies $b=d=0$. Thus,  $(a,b)=(a,0)$ and $(c,d)=(c,0)$ are scalar multiples of each other.
If $e\ne 0$, then, from the above system 
$$
 a=-\textstyle{ f\over e}\, b  \quad \text{and}\quad c=-{f\over e}\,d;
$$
whence, 
$$(a,b)=(\textstyle{- f\over e} \thinspace b, b)=b(\textstyle{-f\over e}, 1)$$and$$(c,d)=(\textstyle{- f\over e} \thinspace d, d)=d(\textstyle{-f\over e},1).$$
This implies that $(a,b)$ and $(c,d)$ are scalar multiples of each other.
