# Can't find the perpendicular vector of a hyperplane

I have been reading this discussion here, and I tried to follow along the steps of the top response.

So, I chose $$4x_1+x_2-3=0$$ as my hyperplane. This means that the weight vector $$w=(4,1)$$ is perpendicular to the hyperplane, right? But, when I plot $$x_2=-4x_1+3$$, I don't see how $$w$$ is perpendicular at all.

Any thoughts?

The vector $$w$$ is perpendicular to vectors parallel to the hyperplane, not points in the hyperplane. For example $$(0,3)$$ and $$(1,-1)$$ are both points in the hyperplane. $$u=(0,3)-(1,-1)=(-1,4)$$ is a vector parallel to the hyperplane and indeed $$\langle u,w\rangle=0$$.
• Simply said, the "tip" of the vector $w$ doesn't necessarily have to sit on the hyperplane? Jul 24 '19 at 9:49
• Definitely not, vectors are always written like they are starting from the origin. If you shift the hyperplane such that it includes the origin, i.e. $4x_{1}+x_{2}=0$ and draw the picture then it might look more intuitive. Jul 24 '19 at 9:53