How many lines can pass through a point with the same distance from two other points?

Question

Consider three points $A$,$B$ and $C$ in the space which are not on the same line. how many line we can pass through $A$ so that it has the same distance from both points $B$ and $C$ ?

$1)\text{infinity}\qquad2)\text{at least two lines}\qquad3)\text{two lines}\qquad4)\text{one line}$

I know that in the space the points on the perpendicular bisector plane of a segment have the same distance from both endpoint of that segment. So I think if $A$ place on the perpendicular bisector of $BC$ then any line on that plane which passes through $A$ has the same distance from $B$ and $C$.

But I don't know how to solve the problem in general.

Answer 1

You reasoning regarding perpendicular bisector plane is valid.
Now, consider vectors $\vec{AB}$ and $\vec{AC}$. The problem boils down to finding vector $\vec{AX}$ such that $\left\lVert\vec{AB} \times \vec{AX}\right\rVert=\left\lVert\vec{AC} \times \vec{AX} \right\rVert$ or $\left\lVert\vec{AB} \right\rVert \left\lVert\vec{AX} \right\rVert \sin \alpha=\left\lVert\vec{AC} \right\rVert \left\lVert\vec{AX} \right\rVert \sin \beta$ where $\alpha$ and $\beta$ are angles between the corresponding vectors. Thus, we have the equation $\left\lVert\vec{AB} \right\rVert \sin \alpha=\left\lVert\vec{AC} \right\rVert \sin \beta$ which has infinite number of solutions as all we need to do is to pick angle $\alpha$ such as $-1 \le \frac{||\vec{AB}||}{||\vec{AC}||}\sin \alpha \le 1$ and we can find angle $\beta$ to satisfy our equation above.
P.S. Try to solve this problem in 2D to see that two solutions will always exist in 2D.