The equation of a sphere circumscribing a tetrahedron whose faces are x=0,y=0,z=0 and x/a+y/b+z/c=1 is
Answers
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
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