How do I calculate the distance between the line joining the two points on a spherical surface and another point on same surface? I have illustrated my problem in the image below.


In the above illustration, the points A, B and X lies on a spherical surface, I need to find the distance between points (A,B) and X. I am not a mathematics guy. If possible please illustrate me the solution as non-mathematics guys could understand. Thanks.

  • $\begingroup$ Does this line lie on the sphere or inside the sphere? $\endgroup$ – Eivind Feb 21 '11 at 15:56
  • $\begingroup$ geodesic distance on the sphere? $\endgroup$ – yoyo Feb 21 '11 at 16:19
  • $\begingroup$ Have you tried to set up the standard parametrized integral for the distance between X and an arbitrary point on the line, then vary the integral to find where it obtains it's minimum? $\endgroup$ – Matt Calhoun Feb 21 '11 at 16:27
  • $\begingroup$ How are A, B, and X specified? For example, by latitude and longitude? And is the distance along the sphere or a straight line through the sphere? $\endgroup$ – Ross Millikan Feb 21 '11 at 18:11
  • 1
    $\begingroup$ @Ross Millikan: the points A, B and X are in Latitudes and longitudes. the distance is along the sphere. $\endgroup$ – brainless Feb 23 '11 at 12:33

This assumes that everything is on the surface of the sphere. Furthermore I assume the sphere has radius $1$.

Change the coordinates so that $A$ and $B$ are both on the equator of the sphere. For definiteness, move $A$ to $(1,0,0)$ and move $B$ to $(\cos \theta, \sin \theta, 0)$ where $\theta$ is the angle between the vectors from the center of the sphere to $A$ and to $B$. This is a linear transformation.

Then what you care about is the latitude of $X$. If $X$ is in the sector of the sphere immediately north or south of the line $AB$, then the answer is just $2\pi \phi$ where $\phi$ is the latitude of $X$. If $X$ is not in that sector then the answer is just the distance to either $A$ or $B$, whichever is closer.

  • $\begingroup$ Isn't what you care about the longitude to determine if X is above or below AB? $\endgroup$ – NetMage Nov 20 '20 at 23:08
  • $\begingroup$ The least practical answer I have ever read...! $\endgroup$ – SeF Apr 29 at 12:34

The question is a little ambiguous: the three previous answers used three different interpretations. If the OP wants the surface distance from the point $X$ to the geodesic line $\overleftrightarrow{AB}$, the answer is straightforward. If the desired distance is between $X$ and the segment $\overline{AB}$, a bit more work is required.

Using longitude ($\theta$) and latitude ($\phi$), let $A=(\theta_A, \phi_A)$, $B=(\theta_B, \phi_B)$, and $X=(\theta_X, \phi_X)$. The direction vectors for these points are $$\hat A = (\cos \phi_A \cos \theta_A, \cos \phi_A \sin \theta_A, \sin \phi_A),$$ $$ \hat B = (\cos \phi_B \cos \theta_B, \cos \phi_B \sin \theta_B, \sin \phi_B), $$ $$\hat X = (\cos \phi_X \cos \theta_X, \cos \phi_X \sin \theta_X, \sin \phi_X).$$

Let $\Phi$ be the distance on the unit sphere between $\hat X$ and the geodesic line passing through $\hat A$ and $\hat B$. Imagine the plane $\mathcal{P}$ passing through $\hat A$, $\hat B$, and the origin, which cuts the unit sphere in half. Then the Euclidean distance of $\hat X$ from plane $\mathcal{P}$ is $\sin \Phi$. Now let $\hat n$ be a unit normal vector for $\mathcal{P}$, and we have

$$\hat n = \hat A \times \hat B$$ $$\sin \Phi = | \hat n \cdot \hat X |$$

So, if the radius of the original sphere is $R$, then the surface distance from the point $X$ to the geodesic line $\overleftrightarrow{AB}$ is $R \Phi$.

To determine the distance to the segment $\overline{AB}$, we need to determine whether or not the point of line $\overleftrightarrow{ A B}$ that $ X$ is closest to is between $A$ and $B$. If the closest point is between $A$ and $B$, then the surface distance to the segment is $R \Phi$. Otherwise, the distance to the segment is the distance to the closest endpoint, which is best resolved though the methods described in the Wikipedia article referenced by Ross Millikan. One way to make this determination is to find the point $\hat{X}_{\textrm{proj}}$, the projection of $\hat X$ onto plane $\mathcal{P}$,

$$\hat{X}_{\textrm{proj}} = \hat X - (\hat n \cdot \hat X) \hat n,$$

and then normalize $\hat{X}_{\textrm{proj}}$,

$$\hat x = \frac{\hat{X}_{\textrm{proj}} }{| \hat{X}_{\textrm{proj}} |},$$

So determining whether the point of line $\overleftrightarrow{AB}$ that $X$ is closest to is between $A$ and $B$ reduces to determining whether $\hat x$ is between $\hat A$ and $\hat B$.

Now consider the mid-point of $\hat A$ and $\hat B$,

$$M=\frac{\hat A + \hat B}{2}$$

If the projection of $\hat x$ on the ray $\overrightarrow{OM}$ is further along than the projection of $\hat A$ or $\hat B$, then $\hat x$ is between $\hat A$ and $\hat B$, that is, if $\; \hat x \cdot M > \hat A \cdot M \; \; (=\hat B \cdot M)$, then $\hat x$ is between $\hat A$ and $\hat B$, otherwise not.

  • $\begingroup$ How would one calculate the interception point $Y$ with $\overleftrightarrow{XY}$ being the shortest distance of $X$ to $\overleftrightarrow{AB}$? $\endgroup$ – oschrenk Jun 21 '12 at 19:48

Assuming you mean the geodesic line $AB$.
Hint: Calculate the angle $\theta =\angle XCP$ where $P$ is a point on the line and $C$ is the center of the sphere. For instance with $\cos \theta =\frac{\vec{CX}\cdot \vec{CP}}{|\vec{CX}|\cdot |\vec{CP}|}=\frac{\vec{CX}\cdot \vec{CP}}{r^2}$, where $r$ is the length of the sphere's radius and $\vec{CX}, \vec{CP}$ are vectors in $\mathbb{R}^3$. This angle is proportional to the geodesic distance between $P$ and $X$.

What you have to do then, is to find the point $P$ for which the angle is smallest.


There is a detailed formula in Wikipedia, with a warning about rounding errors and alternate versions to avoid them. This gets you the distance between any two points, say A and X or B and X. I am not sure what you mean by the distance from (A,B) to X unless it is this.

  • $\begingroup$ The cited great-circle distance is for the distance between two points, not point and line. $\endgroup$ – SeF Apr 29 at 12:34

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