Question

Find the equation of sphere circumscribing tetrahedron whose faces are y/b+z/c=0, z/c+x/a=0,x/a+y/b=0,x/a+y/b+z/c=0

Answer 1

Answer:

Answer:

x² + y² + z² -ax -by -cz = 0

Step-by-step explanation:

Now to find the vertices of the tetrahedron, we need to consider to combinations of 3 planes intersection to get the vertex(Since any 3 faces of tetrahedron meet in a vertex).

Given faces of a tetrahedron as

P₁:x = 0, P₂:y = 0 ,P₃: z = 0 and P₄:x/a+y/b+z/c=1 .

P₁∩P₂∩P₃ , we get O(0, 0, 0)

P₁∩P₂∩P₄, we get C(0, 0, c)

P₁∩P₃∩P₄, we get B(0, b, 0)

P₂∩P₃∩P₄, we get A(a, 0, 0)

Now, we know all the 4 vertices of the tetrahedron,

Equation of a sphere will be in the form x² + y² + z² + lx + my + nz + p = 0

But this equation passes through all the 4 vertices of tetrahedron,

Substituting O(0,0,0) we get p = 0,

Substituting A(a,0,0) we get l = -a,

Substituting B(0,b,0) we get m = -b,

Substituting C(0,0,c) we get n = -c,

On substituting the above values of l,m,n andp in the sphere equation,

we get

  x² + y² + z² -ax -by -cz = 0