On an example of an idempotent which is not a projection I am working through a solution of the following exercise in Conway's functional analysis:

Let $\mathscr{H}$ be the two-dimensional real Hilbert space $\mathbb{R}^2$, $\mathscr{M}=\{(x,0):x\in\mathbb{R}\}\subseteq\mathscr{H}$, and $\mathscr{N}=\{(x,x\tan{\theta}):x\in\mathbb{R}\}\subseteq\mathscr{H}$, where $0<\theta<\frac{\pi}{2}$ is fixed.  Find a formula for the idempotent $E_\theta$ with range $\mathscr{M}$ and kernel $\mathscr{N}$.  Also show that $\|E_\theta\|=(\sin{\theta})^{-1}$.

I am very confused with this question. I will write my try can you please help me why I can't get the solution?
My questions :
1- $Ε_\theta$ is an operator s.t. it transforms all points of $\mathbb R^2$ to the $x$-axis. So intuitively $Ε^2_\theta=Ε_\theta$. But how one write an explicit formula for $Ε_\theta$. Isn't it $Ε_\theta (x,y)=(x,0)$ and if so it doesn't depend on $\theta$ then?
2- $\ker (Ε_\theta)$ is the set of point s.t. $Ε_\theta(x,y)=(0,0)$ then it must be the $y$-axis, then why it is $\{(x,x \tan\theta): x \in \mathbb R\}$?
3- how $\|Ε_\theta\|=(\sin \theta)^{-1}$? Calculating  $\|Ε_\theta\|=\sup Ε_\theta$ depends on two things to know before that : is sup taken on the circle in $\mathbb R^2$? what is explicit formula for $Ε_\theta$ so that to know on what supremum is taken?
4- this exercise is designed to indicate an example of an idempotent which is not a projection? $\ker E$ is the $y$-axis and $\operatorname{ran} E$ is the $x$-axis so how $\ker E = (\operatorname{ran} E)^{\perp}$ does not hold when it holds?
 A: The operator is $E((x, y) ) = ( x - \dfrac{y}{\tan \theta} , 0 ) $
Its kernel is the set $ \{(x, y) | x - \dfrac{y}{\tan \theta} = 0 \}$ which is exactly the
set $ \{ (x, x \tan \theta) \} $
Thus it follows that $ ker E \ne (ran E)^\perp $
The norm of the operator is $\sup  | (\cos \phi - \dfrac{\sin \phi}{\tan \theta } | $
which is $ \sqrt{ 1 + \dfrac{1}{\tan^2 \theta} } = \dfrac{1}{\sin \theta} $
A: Not a direct answer, but some background that's likely to be useful:
Let $u$ and $v$ be non-zero vectors in a Euclidean space, viewed as column matrices, and let $v^{t}$ denote the transpose of $v$.
The inner product $v^{t}u = u^{t}v$ is the dot product $u \cdot v$ viewed as a $1 \times 1$ matrix.
The rank-one matrix $uv^{t}$, also known as the the outer product of $u$ and $v$, has kernel the set of vectors orthogonal to $v$ and image the line spanned by $u$. Further, if $u \cdot v = 1$, then $uv^{t}$ is idempotent.
In case some exercises help:

*

*Show the image of $uv^{t}$ is the set of scalar multiples of $u$, and the kernel is the set of vectors orthogonal to $v$.


 For every vector $w$, we have $(uv^{t})w = u(v^{t}w) = (v \cdot w)u$, which is a scalar multiple of $u$. Further, this product is $0$ if and only if $v \cdot w = 0$, if and only if $w \perp v$.



*Show $uv^{t}$ is idempotent if $u \cdot v = 1$.


 We have $(uv^{t})^{2} = (uv^{t})(uv^{t}) = u(v^{t}u)v = (u \cdot v)uv^{t} = uv^{t}$.



*Find the standard matrix of $E_{\theta}$. (Geometrically, $E_{\theta}$ is a "slant projection", like casting a shadow on the $x$-axis when the sun is not directly overhead.)


 Hint: Take $u = (1, 0)$ and $v = (1, -\cot\theta) = \frac{1}{\sin\theta}(\sin\theta, -\cos\theta)$, noting that $u \cdot v = 1$ and $(\sin\theta, -\cos\theta) \cdot(\cos\theta, \sin\theta) = 0$.

