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I need to find the third vertex of an isosceles triangle in $3D$ space. What I know are coordinates of two other vertices and angles between sides of the triangle:

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$|AB| = |AC|$. Coordinates of vertices $A$ and $B$ are known. What's more I know that this triangle is parallel to a plane given by its normal and a point that lies on it. I need to find coordinates of vertex $C$.

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    $\begingroup$ It seems to me, there is a whole circle of points that satisfy the condition in 3D. Even if you define a plane the triangle lies on, there still will be two possible solutions. $\endgroup$
    – guest
    Aug 20, 2020 at 11:55
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    $\begingroup$ There is not a unique solution but a an infinite set of solutions : a circle (rotate the triangle around line AB). $\endgroup$
    – Jean Marie
    Aug 20, 2020 at 11:56
  • $\begingroup$ I know that this triangle has to be prallel to a plane given by its normal and a point that lies on it [added also to the question] $\endgroup$
    – michalt38
    Aug 20, 2020 at 12:04

1 Answer 1

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$$(2A-C-B)\cdot(B-C)=0$$

$${(A-B)\cdot(C-B)\over |A-B||C-B|} = \cos(a)$$

Let $C$ has coordinate $x_c,y_c,z_c$ and plug in to the above equation.

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