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?
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?
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
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!