# Visualizing the equation for separating hyperplane

I was wondering if I can visualize with the example the fact that for all points $$x$$ on the separating hyperplane, the following equation holds true:

$$w^T.x+w_0=0\quad\quad\quad \text{... equation (1)}$$

Here, $$w$$ is a weight vector and $$w_0$$ is a bias term (perpendicular distance of the separating hyperplane from the origin) defining separating hyperplane. I was trying to visualize in 2D space. In 2D, the separating hyperplane is nothing but the decision boundary. So, I took following example: $$w=[1\quad 2], w_0=\Vert w\Vert=\sqrt{1^2+2^2}=\sqrt{5}$$ and $$x=[0\quad 2.5]$$. (Check at the bottom of the post, how I came up with these.)

But I don't find this example make above equation true: $$\color{red}{\begin{bmatrix} 1 \\2 \end{bmatrix}\begin{bmatrix} 0 & 2.5 \end{bmatrix}-\sqrt{5}=5-\sqrt{5}\neq 0}$$

However, I then realized that making $$w$$ a unit vector makes equation (1) true:

$$\begin{bmatrix} 1/\sqrt{5} \\2/\sqrt{5} \end{bmatrix}\begin{bmatrix} 0 & 2.5 \end{bmatrix}-\sqrt{5}=\sqrt{5}-\sqrt{5}= 0$$

So, I have following few related questions:

Q1. Does equation (1) applies "only to" unit weight vector? (In read some texts which scale this equation to make $$w$$ a unit vector) Is there any way to make equation (1) work for non-unit weight vector?

Q2. Are weight vectors always considered to be unit vectors? (that is even during actual implementation, do they turn out to be unit vector?)

### How I come up with the graph

First, assumed $$w=[1\quad 2]$$. The slope of the vector will be $$2$$. So to plot line passing through the vector and origin, I plotted $$y=2x$$. The slope of the line perpendicular to this line will be negative inverse of slope of this line, that is it will be $$-1/2$$. So to plot a line (separating plane or decision boundary) perpendicular to $$y=2x$$ and not passing through the origin, but through $$y=2.5$$, I plotted a line $$y=-\frac{1}{2}x+2.5$$.

Q1. The equation is still valid if $$\|w\|\neq1$$, but the interpretation of $$w_1$$ as the (signed) distance from the origin is not.
Q2. You haven't specified a learning algorithm, but for example with SVM, the popular libsvm formulates the problem(s) with $$w$$ not a unit vector, instead scaling so that $$\|w\|$$ gives the margin width. But also, quite often under the hood it solves the dual problem instead.