Prove that two parallel hyperplanes in $R^2$ is intersected by a normal line

Question

Given two hyperplanes $\{x\mid a^Tx = b_1, a^Tx = b_2\}$, want to show that the hyerplanes are intersected by line normal to both planes.

Okay, so in $R^2$ the two lines are:

$a_1x_1 + a_2x_2 = b_1 \Rightarrow x_2 = b_1/a_2 + a_1/a_2x_1 $

$a_1x_1 + a_2x_2 = b_2 \Rightarrow x_2 = b_2/a_2 + a_1/a_2x_1 $

Very simple lines and assuming that all scalars are positive we have two downward slanted lines in say quadrant I

Then I claim that the vector perpendicular to each line is $a$, this gives us another line that is perpendicular to both of the lines, namely:

$x_2 = \dfrac{a_2}{a_1} x_1$

So the picture is this:

How can I show that the new line $x_2 = \dfrac{a_2}{a_1} x_1$ is indeed normal to both hyperplanes?

Answer 1

Take a vector $(1, -a_1/a_2)$ tangent to a hyperplane and vector $(1, a_2/a_1)$ contained in your normal line. Their inner product is $$ 1\cdot 1 + \frac{-a_1}{a_2}\cdot \frac{a_2}{a_1} = 0, $$ so they are orthogonal.

Answer 2

Two lines are perpendicular (orthogonal) if the slope of one is the inverse of the other. Therefore, the green line is perpendicular to the first red (b2) plane because its slope (a2/a1) is the inverse of red plane's slope (a1/a2). Since the planes are parallel, the green line will also be perpendicular to the second (b1) red plane.