In three-dimensional space, when the sum of the distances from an unknown point to two known points is constant, what is the trajectory of the point?

Question

The unknown point P (x, y, z) in three-dimensional space is at a constant sum of distances dPA and dPB from two known fixed points A and B, i.e., dPA + dPB = constant. How can we express the trajectory of P? I know that in the two-dimensional plane, the trajectory of P would be an ellipse. In three-dimensional space, it seems to be a shape resembling an ellipse that consistently maintains the same major and minor axes regardless of the angle of the cross-section. I would like to use MATLAB to generate N random points on this shape, but I don't have a good method. Can anyone help me with this?

Answer 1

Yes, you are correct in noting that the surface is that of an ellipse for each cross-section. The shape is essentially that which is formed by considering the 2D ellipse formed with A and B as focii, and rotating it about the line AB.

To randomly sample points on this ellipsoid, you can use this answer, which gives code for the description provided here

Answer 2

From the givens, let $p = (x,y,z)$ and let the two points be $F_1$ and $F_2$, and let the distance between them be $2 c = \| F_1 - F_2 \|$, and let the sum of distances be $2a$, then

$ \| p - F_1 \| + \| p - F_2 \| = 2 a $

Square this equation, you get

$ (p - F_1) \cdot (p - F_1) + (p - F_2) \cdot (p - F_2 ) + 2 \| p - F_1 \| \| p - F_2 \| = 4 a^2 $

Re-arrange the equation as follows

$ 2 \| p - F_1 \| \| p - F_2 \| = 4 a^2 - (p - F_1) \cdot (p - F_1) - (p - F_2) \cdot (p - F_2 ) $

Simplify the right hand side

$ 2 \| p - F_1 \| \| p - F_2 \| = 4 a^2 - 2 p \cdot p + 2 p \cdot (F_1 + F_2) - F_1 \cdot F_1 - F_2 \cdot F_2 $

At this point introduce the midpoint of $F_1$ and $F_2$:

$ C = \dfrac{1}{2} ( F_1 + F_2 ) $

And let

$ U = \dfrac{1}{2} (F_2 - F_1) $

Then

$ 2 \| p - F_1 \| \| p - F_2 \| = 4 a^2 - 2 p \cdot p + 4 p \cdot C - 2 (C \cdot C + U \cdot U) $

Now, replace $F_1$ and $F_2$ on the left hand side, and re-write them in terms of $C$ and $U$, and divide through by $2$:

$ \| p - C + U \| \| p - C - U \| = 2 a^2 - p \cdot p + 2 p \cdot C - (C \cdot C + U \cdot U ) $

Note that $(p - C) \cdot (p - C) = p \cdot p - 2 p \cdot C + C \cdot C $

Therefore,

$ \| p - C + U \| \| p - C - U \| = 2 a^2 - (p - C)\cdot (p - C) - U.U $

Now let $q = p - C $, then

$ \| q + U \| \| q- U \| = 2 a^2 - q\cdot q - U.U $

Squaring both sides,

$ ( (q + U) \cdot (q + U) )( (q - U) \cdot (q - U) ) = 4 a^4 + (q \cdot q)^2 + (U. U)^2 - 4 a^2 (q \cdot q) - 4 a^2 (U.U) + 2 (U.U) (q \cdot q) $

Expanding the left hand side we get,

$ ( q\cdot q + 2 q\cdot U + U\cdot U)( q\cdot q - 2 q\cdot U + U\cdot U ) = RHS $

Simplifying the left hand side, the equation becomes,

$ (q \cdot q)^2 + (U \cdot U)^2 + 2 (q \cdot q) (U \cdot U) - 4 (q \cdot U)^2 = 4 a^4 + (q \cdot q)^2 + (U. U)^2 - 4 a^2 (q \cdot q) - 4 a^2 (U.U) + 2 (U.U) (q \cdot q) $

Cancelling $(q \cdot q)^2 $ and $(U \cdot U)^2$ and $ 2 (U.U) (q \cdot q) $ on both sides of the equation gives

$ - 4 (q \cdot U)^2 = 4 a^4 - 4 a^2 (q \cdot q) - 4 a^2 (U.U) $

Dividing by $4$ through,

$ - (q \cdot U)^2 = a^4 - a^2 (q \cdot q) - a^2 (U.U) $

Using the matrix-vector equivalent of the dot product this last equation becomes

$ - (q ^T U)^2 = a^4 - a^2 (q^T q) - a^2 (U^T U) $

Re-arrange the equation

$ a^2 (q^T q) - (q^T U)^2 = a^2 (a^2 - U^T U) $

Note that $(q^T U)^2 = (q^T U) (q^T U) = (q^T U)(U^T q) = q^T (U U^T) q $

Therefore, our equation becomes

$ q^T ( a^2 I - U U^T ) q = a^2 (a^2 - U^T U ) $

Replacing $q$ with $p - C$ we get

$ (p - C)^T ( a^2 I - U U^T ) (p - C) = a^2 (a^2 - U^T U ) $

And this the equation of the desired locus of points, which is an ellipsoid of revolution. Recall that we defined the focal distance as

$ c = \dfrac{1}{2} \| F_2 - F_1 \| = \| U \| $

Therefore,

$ (p - C)^T ( a^2 I - U U^T ) (p - C) = a^2 (a^2 - c^2 ) $

Let $b^2 = a^2 - c^2 $, then

$ (p - C)^T ( a^2 I - U U^T ) (p - C) = a^2 b^2 $

The unit eigenvectors of $U U^T $ are

$ X_1 = \dfrac{1}{c} U$, because

$ U U^T X_1 = \dfrac{1}{c} U (c^2) = c U = c^2 X_1 $

So the corresponding eigenvalue is $c^2$

And the other two eigenvectors of $U U^T$ are any two unit vectors that are orthogonal to $X_1$, i.e. to $U$, with eigenvalues of $0$.

Therefore, the eigenvalues of $ (a^2 I - U U^T ) $ are $(a^2 - c^2), a^2, a^2$, that is, $b^2 , a^2, a^2$

Remember the right hand side of the equation is $a^2 b^2$, so dividing by this scalar, scales the eigenvalues by $ \left(\dfrac{1}{a^2 b^2}\right) $

So that the eigenvalues of

$Q = \dfrac{a^2 I - U U^T} { a^2 b^2 } $ are $\dfrac{1}{a^2}, \dfrac{1}{b^2}, \dfrac{1}{b^2} $

And our final form of the equation of this ellipsoid becomes

$ ( p - C )^T Q ( p - C) = 1 $

And as we have demonstrated, it is an ellipsoid of revolution with semi-minor axis lengths of $b$ and semi-major axis length $a$. The major axis is along the vector $U$.