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

  • 1
    $\begingroup$ 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. $\endgroup$
    – Daryl
    Oct 21, 2015 at 20:57

1 Answer 1


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

enter image description here

  • $\begingroup$ why do you set b equal to -n dot x_0 ? Can you give a spatial explanation? $\endgroup$
    – bernie2436
    Oct 21, 2015 at 21:06
  • $\begingroup$ In fact $ - {\bf{n}}.{{\bf{x}}_0}$ is set to ${\bf{b}}$ just for simplifying notations and using some abbreviation. That's all. $\endgroup$ Oct 21, 2015 at 21:11
  • $\begingroup$ see the added picture. :) $\endgroup$ Oct 21, 2015 at 21:35
  • $\begingroup$ so you can think of X_0 going to a special point. There could be infinite planes pivoting from the tip of X_0 in that special point (by rotating/twising the plane in any direction). However, the dot product between the normal and X_0 must equal some value, b. I'm almost there I think -- but why negative b? $\endgroup$
    – bernie2436
    Oct 21, 2015 at 22:06
  • $\begingroup$ Take a look at the answer again $\endgroup$ Oct 21, 2015 at 22:19

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