# Intersection of two lines in 3D

The two points $A=(x_{0},y_{0},z_{0})$ and $B=(x_{1},y_{1},z_{1})$ are given. I want to find the coordiantes of the point $C=(x,y,z)$. The line segments $AC$ and $BC$ make equal angle $\alpha$ with the horizontal plane through $C$. The angle $\alpha=\arctan(m)$ which is known.

The question is how to find the location of $C=(x,y,z)$ in terms of $x_{0},y_{0},z_{0},x_{1},y_{1},z_{1}$ and $m$?

• What plane is the angle bisector in? – Mastrel Aug 6 '14 at 1:15
• @Mastrel I don't know the equation of that plane. – Harry Aug 6 '14 at 1:16
• Do you want to find one point C? Because there seems that there will be infinite such C. – Mastrel Aug 6 '14 at 1:21
• @Mastrel But this can be solved for 2D. I was thinking for 3D? – Harry Aug 6 '14 at 1:22
• In 2D there are two solutions (complete the parallelogram whose three vertices are $A$, $B$, and $C$). In 3D there are infinitely many solutions. – Rahul Aug 6 '14 at 1:26

First, let us simplify and assume $z = z_0$. Then we would have
$$\frac{y-y_0}{x-x_0} = m \quad \mbox{or} \quad y = m(x-x_0)+y_0$$
It remains to determine $x$
We know that $AC \cdot BC = |AC||BC| \cos(2\alpha)$ so this will give us a quadratic in $x$. Solving this we will get two possible points for $C$.
I hope you'll excuse me for not working out all the details as the quadratic equation you get in $x$ appears to be quiet messy to work with.