problem with construction of projection

Question

Let $E=\left\{(x,v): x\in\mathbb{R}^2, v\in \mathbb{R}\cdot\langle x\rangle \right\}$ and let $p:E\rightarrow \mathbb{R}^2$ be projection onto the first coordinate. Why for $x\neq 0$ the fiber $p^{-1}(x)$ is one-dimensional?

My question comes from the fact that I don't know what exatcly $\mathbb{R}\cdot\langle x\rangle$ means. If I take a scalar $\lambda\in\mathbb{R}$ and just multiply $\lambda\cdot\langle x\rangle$, I'll get something which already is in $\langle x\rangle$ so it has to be something different.

Answer 1

The notation that you don't understand means all (real) scalar multiples of the fixed vector $x$. So $E$ consists of pairs where first item is a vector from the plane and the second item is every possible scalar multiple of the first item. Then it is now clear $p^{-1}(x)$ is what it is claimed.

Answer 2

If I understand, let's write $p^{-1}(x)$: $$p^{-1}(x) = \{ v\in \mathbb{R}: p(x,v) = x \} $$

So, $p^{-1}(x)$ is one-dimensinal.