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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).

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    Whatever method you use, it *must* somehow and very intrinsically use the fact that you are working in a 2-dimensional space, since the claim is false in 3 or more dimensions (just take three elements of any orthonormal basis); but I don't see how one could try to leverage the dimension into the Cauchy-Schwarz Inequality, so I would have little hope of being able to solve it along your approach (doesn't mean it can't be done, just that I can't see how *I* would make it work...)2012-02-09

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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.

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    This is most likely the solution (and probably the only possible one without the use of dimension and basis) the book is after (this section of the book doesn't have answers). Thank you for your contribution.2012-02-09
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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?

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    @user22678: Just that I can't think of any way to use the abstract Cauchy-Schwarz inequality (which is applicable and identical in a very large number of settings) in a way that takes into account what dimension we're in. Again, *I* can't think of a way, doesn't mean it's impossible.2012-02-09