12
$\begingroup$

I have two parallel hyper planes $$a^Tx=b_1,a^Tx=b_2$$ where $a \in \mathbb{R}^n, x \in \mathbb{R}^n ,b \in \mathbb{R}$ and I want to find the distance between the two. I have read that the distance between the two hyperplanes is also the distance between the two points $x_1$ and $x_2$ where the hyperplane intersects the line through the origin and parallel to the normal vector $\vec a$. These points are given by $$x_1=\frac{b_1}{\|a\|^2_2}a$$ and $$x_2=\frac{b_2}{\|a\|^2_2}a$$ Then the distance is $|x_1-x_2|$ but I don't really understand how we got $x_1$ and $x_2$.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ what is $\alpha$? $\endgroup$ – uniquesolution Oct 17 '15 at 10:42
  • $\begingroup$ So what you have is an equation of a hyperplane in $\mathbb{R}^n$, that is, a translation of an $n-1$-dimensional subspace. There are lots of solutions. $\endgroup$ – uniquesolution Oct 17 '15 at 10:47
  • $\begingroup$ I have not idea what you trying to do. You keep adding new information every time. You should take more care in formulating your problems. $\endgroup$ – uniquesolution Oct 17 '15 at 11:07
  • $\begingroup$ @Marc van Leeuwen. Is there any difference between distance and the shortest distance between the two hyperplanes? $\endgroup$ – holala Nov 28 '20 at 3:41
17
$\begingroup$

Let $x_1$ be any point in the first hyperplane and consider the line $L$ that passes through $x_1$ in the direction of the normal vector $a$. An equation for $L$ is given by $x_1 + at$ for all $t\in\mathbb{R}$. Now find the intersection of $L$ and the second hyperplane:

$$ a^T(x_1 + a t) = b_2 \iff t = (b_2 - a^T x_1)/a^Ta = (b_2 - b_1)/a^T a $$

Therefore the intersection point is $x_2 = x_1 + a(b_2 - b_1)/a^Ta$. The distance between these two points is the distance between the hyperplanes:

$$ \|x_1 - x_2\| = \frac{|b_2 - b_1|}{a^Ta}\|a\| = \frac{|b_2-b_1|}{\|a\|} $$

Improve this answer
$\endgroup$
  • $\begingroup$ Just a small technical note, I think you meant to write $\vert \vert a \vert \vert^2$, since you need to take the square root to obtain the norm out of an inner product.... $\endgroup$ – Louis May 10 '19 at 20:42
  • 1
    $\begingroup$ Hi Louis, in this case $a^Ta = \|a\|^2$. $\endgroup$ – K. Miller May 13 '19 at 19:21
  • $\begingroup$ @K.Miller Is there any difference between distance and the shortest distance between the two hyperplanes? $\endgroup$ – holala Nov 28 '20 at 3:42
  • $\begingroup$ @holala Usually when one speaks about the distance between two geometric objects it is the shortest distance. $\endgroup$ – K. Miller Nov 28 '20 at 16:44
0
$\begingroup$

To try an other angle : $x_1 \in \mathbb R^n, a \in \mathbb R^n $

$$\left\| x_1 - a\right\| = \frac {b_1}{\left\| a \right\|} $$

If we consider a as the origin of $\mathbb R^n$ seen as an orthogonal plan we should be able to write $$ x_1 = \frac {b_1}{\left\| a \right\|} . \frac {a}{\left\| a \right\|} = \frac {b_1}{\left\| a \right\|^2}.a $$ and then your $x_1$.

Improve this answer
$\endgroup$
  • $\begingroup$ @Mc Cheng. Is there any difference between distance and the shortest distance between the two hyperplanes? $\endgroup$ – holala Nov 28 '20 at 3:43
-1
$\begingroup$

Points $x_1$ and $x_2$ are the images of the coordinate system's origin in an orthogonal projection on the two given planes.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.