Equation Of Plane Tangent To Two Functions

dagr8est

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Nov 2, 2004
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"The plane L is tangent to the x^2+y^2+z^2 = 1 at point (1/3, sqrt(8)/3, 0). The plane L is also tangent to the sphere (x-a)^2+y^2+(z-c)^2 = 4 at the point Q = (-7/3, 2sqrt(8)/3, 0). Find a and c."

So I did:

f(x,y,z) = x^2+y^2+z^2-1 = 0
g(x,y,z) = (x-a)^2+y^2+(z-c)^2-4 = 0

grad(f) = <2x, 2y, 2z>
grad(g) = <2x-2a, 2y, 2z-2c>

grad(f) @ (1/3, sqrt(8)/3, 0) = <2/3, 2sqrt(8)/3, 0>
grad(g) @ (-7/3, 2sqrt(8)/3, 0) = <-14/3-2a, 4sqrt(8)/3, -2c>

L = 2/3(x-1/3)+2sqrt(8)/3(y-sqrt(8)/3) = 0
L = (-14/3-2a)(x+7/3)+4sqrt(8)/3(y-2sqrt(8)/3) = 0

I tried equating those 2 equations but I think I'm doing something wrong because the coefficient of y is independent of a and c so it should be the same for both but in one it is 2sqrt(8)/3y and in the other it is 4sqrt(8)/3y. I can't see what I am doing wrong though. Any help is appreciated. :D
 
dagr8est said:
"The plane L is tangent to the x^2+y^2+z^2 = 1 at point (1/3, sqrt(8)/3, 0). The plane L is also tangent to the sphere (x-a)^2+y^2+(z-c)^2 = 4 at the point Q = (-7/3, 2sqrt(8)/3, 0). Find a and c."

So I did:

f(x,y,z) = x^2+y^2+z^2-1 = 0
g(x,y,z) = (x-a)^2+y^2+(z-c)^2-4 = 0

grad(f) = <2x, 2y, 2z>
grad(g) = <2x-2a, 2y, 2z-2c>

grad(f) @ (1/3, sqrt(8)/3, 0) = <2/3, 2sqrt(8)/3, 0>
grad(g) @ (-7/3, 2sqrt(8)/3, 0) = <-14/3-2a, 4sqrt(8)/3, -2c>
You can solve for a and c from here.
 
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