# Thread: "z" graph of a tetrahedron

1. ## "z" graph of a tetrahedron

How do you find the "z" graph of a tetrahedron when you are only given the vertices?

For example....

What's the "z" graph of a tetrahedron when the xy-plane vertices are (0,2), (0,0), and (1,0) and when z=3 as long as x=0 and y=0?

2. What do you mean by graph? What I've done is just plotting the points and join them with lines, first the triangle in the x-y plane and then, with the last point at (0,0,3)

3. Every tetrahedron is bounded by a 3D plane, akin to a z-based graph z(x,y). I was wondering how you would find out the exact z-equation based on the vertices. We know the vertices and the shape, but what is the equation of the 3D plane?

4. Ok.

For example take the vectors (0-0,0-2,3-0) and (0-1,0-0,3-0), do the cross product and you will have a normal vector, then the plane can be written as $\displaystyle Ax+By+Cz=0$ where $\displaystyle (A,B,C)$ is a normal vector and $\displaystyle (x,y,z)$ it's a point on the plane.

I think this should work

5. Well, my answer is that a normal vector is $\displaystyle (6,3,2)$. Then $\displaystyle 6x+3y+2z=0$ and so $\displaystyle z= -2x-y$

Hope it will be right!

6. No that's not the right one. I used my graphing software and it doesn't fit in with the vertices.

7. In fact it's wrong: I divided by 3 instead of 2. $\displaystyle z=-3x - \frac{3}{2}y$, check this one.

8. Nope lol.

Keep trying.

Fyi.... I have a very limited knowledge of vectors (cross product, dot product, etc.) right now so I kinda can't figure out how you're deriving the equations but that's for another time.

9. Ok no problem! Then a brief explanation

You gave some points in the space, $\displaystyle (0,0,3), (0,2,0)$ and $\displaystyle (0,1,0)$. it's possible to form vectors by substracting the coordinates of two given points, with three points we form the vectors $\displaystyle (0,-2,3)$ and $\displaystyle (-1,0,3)$ As those vectors are the sides of tetrahedron, they sall belong to the plane we are looking for

From cross product theory, the vector $\displaystyle u \times v = w$ its perpendicular to both $\displaystyle u$, and $\displaystyle v$, that's it, to the plane that belong to those vectors.

Now from dot product theory, if we take some vector with coordinates $\displaystyle (x,y,z)$ wich has its end (thinking in vectors as oriented segments) in the point $\displaystyle (x,y,z)$ and another vector $\displaystyle (A,B,C)$ wich is perpendicular to the plane, the dot product is zero. Explicitly: $\displaystyle (A,B,C) \cdot (x,y,z) = 0$. Then we have $\displaystyle Ax + By + Cz = 0$. Our $\displaystyle (A,B,C)$ is $\displaystyle (6,3,2)$ so this is were my equation comes from.

You're saying that there's something wrong, so there should be some fail in derivation or computation, but at least we're trying!