Question: If $k \geq 2$ and $x \in R^k$ ,prove that there exists $y \in R^k$ such that $y \neq 0$ but $x \cdot y = 0$. Is this also true if k = 1?
Proof:
$$|x+y|^2 = (x \cdot y)(x\cdot y) = |x|^2 + |y|^2 +2(x\cdot y)$$
$$\text{As} \space (x \cdot y)=0$$
$$\text{Then} \space|x|^2 + |y|^2 =|x+y|^2$$
So any $x$ and $y$ that fulfils this this equation will form a right triangle and $x \cdot y = 0$
For the case where $k=1$, suppose $x$ and $y$ are non zero, then their dot product is $xy$, which must be nonzero as $R^1$ is an ordered field.
Does this suffice as a proof? Is it missing anything?
You did not complete the proof of the first part. You cannot end the proof by saying 'any $x$ that fulfils....'.
When $y=0$ you can take $x=(1,1,..,1)$. For $y \neq 0$ there is at least one vector $z$ linearly independent of $y$ because $k \geq 2$. Now take $x=z-ay$ where $a=\frac {y.z} {\|y\|^{2}}$. Then $x.y=0$ and $x \neq 0$.