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I have to show the following:

If $v, w \in \mathbb{R}^3 \setminus \{(0,0,0)\}$ such that the set of vectors orthogonal to both of them is a plane through the origin, then each is a scalar multiple of the other.

I've proved that if $u^\perp = \{x \in \mathbb{R}^3 : x \perp u\}=\{x \in \mathbb{R}^3 : \langle x , u \rangle=0\}$, the set of all vectors which are ortogonal to $u$, then $v^\perp=w^\perp$ [since $v^\perp \cap w^\perp$ is a plane through origin (given) then dim$(v^\perp \cap w^\perp)$=dim$(v^\perp)$=dim$(w^\perp)$=2 and $v^\perp \cap w^\perp \subseteq v^\perp$,$v^\perp \cap w^\perp \subseteq w^\perp$]. Will this be of any help? I can't go any further. Many thanks.

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The answer to this question is evident from the uniqueness of dot product. Since the dot product of orthogonal vectors is zero, and we have the two vectors having dot product zero with SAME PLANE. Hence, they ought to be scalar multiples of each other. –  Nikhil Bellarykar Dec 15 '11 at 17:54
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2 Answers

up vote 2 down vote accepted

Take an orthogonal basis of the plane $\{{\bf v}_1,{\bf v}_2\}$. The set $\{{\bf v},{\bf v}_1,{\bf v}_2\}$ must be independent.

Indeed if $c_1{\bf v} +c_2{\bf v}_1+c_3 {\bf v}_2=\bf 0$, then $${\bf v}_1\cdot ( c_1{\bf v} +c_2{\bf v}_1+c_3 {\bf v}_2)=0 \quad\Rightarrow \quad c_2=0 $$ and $${\bf v}_2\cdot ( c_1{\bf v} +c_2{\bf v}_1+c_3 {\bf v}_2)=0 \quad\Rightarrow \quad c_3=0. $$

we then must have $c_1=0$.

So, $\{{\bf v},{\bf v}_1,{\bf v}_2\}$ is an independent set and thus a basis of $\Bbb R^3$.

Now write $\bf u$ in terms of this basis. We have for some constants $c_1$, $c_2$, and $c_3$

$${\bf u}=c_1{\bf v}+c_2 {\bf v}_1+c_3 {\bf v}_2.$$

Now

$$0={\bf v}_1\cdot {\bf u}= c_1 {\bf v}_1\cdot{\bf v} +c_2 {\bf v}_1\cdot {\bf v}_1+c_3 {\bf v}_1\cdot {\bf v}_2=c_2 ||{\bf v}_1||^2;$$

thus, $c_2=0$. Similarly, taking dot products with ${\bf v}_2$, it follows that $c_3=0$. Thus ${\bf u}=c_1{\bf v} $.

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Hint: For finite dimensional vector spaces, if you have subspaces $A,B$ such that $\text{dim}(A) = \text{dim}(B)$ and $A \subseteq B$, then $A = B$.

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I don't really understand your hint. Are you pointing towards proving $\mathbb{R}v=\mathbb{R}w$? In that case how shall I show either $\mathbb{R}v \subseteq \mathbb{R}w$ or $\mathbb{R}w \subseteq \mathbb{R}v$? I can't explicitly find any non-zero point common in them. –  Sayantan Dec 15 '11 at 17:36
    
Using what you already have above, my hint gets you that $v^\perp = w^\perp$. Which implies that $\text{span}(v)=\text{span}(w)$. –  Brandon Carter Dec 15 '11 at 17:45
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