# Intuitive understanding of the -b in this definition of a plane

I am learning about SVMs in computer science. The book I'm reading defines a hyperplane with an "intercept term", b

$\vec{w}^T \vec{x}= -b$

What does this intercept mean, intuitively? From Khan Academy, I understand how the dot product of a vector on a plane, $\vec{a}$ and a normal vector to that plane $\vec{n}$, i.e. I see how $\vec{a}\cdot\vec{n}^T=0$ would define a plane. (because the normal dotted with any vector on the plane is zero).

But what is the $-b$, intuitively, in $\vec{w}^T \vec{x}= -b$?

• Hint: A plane that doesn't pass through the origin is written $n\cdot (x-p)=0$, where $p$ is any point on the plane. Oct 21, 2015 at 20:57

That's not too complicated. Let us imagine $n=3$. Suppose we are looking for plane whose unit normal is ${\bf{n}}$ and passes through the point ${{{\bf{x}}_0}}$. Now the equation of all point lying on this plane will be
\eqalign{ & {\bf{n}}.\left( {{\bf{x}} - {{\bf{x}}_0}} \right) = 0 \cr & {\bf{n}}.{\bf{x}} - {\bf{n}}.{{\bf{x}}_0} = 0 \cr & \left\{ \matrix{ {\bf{n}}.{\bf{x}} + b = 0 \hfill \cr b = - {\bf{n}}.{{\bf{x}}_0} \hfill \cr} \right. \cr}
and hence your $b$ is minus of the dot product of unit normal and position vector of a special point on the plane. I emphasize that it is just a matter of notation and nothing more. See the below picture
• In fact $- {\bf{n}}.{{\bf{x}}_0}$ is set to ${\bf{b}}$ just for simplifying notations and using some abbreviation. That's all. Oct 21, 2015 at 21:11