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$\def\vec{\overrightarrow}$The full question reads as following:

Prove that if two vectors $\vec{u}$ and $\vec{v}$ in $\mathbb{R}^2$ are orthogonal to a nonzero vector $\vec{w}$ in $\mathbb{R}^2$, then $\vec{u}$ and $\vec{v}$ are scalar multiples of each other.

The entirety of this chapter has been finding an arbitrary point provided we know a point. Naturally I would think that they imbed this theory in the exercises.

Initially I wanted to apply the formula for finding a point $x$, which is

$$x= x_{0} + su + tv.$$

However this is does't seem to get me anywhere.

Any hints to how I could prove this?

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  • $\begingroup$ The first thing I would say is that the theorem isn't true in$\mathbb R^3,$ since both the $x-$ and $y-$axes are orthogonal to the $z-$ axis, so it looks like you're going to have to use coordinates. At any rate, you have to use the dimension somehow. $\endgroup$
    – saulspatz
    Commented Feb 10, 2018 at 2:33

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We first take note that $w \neq 0.$ If either $u$ or $v$ equal $0$, they must be scalar multiples of each other.

We prove for when $u$ and $v$ are not zero vectors, they are scalar multiples of each other.

Let $u = \binom{u_1}{v_1}, v = \binom{v_1}{v_2},$ and $w = \binom{w_1}{w_2}.$ Since $u$ and $v$ are both orthogonal to $w,$ we have $u \cdot w = 0$ and $v \cdot w = 0.$

In other words,

$$u_1w_1 + u_2w_2 = 0$$ $$\text{and}$$ $$v_1w_1 + v_2w_2 = 0.$$

Rearranging both equations gives us $\frac{u_1}{u_2} = -\frac{w_2}{w_1} = \frac{v_1}{v_2}.$ Therefore, for a value $k \in \mathbb{R}$ we have $u_1 = kv_1$ and $u_2 = kv_2$.

$u$ and $v$ are scalar multiples of each other.

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  • $\begingroup$ so for $u_{1}$ k is equal to the reciprocal of $v_{2}$ multiplied with $u_{2}$? $\endgroup$ Commented Feb 10, 2018 at 2:45
  • $\begingroup$ For the case where $u \neq 0$ and $v \neq 0,$ we have $k = \frac{u_2}{v_2}.$ $\endgroup$ Commented Feb 10, 2018 at 2:49
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Hint:  if either $\,u\,$ or $v\,$ are $\,0\,$ then the proposition holds trivially. Otherwise, $\,(u, w)\,$ forms a basis of $\,\mathbb{R}^2\,$, so $\,v = a u + b w\,$ for some real $\,a,b \in \mathbb{R}\,$. Since $\,v \cdot w = 0$ $\require{cancel}\iff \cancel{a (u \cdot w)} + b (w \cdot w) = 0\,$ it follows that $\,b = 0\,$, so in the end $\,v = a u\,$.

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  • $\begingroup$ +1 for the diagonal strikethrough. I looked at the MathJax to see if I could learn to do that, but I'm afraid it's just one of those things that will be forever beyond me, like cohomology. $\endgroup$
    – saulspatz
    Commented Feb 10, 2018 at 2:41
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    $\begingroup$ @saulspatz Never vote on style alone, please. That said, just put \require{cancel} somewhere first, then do \cancel{a} $\,\cancel{a}\,$, or \bcancel{b} $\,\bcancel{b}\,$, or \xcancel{c} $\,\xcancel{c}\,$ happily ever after ;-) $\endgroup$
    – dxiv
    Commented Feb 10, 2018 at 2:46

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