Thread: Find coordinates of 3rd vertex(x,y,z) of triangle(3D)

1. Find coordinates of 3rd vertex(x,y,z) of triangle(3D)

I have a triangle , coordinates of two vertices are known [(x1,y1,z1) and (x2,y2,z2)] . I know all the side lengths and angles of the triangle. I want to find the coordinates (x3,y3,z3) of 3rd vertex.

Hope I will get a solution from Math Help Forum.

2. Originally Posted by geoleo
I have a triangle , coordinates of two vertices are known [(x1,y1,z1) and (x2,y2,z2)] . I know all the side lengths and angles of the triangle. I want to find the coordinates (x3,y3,z3) of 3rd vertex.

Hope I will get a solution from Math Help Forum.
Correct me if I'm wrong, but isn't there an infinite number of solutions here? Suppose we restrict the solution to a certain plane, then we can find a point that gives the desired side lengths and angles in that plane, in fact we can find two of them, one on either side of the line defined by the first two vertices. But just pick one of the solutions and imagine rotating the triangle in 3-space about the line defined by the first two vertices. The location of the third vertex will trace a circle, and all points on this circle will be valid solutions to the problem.

3. mmmm, So there is no unique solution. . Ok then. Thank you .

4. Hello, geoleo!

I have a triangle, coordinates of two vertices are known: . $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$
I know all the side lengths and angles of the triangle.

I want to find the coordinates $C(x_3,y_3,z_3)$ of 3rd vertex.

undefined is absolutely correct!

Let the sides be $a,b,c$ and the angles $\alpha, \beta, \gamma$

Place $\Delta ABC$ on the "floor" $(xy\text{-plane).}$

Code:
            * - - - - - - - - - - - - - - - - *
/                B                /
/             ..o.                /
/          ..*:::β:*.             /
/       ..*:::::::::::*.          /
/    ..*:α:::::::::::::::*.       /
/   o'''*'''*'''*'''*'''*'''o C   /
/  A                              /
* - - - - - - - - - - - - - - - - *

Then vertex $C$ can be revolved about $AB$ to any position $C\:\!'$.

Code:
    C'
o--
----*--
*--------*--
-------------*-     B
*---------------..o.
-----------..*::β::*.
*------..*:::::::::::*.
--..*:α:::::::::::::::*.
o'''*'''*'''*'''*'''*'''o C
A

Therefore, there are a brizillion locations for point $C.$

5. Originally Posted by Soroban

Code:
    C'
o--
----*--
*--------*--
-------------*-     B
*---------------..o.
-----------..*::β::*.
*------..*:::::::::::*.
--..*:α:::::::::::::::*.
o'''*'''*'''*'''*'''*'''o C
A

Did you type these diagrams out manually or use some sort of program/script? They're nifty!

6. Hello, undefined!

I invented my own system for typing these diagrams
and getting the symbols to line up.

I described the system a few years ago.
So far, no one has tried my system, though.

7. Can we find a unique vertex using additional information?

Hello Thanks for the useful diagrams and solutions.
1) The Triangle is on a sphere whose center and Radius is known.
2) One end point of triangle is center of sphere ie, (0,0,0)
3) Second end point also I have.
4) I have to find the third end point.
5) I have a line that is perpendicular to one side of triangle, connecting the unknown vertex and (0,0,0) vertex. So using the direction cosines l1l2+m1m2+n1n2=0 property, I can check out of all vertices which is the valid vertex.

So I would like to know the method of finding the third vertex (eventhough infinite many solutions) from the available sides and angles and two vertices known,also the triangle is on a sphere .

8. Originally Posted by geoleo
Hello Thanks for the useful diagrams and solutions.
1) The Triangle is on a sphere whose center and Radius is known.
2) One end point of triangle is center of sphere ie, (0,0,0)
3) Second end point also I have.
4) I have to find the third end point.
5) I have a line that is perpendicular to one side of triangle, connecting the unknown vertex and (0,0,0) vertex. So using the direction cosines l1l2+m1m2+n1n2=0 property, I can check out of all vertices which is the valid vertex.

So I would like to know the method of finding the third vertex (eventhough infinite many solutions) from the available sides and angles and two vertices known,also the triangle is on a sphere .
I am very confused by your description. What does it mean for a triangle to be "on" a sphere? Normally I would think that means, for example, the triangle connecting Mexico City, Toronto and Portland, for which the sides of the triangle are curved and the sum of the internal angles is greater than 180 degrees; this is part of spherical geometry. Anyway the point is that I can't tell what you're talking about. Did you copy this problem from somewhere? Could you make sure the wording is correct?

And for example "3) Second end point also I have." This is vague and doesn't really help us answer your question. If you have it, why don't you tell us what it is?

9. Actually, this is not a text book problem. I am doing Space research for my Mtech programme. Sorry for the inconvenience caused. Since it is space Research related and I don't have the coordinates as such. If I have proper method only I can go for simulation. I am just designing the concept now. The Sphere is nothing but Earth. The Problem which I have mentioned here is part of the Global Problem. Yah what you said is right, Spherical Geometry can also be considered here. A triangle on the sphere relates to Spherical geometry, I read somewhere. But we cannot consider normal triangle on a sphere? I have put only the Geometry problem in this forum. For the time being I can ignore the sphere part. Can I get the method to find the third vertex. I am Sorry once again. So I will state my problem clearly: 1) I have a triangle whose all sides and angles are known. 2) Two vertices are known. 3) There is a line whose direction cosines are known, which is perpendicular to one side of triangle having unknown vertex and one known vertex. 4)I require to find the third vertex. Thank you friends.

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What is the formula used to find third vertex of equilateral triangle in 3d geometry

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