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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?

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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 $\langle u,w\rangle=0$.

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  • $\begingroup$ Simply said, the "tip" of the vector $w$ doesn't necessarily have to sit on the hyperplane? $\endgroup$
    – zafirzarya
    Jul 24 '19 at 9:49
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    $\begingroup$ 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. $\endgroup$ Jul 24 '19 at 9:53

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