Question

find the equation of sphere circumscribing a tetrahedron ,bounded by the planes X=0,y=0,z=0 and X+y+z=1​

Answer 1

Answer:

I think there is some misprint in writing the equation of the last the four planes i.e. (x+y+z) = 0 . It should have been like (x+y+z) = 1 or k, not equal to zero. Because if x+y+z = 0, then the four planes do not form a tetrahedron. So taking (x + y + z) = 1 as the equation of the last plane, we may proceed as follows;

The required equation of the sphere passes through the four vertices of the tetrahedron formed by the intersection of any three of the four given planes viz x + z = 0, y + z = 0, x + y= 0 & x + y + z = 1. Solving, taking any three of these eqns. together, we get, the four vertices of tetrahedron as O(0,0,0), A(1,-1, 1),B(1,1,-1) & C(-1,1,1). Now the equation of the sphere passing through these four points can easily be found to be ;

x^2 + y^2 + z^2 - 3(x + y + z) = 0

Step-by-step explanation:

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