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Math Help - Finding common plane

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    Finding common plane

    Given the surfaces x+2y+4=\ln z and x^2-xy+z+5=8x, they have a common plane which passes through (2,-3,1), how can I get the common plane?
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    Quote Originally Posted by Soldier View Post
    Given the surfaces x+2y+4=\ln z and x^2-xy+z+5=8x, they have a common plane which passes through (2,-3,1), how can I get the common plane?
    Please, please, please, state the problem as it is actually given to you! I am absolutely cerrtain the problem does NOT say "they have a common plane" because there is no such thing as a "common plane" for two surfaces or even "a plane" for a given surface. I can guess that you mean "a common tangent plane" but that missing word is very important!

    Assuming you really mean "common tangent plane", write the first surface as f(x,y,z)= x+ 2y- ln(z)= -4 so that we can think of the surface as "level surface" for f. The gradient of f, \vec{i}+ 2\vec{j}- \frac{1}{z}\vec{k} is a normal vector to that surface and so to its tangent plane, at any given x, y, z. Taking the point at which the plane is tangent to the surface as (x_0, y_0, z_0) then any tangent plane to the surface at that point and that passes through (2, -3, 1), is of the form (x- 2)+ 2(y+ 3)+ \frac{1}{z_0}(z- 1)= 0.

    Similarly,we can write the second surface as g(x,y,z)= x^2- xy+ z- 8x= -5 so that grad g= (2x-y- 8)\vec{i}- x\vec{j}+ \vec{k} is normal to the surface and tangent plane. Taking (x_1, y_1, z_1) as point of tangency, the tangent plane that contains (2, -3, 1) is of the form (2x_1- y_1-8)(x- 2)- x_1(y+ 3)+ (z- 1)= 0.

    The fact that this is a common tangent plane means that we must have ](x- 2)+ 2(y+ 3)+ \frac{1}{z_0}(z- 1)= (2x_1- y_1-8)(x- 2)- x_1(y+ 3)+ (z- 1)= 0 for all x, y, z and so that "corresponding coefficients" are equal: we must have 1= 2x_1- y_1- 8, 2= -x_1, and \frac{1}{z_0}= 1.
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