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    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?
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    (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

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