How to find expression for where vector intersects a hyperplane?

Question

I have a tutor who write this on the board:

"vector $r$ intersects hyperplane $r^Tx=c$ at the point $\frac{c}{\vert\vert r \vert\vert^2}r$

I am not able to figure out how this is the case.

Somethings I do understand are that the vector $r$ is orthogonal to the $x$ vectors satisfying the equation $r^Tx=c$. But this does not seem to help me derive the expression $\frac{c}{\vert\vert r \vert\vert^2}r$ ?

One thing that wud help me is if I can find the expression for the line expressed by the vector $r$. I know that the line goes through the origin, and is in the direction of $r$, but how do I express the line generally, without a vector equation?

If I knew the equation for that line, then obviously I could just solve thoes two equations simultaneously: the equation for that line, and the equation for the hyperplane $r^Tx=c$. That would be my approach if I could get an equation for that line.

Can someones help me with this?

Answer 1

That particular point in the direction of $r$ can be expressed as $x=kr$.

Hence $$r^T(kr) = c$$ $$k(r^Tr)=c$$

$$k = \frac{c}{\|r\|^2}$$

Hence the point of interest is $x= \frac{c}{\|r\|^2}r$.

Answer 2

What the equation is saying is that $\mathbf{x}$ is on the plane if the projection of $\mathbf{x}$ on the normal vector $\mathbf{r}$ has the length $c$. I.e. if the part of $\mathbf{x}$ that is parallel to $\mathbf{r}$ has the length $c$.

So let $\mathbf{x} = \lambda \frac{\mathbf{r}}{||r||} $ be a vector of length $\lambda$ into the direction of $\mathbf{r}$. Then you will find that $\lambda = \frac{c}{||r||}$. I.e. $\mathbf{x} = \mathbf{r} \frac{c}{||r||^2}$. In other words the plane defined by $\mathbf{r}^T \mathbf{x} = c$ intersects its normal vector $\mathbf{r} $ at $\mathbf{r} \frac{c}{||r||^2}$.