Finding a 4th point in 3D space knowing 3 other points and 2 distances to the 4th point from them I have 3 points in space A, B, and C all with (x,y,z) coordinates, therefore I know the distances between all these points. I wish to find point D(x,y,z) and I know the distances BD and CD, I do NOT know AD.
The method I have attempted to solve this using is first saying that there are two spheres known on points B and C with radius r (distance to point D). The third sphere is found by setting the law of cosines equal to the formula for distance between two vectors ((V1*V2)/(|V1||V2|)) = ((a^2+b^2-c^2)/2ab). 
Now point D should be the intersection of these three spheres, but I have not been able to calculate this or find a way to. I either need help finding point D and I can give numeric points for an example, or I need to know if I need more information to solve (another point with a distance to D known).
Ok, now assuming that distance AD is known, how do I calculate point D?
This is what it looks like when I graph the two spheres and the one I calculated, as you can see it intersects on point F. (D in this case)
 A: Knowing $|BD|$ and $|CD|$ is only enough to know that $D$ is on the intersection of two spheres, one around $B$ and one around $C$.  Assuming they intersect at all, the intersection is likely a circle and $D$ could be anywhere on that circle.  $A$ is not giving you anything.  Even if you get $|AD|$ you will still have an ambiguity between two points unless two of the spheres are tangent.
A: Suppose the known distances are
$$d(B,C)=d(A,C)=d(C,D)=1$$
and
$$d(A,B)=d(B,D)=\sqrt{2}$$
For concreteness, we can place $A,B,C$ ae
$$C=(0,0,0),\;\;B=(1,0,0),\;\;A=(0,1,0)$$
Then if $D$ is any point on the circle in the $yz$-plane, centered at the origin, with radius $1$, all of the distance specifications are satisfied.

But since $D$ can be any point on that circle, it follows $D$ is not uniquely determined.

Note that as $D$ traverses the circle, $d(A,D)$ varies from a minimum of $0$ (when $D=A$), to a maximum of $2$ (when $D=(0,-1,0))$, so $d(A,D)$ is also not uniquely determined.

If $d(A,D)$ is also given, say $d(A,D)=t$, where $0\le t\le 2$, then $D=(x,y,z)$ is determined by the system
$$
\begin{cases}
x=0\\[4pt]
y^2+z^2=1\\[4pt]
x^2+(y-1)^2+z^2=t^2\\
\end{cases}
$$
which yields
$$D=\left(0,\,1-{\small{\frac{t^2}{2}}},\,\pm{\small{\frac{t}{2}}}\sqrt{4-t^2}\right)$$
hence,


*

*If $t=0$, we get $D=(0,1,0)$.$\\[4pt]$

*If $t=2$, we get $D=(0,-1,0)$.$\\[4pt]$

*If $0 < t < 2$, there are two choices for $D$, as specified above.

