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In a 3D cartesian coordinate system, the coordinates of start point and end point have been given as $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. If the central angle of the two points (the one smaller than 180 degrees) is known as $\theta$, and the plane on which the arc lies is perpendicular to $XY$ plane, how to find the parametric expression $(x, y, z)$ of the arc? I mean, each coordinate of $(x, y, z)$ needs to be represented by variables given above. Can anyone help? Thanks!

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Your specs will still give you at least two solutions, one concave and one convex relative to the xy plane. Is that acceptable? –  John Moeller Feb 28 '13 at 7:36
    
Any give arc can have an infinite number of parametric equations. Do you have any further conditions that would help us to pick the right one? –  bubba Feb 28 '13 at 8:09
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1 Answer 1

First you can assume $(x_2,y_2,z_2) = (0,0,0)$. If it isn't, you can just add $(x_2,y_2,z_2)$ back into all the parametric equations at the end. So we'll work with $(x_1',y_1',z_1') = (x_1-x_2,y_1-y_2,z_1-z_2)$.

Next let $(x_1',y_1',z_1') = (r_1\cos\phi_1,r_1\sin\phi_1,z_1')$. This simplifies things because now your $(r,z')$ plane is perpendicular to the $(x,y)$ plane and we can just work with two coordinates for now.

Now we need to find the center $(\overline r,\overline z')$. This will be one of two points on either side of the midpoint $(r_1/2,z_1'/2)$, perpendicular to the line that goes through the origin and $(r_1,z_1')$. We can just find a constant $c$ such that $(\overline r,\overline z') = (r_1/2,z_1'/2) + c(-z_1',r_1)$. We just need to be sure that the angle is right. We can do that by computing the dot product between the segments from the center to each point, or:

$$\begin{align*} \langle(r_1/2 - c z_1',z_1'/2 + cr_1),(-r_1/2 - c z_1',-z_1'/2 + cr_1)\rangle &= \cos\theta \ell_1 \ell_2 = \cos\theta \ell_1^2 \\c^2(z_1')^2 - r_1^2/4 + c^2r_1^2 - (z_1')^2/4 &= \cos\theta (r_1^2/4 + (z_1')^2/4 + c^2(z_1')^2 + c^2r_1^2) \\(c^2 - 1/4)(r_1^2 + (z_1')^2) &= \cos\theta (c^2 + 1/4)(r_1^2 + (z_1')^2) \\(c^2 - 1/4) &= \cos\theta (c^2 + 1/4) \\c^2(1-\cos\theta) &= (\cos\theta + 1)/4 \\4c^2 &=\cot^2(\theta/2) \\c &=\pm\cot(\theta/2)/2 \end{align*}$$

Once you have the center, one possible parametric equation for the arc is an easy matter of letting $\gamma_0$ be the angle in the $(r,z')$ plane where you find $(\overline r,\overline z')$ (i.e., $\gamma_0 = \arctan(\overline z'/\overline r)$). Then you can use a parametric form like this:

$$ (r,z') = (\overline r,\overline z') - \ell_1(\cos(\gamma_0 \mp \gamma),\sin(\gamma_0 \mp \gamma)) $$

The $\gamma$ parameter will trace out your arc in the range $[0,\theta]$.

So to sum up: $$ (r,z') = (r_1/2,z_1'/2) \pm\cot(\theta/2)(-z_1'/2,r_1/2) - \ell_1(\cos(\gamma_0 \mp \gamma),\sin(\gamma_0 \mp \gamma)) $$ Where $\ell_1^2 = (r_1^2 + (z_1')^2)/(4\sin^2(\theta/2))$, $r_1^2 = (x_1')^2+(y_1')^2$, and $(x_1',y_1',z_1') = (x_1-x_2,y_1-y_2,z_1-z_2)$.

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