# Intuitive Way of Thinking about $\boldsymbol{w}^T\boldsymbol{x} + b = 0$.

I can understand $y = ax + b$ as $x$ represents how much you go up and to the right per increment of $b$, $b$ represents the $y$-intercept (due to solving for $y$ given $x = 0$ yielding $b$). However, I cannot understand the $\boldsymbol{w}^T\boldsymbol{x} + b = 0$. The math checks out, given that $x$ is $(x, y)$ and $w$ stores the coefficients with some modifications. Can someone offer an intuitive explanation of how a hyperplane can be described by it's perpendicular vector, and why multiplying x by a perpendicular vector will yield a point on a line that is perpendicular to the vector it was inner-product-ed with?

## 1 Answer

Let us refer to the normal vector as $\boldsymbol{w}$. Additionally, we define $\boldsymbol{a}$ to be an arbitrary constant point in the hyperplane and $\boldsymbol{x}$ to be an additional arbitrary variable point in the plane.

If we set ut the vector from $\boldsymbol{a}$ to $\boldsymbol{x}$ (both points in the plane) by $\boldsymbol{x}-\boldsymbol{a}$, then we know that this vector (or any Vector inside the hyperplane) is orthogonal to the normal vector. Orthogonality can be checked by the dot product. The dot product of orthogonal vectors is equal to zero.

Hence we can write (by using the dot product in vector notation)

$$\boldsymbol{w}^T\left[\boldsymbol{x}- \boldsymbol{a}\right]=0 \qquad \text{or} \qquad \boldsymbol{w}^T\boldsymbol{x}-\boldsymbol{w}^T\boldsymbol{a}=0 .$$

In the last expression substituting $b=-\boldsymbol{w}^T\boldsymbol{a}$ results in your representation.

• If we are using w as only a vector to determine all perpendicular vectors to itself (by taking the dot product of w with all possible values of x), then the magnitude of w does not matter, correct? Rather, it is the ratio between its y and x component that determines the slope of the hyper-plane. In addition, B would be a scalar in both forms (y-intercept form, this form) that determines the y-intercept of the hyperplane, correct? If both of my thoughts are confirmed correct, I'll have understood your answer, and I will accept it. – Mario Ishac Jun 1 '18 at 1:05
• The magnitude of $\boldsymbol{w}$ is not relevant. Often it is scaled in such a way that it is equal to the unit normal. The value of $b$ is a measure of the distance to the origin. If the unit normal vector is used then $b$ is exactly equal to the distance to the origin. This case is sometimes called the Hessian normal form of the plane. – MrYouMath Jun 1 '18 at 8:43