How to find triangles third vertex C(x,y,z) if we known:
1. other two vertices A(x,y,z), B(x,y,z)
2. lengths of all triangles sides
3. the triangle's normal
I understand how to solve it in a graph, but can't find an analytical solution.
How to find triangles third vertex C(x,y,z) if we known:
1. other two vertices A(x,y,z), B(x,y,z)
2. lengths of all triangles sides
3. the triangle's normal
I understand how to solve it in a graph, but can't find an analytical solution.
I have no idea what is meant by #3. Normal?
If #2 means that we know the lengths, $\displaystyle \overline{AC}~\&~\overline{BC}$ then the point C is either of the intersections of the circle centered at A with radius $\displaystyle \overline{AC}$ and the circle centered at B with radius $\displaystyle \overline{BC}.$
Thanks, for the answer. Yes #3 is normal or the triangle. We can't use solution with circle because i need three coordinates for C point. I have found one solution, but it is not very elegant:
1. calculate angle between (AC and AB) and (BA and BC) in 2D - where we can easily calculate C coordinates (by the low of cosines)
2. rotate AB vector about normal with two angles - which give us two vectors(lines)
3. find point of intersection on these two lines - it will be our C point.
You did not understand Plato's question. What do you mean by "the normal to a triangle". Do you mean the normal to the plane determined by the triangle? I think Plato was assuming you were working in a given plane. Of course, being given the normal to the plane means you are given a plane.
But what I would do is this: suppose we are given points $\displaystyle A(a_1, a_2, a_3)$ and $\displaystyle B(b_1, b_2, b_3)$ as well as normal vector <X, Y, Z> and are told that the length of AC is $\displaystyle r_a$, the length of BC is $\displaystyle r_b$. Let C be the unknown point $\displaystyle (c_1, c_2, c_3)$.
Then we seek the intersection of three surfaces: the sphere $\displaystyle (x-a_1)^2+ (y- a_2)^2+ (z- a_3)^2= r_a^2$, the sphere $\displaystyle (x- b_1)^2+ (y- b_2)^2+ (z- b_3)^2= r_b^2$, and the plane $\displaystyle X(x- a_1)+ Y(y- a_2)+ Z(z- a_3)= 0$. That gives three equations to solve for x, y, and z.