I have a triangle in space represented by the points $A, B,$ & $C$. I know the coordinates for $A$ & $B$, along with the segment lengths of all $3$ legs ($AB, AC,$ & $BC$).
Not sure this is useful but point $A$ happens to be located at the origin of my coordinate system and point $C$ is constrained to lie in the $XZ$ plane. Point $B$ has $+X$ and $+Y$ and $-Z$ components. When I represent this in $CAD$ with my exact coordinates, I know that there are $2$ solutions for the coordinates of point $C$ but I need to be able to calculate the location(s) of point $C$ in excel.
Can someone help me figure out how to identify the coordinates for point $C$?
Clearly, point $C$ lies on a circle whose center is on $v = AB$ , and radius $r = AC \sin \theta $ where
$\theta = \angle CAB = \cos^{-1} \left( \dfrac{ AB^2 + AC^2 - BC^2 }{ 2 (AB) (AC) } \right) $
The center $O$ is given by
$ O = A + (AC \cos \theta) \dfrac{ v }{ \| v \| } $
The plane of the circle is spanned by two unit vectors $u_1, u_2$ orthogonal to $v$, and if
$ v = (v_x, v_y, v_z) $
Then we can express $v$ in spherical coordinates as follows
$ v = \| v \| ( \sin \theta_1 \cos \phi_1 , \sin \theta_1 \sin \phi_1 , \cos \theta_1 ) $
The standard unit vectors that are perpendicular to $v$ are given by
$ u_1 = ( \cos \theta_1 \cos \phi_1 , \cos \theta_1 \sin \phi_1 , - \sin \theta_1 ) $
and
$ u_2 = ( - \sin \phi_1 , \cos \phi_1 , 0 ) $
Now we can express possible positions of vertex $C$ parametrically as follows:
$ C = O + r \cos t \ u_1 + r \sin t \ u_2 $
This is the parametric equation of the circle on which $C$ lies.
Since we want $C$ to lie on the YZ plane, then its $x$ coordinate is zero.
Hence
$ 0 = O_x + r \cos t \ u_{1x} + r \sin t \ u_{2x} $
This is a classical trigonometric equation in the parameter $t$, and its solution is given by
$ t = \beta \pm \cos^{-1} \left(\dfrac{ - O_x }{\sqrt{u_{1x}^2 + u_{2x}^2} } \right) $
where $ \beta $ satisfies,
$\cos(\beta) = \dfrac{u_{1x}}{\sqrt{u_{1x}^2 + u_{2x}^2} } $ and
$\sin(\beta) = \dfrac{u_{2x}} { \sqrt{u_{1x}^2 + u_{2x}^2} } $,
i.e. $\beta = \text{atan2}( u_{1x}, u_{2x} ) $
The number of solutions for the parameter $t$ can be $0, 1$, or $2$.
Once we have the solutions for $t$, we can substitute immediately in the above equation to obtain $C$. This is repeated below for convenience,
$ C = O + r \cos t \ u_1 + r \sin t \ u_2 $