This is a statement I came up on my own and it makes intuitive sense but I'm not sure how to go about proving it. (Also I'm sure this exists already but couldn't find it online for odd reasons.)
Claim: Let $v,x,y \in \mathbb{R^n}$ such that $v \cdot x=0$ and $v \cdot y=0$. Then prove that there exists a $c\in \mathbb{R}$ such that $x=cy$.
So of the claim above it makes intuitive sense to me because $x,y$ are both vectors that is perpendicular to the vector $v$ so which means $x,y$ must be a scalar multiple of each other. But I'm not sure how to prove it... Here is my attempt.
Proof: Let $v=(v_1,...,v_n),x=(x_1,...,x_n),y=(y_1,...,y_n)$
Suppose $v \cdot x=0$ and $v \cdot y=0$
Then by definition of dot product we have the following:
$v_1x_1+...+v_nx_n=0$ and $v_1y_1+...+v_ny_n=0$
And we also have $x=cy$ which is equivalent to $(x_1-cy_1,...,x_n-cy_n)=(0,...,0)$.
This is where I came stuck but I'm not even sure whether if the claim statement is correct or not... Anyways, thanks in advance!
This is false in dimension $3$ and higher because there are multiple perpendicular directions. So, for instance, $\mathbf{i} \cdot \mathbf{j}=0$ and $\mathbf{i} \cdot \mathbf{k}=0$ and yet there is no way to scale $\mathbf{j}$ into $\mathbf{k}$.
It is nearly true in $\mathbb{R}^2$, but you need to specify that none of the vectors are $\mathbf{0}$, otherwise you get infinitely many counterexamples. But in $\mathbb{R}^2$ if none of the vectors are $\mathbf{0}$ and $\mathbf{v} \cdot \mathbf{x} = 0 = \mathbf{v} \cdot \mathbf{y}$ then yes, $\mathbf{x}$ is a scalar multiple of $\mathbf{y}$. This is because being perpendicular to $\mathbf{v}$ can only occur in two directions (along the same line) and you have two possible signs (positive or negative) to attach to a scalar.