Perpendicular distance from a hyperplane Let the hyperplane equation be $\theta^Tx + \theta_0 = 0.$ Let p be any point. Find the signed perpendicular distance between the point and the hyperplane. (Answer in terms of $\theta^Tx$, $\theta_0$, and $p$; you may not answer using $x$)
My attempt: 
$$(p-x)*\theta^T*(\frac1{len\theta^T})$$ 
Then I distribute the $\theta^T$:
$$ p*\theta^T - x*\theta^T*(\frac1{len\theta^T})$$
The middle term can be re-written as $\theta_0$ from the equation we of the hyperplane.
We'll end up with $$ p*\theta^T + \theta_0*(\frac1{len\theta^T})$$
(Btw, $len\theta^T$ is the lengths of $\theta^T$)
 A: Geometrically $\theta$ is a normal vector to the hyperplane $H$. Then we must draw a line $l$ from $p$ in the direction of $\theta$ and compute the distance between $p$ and $q \in l\cap H$.
Every point in $l$ is written as  $p+ t\theta$ for some $t\in \mathbb{R}$. To find the point of intersection with $H$ we just set the equation
$$
0 = \theta^T( p + t\theta) + \theta_0 = \theta^T\cdot p + t\theta^T\cdot \theta + \theta_0
$$
Solving for $t$ we get  $t = \frac{-\theta_0 - \theta^T\cdot p}{\|\theta\|^2}$. Hence 
$$
q = p +  \frac{-\theta_0 - \theta^T\cdot p}{\|\theta\|^2}\theta
$$
Now we head to the signed distance. We have that 
$$(p-q)^T\cdot \frac{\theta}{\|\theta\|} = \frac{\theta_0 + \theta^T\cdot p}{\|\theta\|}$$ 
Note that
$$
\|q-p\| = \left\| \frac{-\theta_0 - \theta^T\cdot p}{\|\theta\|^2}\theta \right\| =  \frac{|\theta_0 + \theta^T\cdot p|}{\|\theta\|}
$$
which is the absolute value of the signed distance.
A: Hi I am showing with one example.
For finding the distance between Plane($\theta$ and $\theta_0$) and point $x$.
$D=\dfrac{||\theta x+\theta_0||}{||\theta||}$
Signed distance=$\dfrac{(\theta x+\theta_0)}{||\theta||}$
Reference:
http://mathworld.wolfram.com/Point-PlaneDistance.html
