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I have two intersecting planes and two arbitrary points $\mathbf{p}_1$ and $ \mathbf{p}_2$, one on each plane. I would like to calculate the minimum distance of a path from one point to the other with the path constrained to the planes. This is my current method for calculating the minimum distance:

Find an equation for the intersection of the two planes, let's call it $\mathbf{\ell}(t)$. Then by minimizing the following distance function I should get the minimum distance:

$\| \mathbf{\ell}(s)-\mathbf{p}_1 \| + \|\mathbf{\ell}(s)-\mathbf{p}_2 \|$.

We differentiate to get

$\frac{(\mathbf{\ell}(s) - \mathbf{p}_1) \cdot \nabla \mathbf{\ell}}{\| \mathbf{\ell}(s) - \mathbf{p}_1 \|}+\frac{(\mathbf{\ell}(s) - \mathbf{p}_2) \cdot \nabla \mathbf{\ell}}{\| \mathbf{\ell}(s) - \mathbf{p}_2 \|}=0$

I can solve this equation; however, the formula is rather unruly. I was wondering if there is a simple solution to this or a better approach. For instance is the shortest path the projection of the straight line (in $\mathbb{R}^3$) between $\mathbf{p}_1$ and $\mathbf{p}_2$ onto the two planes? Thanks in advance.

John
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1 Answers1

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If $l$ is intersecting line then $$ h_i:= d(l,p_i)=d(q_i,p_i),\ q_i\in l $$ where $d=|\ \ \ |$ is a canonical metric on ${\bf R}^3$

Then $$ d_P(p_1,p_2)=\sqrt{d(q_1,q_2)^2+ (h_1+h_2)^2 }$$ where $d_P$ is a metric induced from path on planes

HK Lee
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  • This looks great. Could you explain why $d_p(p_1,p_2)$ is the shortest distance between paths? – John Feb 27 '16 at 05:50
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    Note that $d_P$ is metric induced from length of path so that it is independent of angle between two plane That is we consider when the angle is $\pi$ That is two planes are equal So problem is to find a distance between two points in a plane – HK Lee Feb 27 '16 at 05:55
  • I see. Thanks a lot. – John Feb 27 '16 at 06:28