I was just watching this tutorial about Support Vector Machines, and I came to a halt because of the following problem.

Given that $\vec{w}$ is a vector perpendicular to a hyperplane separating two classes, does the equation $\vec{u} \cdot\vec{w}+b=1$ or ($\vec{u}\cdot\vec{w}+b=-1$) represent another hyperplane parallel to this one? Where $\vec{u}$ is a some vector.

That is, in the case of a 2D plane, does $\vec{u}\cdot\vec{w}+b=1$ represent a line, that is parallel to the original hyperplane?


1 Answer 1


Yes it does.

We can directly see the relation to the equation of a line segment $mx+b=y$.

Anyways, let's get the equation of the hyperplane given here. We're given that $\vec{w}$ is a perpendicular vector to it (from the origin). Let's assume that $\vec{x}$ is a vector that represents a point on the hyperplane given. Therefore, since $\vec{x}-\vec{w}$ is vector perpendicular to $\vec{w}$ therefore, it can be represented by $$(\vec{x}-\vec{w})\cdot\vec{w}=0$$

Now, consider a hyperplane that is parallel to the original one. Let that be represented by $\lambda\vec{w}$. So, it's equation would be $(\vec{x}-\lambda\vec{w})\cdot\lambda\vec{w}=0$. That is $$\vec{x}\cdot\vec{w}-\lambda|w|^2=0$$

As we can see, that if we change $b$ to $1-\lambda|w|^2$ (since $b$ is an arbitrary constant), then $\vec{x}\cdot\vec{w}+b=1$ changes to $\vec{x}\cdot\vec{w}-\lambda|w|^2=0$, which is an equation of an hyperplane parallel to the original one.


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