I have the points: $A=(1,2,0)$, $B=(-2,0,3)$, $C=(3,-4,-1)$, and I want to find the coordinate of a fourth point $D$ such that $ABCD$ is a parallelogram.
In the previous part of the question I worked out the angle between $ABC$ using the vector dot product.
I am quite confused because how I do know which sides of my parallelogram are parallel to each other? Is the solution unique? How do I know this? I can't picture the geometry. Any help would be really appreciated.
Hint -
Find mid point of AC and then using this mid point and point B find point D. As in parallelogram mid point of AC and BD is same.
There are three possible parallelograms that include three given non-colinear points as vertices, with the additional point found by various combinations of:
$$\mathbf P_1+\mathbf P_2-\mathbf P_3$$
This can be justified in various ways - one way is to look at the need to have the midpoints of the diagonal coincide, see this answer.
If you know which two points are not connected on the perimeter of the parallelogram, you would use those as $\mathbf P_1$ and $\mathbf P_2$ in the above formula. In this case assuming $A$ and $C$ are separated, we could calculate:
$$\begin{align} D&=A+C-B \\ &= (1,2,0) + (3,-4,-1) - (-2,0,3) \\ &= (6,-2,-4) \end{align}$$
The three parallelograms correspond to adding the different coloured triangles to the original black triangle in this diagram:
we have $$\vec{BA}=[3,2,-3]$$, the searched point $D$ has the coordinates $(x,y,z)$ thus $$\vec {CD}=[x-3,y+4,z+1]$$ since $ABCD$ is a parallelogram we get $$x-3=3$$, $$y+4=2$$ and $$z+1=-3$$