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This question already has an answer here:

I want to draw an arc in 3D.

I viewed circles,ellipses and arcs in 2D with Circle[...] but when I want to draw and arc in 3D I can´t find any instruction for taht purpose.

The question is: Are there any Mathematica instruction to draw an arc in 3D?

I see this posts with solutions creating new function:

How to draw an ellipse arc in 3D?

An efficient circular arc primitive for Graphics3D

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marked as duplicate by Sjoerd C. de Vries, Yves Klett, Karsten 7., ubpdqn, Kuba Mar 22 '15 at 12:35

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    $\begingroup$ No, there isn't such a function built-in in Mathematica. That is why I wrote the function you referenced above. We're several version further now and several graphical primitives were added, but the 3D circle or circular arc is not among them $\endgroup$ – Sjoerd C. de Vries Mar 22 '15 at 9:52
  • $\begingroup$ @SjoerdC.deVries ok, thank you, the solutions in the references post are not suitable to use with Manipulate[], I have errors with Arc3D[...], and with arc[...] $\endgroup$ – Mika Ike Mar 22 '15 at 15:33
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    $\begingroup$ My solution does work with Manipulate. In my post I give various examples using it. You don't seem to be using my solution as it has a different name. $\endgroup$ – Sjoerd C. de Vries Mar 22 '15 at 15:38
  • $\begingroup$ @SjoerdC.deVries As I understand the routine splineCircle[....] it´s only for an arc in the plane z=0. But What I´m looking for is to mark the angles for a parametrization of the sphere. Could I write you in private?, to avoid extende this comments. Supose I want to draw an arc between (0,1,1) and (-1,0,-1). centered in (0,0,0) $\endgroup$ – Mika Ike Mar 23 '15 at 6:53
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    $\begingroup$ You can use GeometricTransformation as I did in the last demo. Circles are translated and rotated everywhere. If you want to specify this in the function call itself, it is easy to wrap that all in a function. If you update this question adding this as specific requirements (making it a real new question instead of a duplicate) I can try to come up with an appropriate answer. BTW Note that there are constraints with respect to the three points you mention. The distance between the center and the first and between the center and the second should be exactly the same. $\endgroup$ – Sjoerd C. de Vries Mar 23 '15 at 13:17

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