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enter image description here In the figure, I tried to indicate a straight line as a hyperplane which is denoted by pi. And the equation of the hyperplane is w^t.x = 0. Here hyperplane is passing through the origin point.

Could anyone help to prove how the minimum distance d between the point and the plane is?

d = w^t.P / ||w||

What I tried is written below:

I got from a tutorial that vector w = w^t. I am not sure how both would be equal to each other. Please help me here, How they are equal?

Now, if w = w^t, then we can write

w^t.P = ||w||.||P||.cos(theta)

=> ||P|| = w^t.P / ||w||.cos(theta)

As ||P|| = d then, => d = w^t.P / ||w|| cos(theta)

cos(theta) remain in the equation. Which I don't know how to eliminate.

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  • $\begingroup$ I assume you mean the minimum distance between a point and a plane? $\endgroup$
    – user277126
    Jan 11, 2022 at 11:40
  • $\begingroup$ Yes, I also edited my question. Thank you for pointing out $\endgroup$
    – F.C. Akhi
    Jan 11, 2022 at 13:16

3 Answers 3

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Let's denote a vector from the origin $0$ to the point $P$ as $p$, i.e. $p = \overrightarrow{0P}$. Let's denote as $P_{\pi}$ the projection (point) of $P$ onto the plane $\Pi$ (and $p_{\pi}$ it's corresponding vector from the origin).

What do we know?

  1. The vector from $P_{\pi}$ to $P$ is parallel to $w^T$ so $p - p_{\pi} = \lambda w^T \iff p_{\pi} = p - \lambda w^T$ for some constant $\lambda$. Note: if $|| w^T || =1$ then $\lambda$ is your distance $d$.
  2. $p_{\pi}$ is orthogonal to $w^T$ ($w^T$ is a vector normal to the plane $\Pi$). This means: $$ \langle p_{\pi}, \lambda w^T \rangle = 0 $$
  3. You can assume the inner product to be the dot product, i.e. $\langle a, b \rangle = a^Tb $

Use (1) and (2) to find $\lambda$. Assuming $|| w^T || = 1$ it is your answer. Otherwise: $$ d = || \lambda w^T || = \lambda || w^T || $$

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  • $\begingroup$ Could you please make me understand in a more easy way maybe with an image? $\endgroup$
    – F.C. Akhi
    Jan 11, 2022 at 13:21
  • $\begingroup$ What are you struggling to understand? Can you identify vectors $p, p_{\pi}, \lambda w^T$ in your image? If so - you should be able to see which vectors are parallel/perpendicular to each other (point 1). If you can visualize the vectors - do you understand the concept of an inner/dot products? If so - you should know that for two vectors $a, b$, if they are orthogonal (perpendicular) then $a^Tb = 0$ (points 2, 3) $\endgroup$ Jan 11, 2022 at 13:59
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    $\begingroup$ At point 1 why p−pπ = λw^T. Why are you writing constant terms here? And the pπ term is along with the plane pi, right? $\endgroup$
    – F.C. Akhi
    Jan 11, 2022 at 15:18
  • $\begingroup$ You know that $w^T$ is perpendicular (normal) to the plane so it's parallel to $p - p_{\pi}$. But you don't know if length of $w^T$ is exactly equal to the length of $p - p_{\pi}$ (blue dotted line in your image). So you need to account for a scaling factor (here - $\lambda$). $\endgroup$ Jan 11, 2022 at 15:43
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    $\begingroup$ Thank you. I understood your explanation now $\endgroup$
    – F.C. Akhi
    Jan 11, 2022 at 16:32
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we can write

w^t.P = ||w||.||P||.cos(theta)

...

cos(theta) remain in the equation. Which I don't know how to eliminate.

You know that $\Vert P \Vert \text{cos}(\theta) = d$ because the the three points (origin, the point $P$ and the projection of $P$) make a right-angled triangle.

(this is a bit indirect argument, you will have to prove that the shortest vector from $P$ to the plane, is perpendicular to the plane)

$$w^t \cdot P = \Vert w \Vert \underbrace{\Vert P \Vert \text{cos}(\theta)}_{=d} = \Vert w \Vert d $$

Btw, the image below (from this question here, the same as yours but a 2d hyperplane) might show the situation more clearly. It depicts the cases for multiple points $P$ and each time the shortest vector from P to the plane points into the same direction and is perpendicular to the plane.

illustration for a small sample size

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  • $\begingroup$ vector w and its transpose directs in the same direction, that's why they are equal vector? $\endgroup$
    – F.C. Akhi
    Jan 11, 2022 at 14:37
  • $\begingroup$ Thank you for the answer. I could do prove by myself after you point the value of d = ||P||.cos(theta) $\endgroup$
    – F.C. Akhi
    Jan 11, 2022 at 16:30
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enter image description here I solved the equation and add an image with the solution. I hope people who were answering my question could give a glance, whether I did any mistake or not.

Thank you

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