How to show that if $x, y, z\in\mathbb R^n$ and $\langle x, z\rangle=\langle y, z\rangle=0\implies x=\lambda y$?

Original Question. Let $\mathbb R^n$ be endowed with the standard inner product

$$\langle x, y\rangle=x_1 y_1+\dots + x_ny_n$$

where $x=(x_1, \ldots, x_n)$ and $y=(y_1, \ldots, y_n)$. How to show that if $x, y, z\in \mathbb R^n$ are such that $z\neq 0$ and

$$\langle x, z\rangle=\langle y, z\rangle=0$$ then there is $\lambda\in\mathbb R$ such that $x=\lambda y$?

As pointed out in the answers, the previous assertion is not valid, so I'll propose a related problem.

Updated Question. If $x, y\in \mathbb R^n$ are such that

$$\forall z\in W, \langle x, z\rangle=0\implies \langle y, z\rangle=0$$

then there is $\lambda\in\mathbb R$ such that $$y=\lambda x?$$ That is, if $y$ is orthogonal to every vector which is orthogonal to $x$ then $y$ is a multiple of $x$?

Solution 1:

We don't know that (unless $n=2$). For instance, if $x=(1,0,0)$, $y=(0,1,0)$, and $z=(0,0,1)$, then (with respect to the usual inner product) $\langle x,z\rangle=\langle y,z\rangle=0$, but there is no $\lambda\in\Bbb R$ such that $x=\lambda y$.