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).
Best Answer
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.