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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.

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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 $m$?

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

Now the complete solution will be the circle that has the two points already found as endpoints of a diameter.

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.

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