1

I am currently working through a textbook, and I am having some problem with the following problem:

Define a hyperplane to be an $(n-1)$-plane of $E^n$. Prove that $P$ is a hyperplane if and only if $$P = \{x \in E^n \: | \: a \cdot (x-x_0) = 0 \}$$ for a unit vector $a$ (unique up to sign) and $x_0 \in P$.

Recall that an $(n-1)$-plane of $E^n$ is a coset of an $(n-1)$-dimensional vector subspace $V$ of $E^n$. I tried to follow my intuition and set $a = b/|b|$ where $P = b + V$, but to no avail.

Andrew Thompson
  • 4,431

1 Answers1

1

Hmm, looks like your intuition needs some more work. Since $b$ is totally unrelated to the normal vector, this looks like a stab in the dark. The geometric picture here is actually pretty easy to describe.

If $P=b+V$, then to move $P$ to the origin, we'd need to translate by subtracting $b$ from all its points, so $x_0=b$.

Now you are dealing with an $n-1$ dimensional subspace through the origin. Its orthogonal complement is $1$ dimensional, and any normal lies in that complement. So, there are two normalized choices for unit vectors in that complement. You can pick either one and call it $a$.

So there you are: the original hyperplane has been described as $ \{x \in E^n \: | \: a \cdot (x-x_0) = 0 \}$.

rschwieb
  • 153,510