[helpmee1234567890]

Find the vertices of the tetrahedron whose faces lie in the planes x − y = 0,
y − z = 0, x + y = 1 and z = 0

Answer: Vertices (0, 0, 0), (1, 0, 0), (1/2, 1/2, 0), (1/2, 1/2, 1/2)
anyone know how i would do this?

[Stonebridge]

Post maths questions in the maths section for a better chance of reply.

[ztibor]

(Original post by helpmee1234567890)
Find the vertices of the tetrahedron whose faces lie in the planes x − y = 0,
y − z = 0, x + y = 1 and z = 0

Answer: Vertices (0, 0, 0), (1, 0, 0), (1/2, 1/2, 0), (1/2, 1/2, 1/2)
anyone know how i would do this?

Let be the base face the z=0 plane
Find the intersections of pairs of the plane from the other 3 planes.
THese will be the equations of 3 lines.
F.e
x-y=0 and y-z=0 planes
normal vektors: (1,-1,0) and (0,1,-1)
The vector product:
i j k
1 -1 0 -> i+j+k ->(1,1,1)
0 1 -1
A common point of the 2 planes:
solve simoultaneously
x-y=0
y-z=0
adding the 2 equations x-z=0 ->x=z let x=1->z=1->y=1
So the equation of the line
x=1+t y=1+t z=1+t
This line intersect the z=0 plane: (substituting x=1+t, y=1+t, z=1+t)
1+t=0 so t=-1
the vertex (with t=-1) x=0, y=0, z=0 -> (0,0,0)

the x-y=0 and x+y=1 planes
normal vectors? (1,-1,0) and (1,1,0)
Vector product
i j k
1 -1 0 -> 0*i -0*j+2k ->(0,0,2)
1 1 0
A common point:
x-y=0
x+y=1 ->2x=1 ->x=1/2 ->y=1/2 perpendicular to z=0
x=1/2 y=1/2 z=2t the equation of teh line ->2t=0 ->t=0
so this intersect z=0 in (1/2,1/2,0)

An so for y-z=0 and x+y=1

The 3 line meet in the 4th point