Find the co-ordinates of a 3D triangle $ABC$ given 2 unit vectors, a side and a property

Question

There is a triangle $ABC$ in 3D space. I am given the following:

1. $\vec{u}_{AC}:$ The unit vector along the direction of $AC$.
2. $\vec{u}_{BC}:$ The unit vector along the direction of $BC$.
3. point $A$ is $(0,0,0)$.
4. $AC+ BC = AB + k$, where $k$ is known.

My aim is to locate points $B$ and $C$.

It seems to me that there is enough information to identify co-ordinates of $B$ and $C$. Do you agree? If yes, how exactly can I proceed to find those co-ordinates?

My attempt: I can use dot product between $\vec{u}_{AC}$ and $\vec{u}_{BC}$ to identify $<ACB$. But, how do I proceed further. I am not sure.

Answer 1

Consider the last equation of

$$ \| C - A \| + \| B-C \| = \| B- A \| + k $$

but with $A=(0,0,0)$, the above is

$$ \| C \| + \| B-C \| = \| B \| + k $$

We can freely rotate our coordinate system around it will not change the nature of the problem, and so I chose a system where $B_y=C_y = y$, yielding

$$ \sqrt{C_x^2+y^2} + (B_x-C_x) = \sqrt{B_x^2+y^2} + k $$

which has three unknowns, $y$, $B_x$ and $B_y$. So there is information missing to solve this geometry.

If you make assumptions about $B_x$ and $C_x$ then you can use the above to set

$$ y^2 = \frac{ k (2 C_x + k) (2 B_x -k) (2 B_x -2 C_x-k)}{4 ( B_x-C_x-k)^2 }$$

Once you solve the 2D planar triangle (if you can), then use the lengths of the sides and the direction vectors to get the 3D coordinates of the vertices.