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)