# Understanding why the hyperplane is perpendicular to the vector

I am following along Stephen Boyd's lectures on convex optimization and am having trouble understanding the diagram in this screenshot.

I have read through a few answers such as this one and this one.

My question is I am having trouble understanding why $$a^Tx = b$$ implies that a and x are perpendicular.

One way I understood it, following this post's answer is $$a^Tx = b \implies a^T x - a^T x_b = 0$$ for some $$x_b$$ and so $$a^T (x - x_b) = 0$$. I understand why $$a$$ will be perpendicular to $$x - x_b$$ but am having trouble seeing why $$a$$ will be perpendicular to $$x$$.

One way I think about it is $$x_b$$ is a particular solution to the equation $$a^Tx = b$$ and since this is a linear program, any $$a' = ka$$ and $$x' = x_b/k$$ for non-zero $$k$$ will solve for $$a'^T x' = b$$. Hence, $$x_b$$ is just one of the solutions, and the other solutions are of the form $$x_b/ k$$, for non-zero $$k$$. Note, the other solutions are all parallel to $$x_b$$ and I think the linear combination of parallel lines is parallel to any of of them.

Hence, since $$x$$ and $$x_b$$ are parallel, $$x - x_b$$ is parallel to $$x$$ and so $$a^T$$ which is perpendicular to $$x - x_b$$ is perpendicular to $$x$$.

However, I'm having trouble seeing this for fixed $$a$$. In particular, is $$a$$ fixed?

The diagram in the screenshot is not showing that $$a$$ and $$x$$ are perpendicular. Indeed, by definition two vectors are perpendicular (aka orthogonal) iff their dot product is zero. Rather, it is showing that if $$x_0$$ satisfies $$a^\top x_0 =b$$, then every element $$y$$ in the hyperplane $$H = \{y: a^\top y = b\}$$ may be written in the form $$y = x_0 + x$$, where $$a^\top x = 0$$.