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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 $x_2$.

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    $\begingroup$ what is $\alpha$? $\endgroup$ Oct 17, 2015 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. $\endgroup$ Oct 17, 2015 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. $\endgroup$ Oct 17, 2015 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, 2020 at 3:41

4 Answers 4

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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:

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

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  • $\begingroup$ 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, 2019 at 20:42
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    $\begingroup$ Hi Louis, in this case $a^Ta = \|a\|^2$. $\endgroup$
    – K. Miller
    May 13, 2019 at 19:21
  • $\begingroup$ @K.Miller Is there any difference between distance and the shortest distance between the two hyperplanes? $\endgroup$
    – holala
    Nov 28, 2020 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, 2020 at 16:44
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If you take two arbitrary points $y_1, y_2$, one on each hyperplane and you consider the projection of their difference on the normalized normal vector $a$, you get $\frac{a \cdot (y_2-y_1)}{\|a\|} \frac{a}{\|a\|} = \frac{b_2-b_1}{\|a\|} \frac{a}{\|a\|} = \frac{b_2 }{\|a\|^2}a - \frac{b_1 }{\|a\|^2}a$, which correspond to your definition of $x_1, x_2$. To obtain the distance, you can take the norm on the second equality.

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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 $x_1$.

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  • $\begingroup$ @Mc Cheng. Is there any difference between distance and the shortest distance between the two hyperplanes? $\endgroup$
    – holala
    Nov 28, 2020 at 3:43
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Points $x_1$ and $x_2$ are the images of the coordinate system's origin in an orthogonal projection on the two given planes.

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