1. ## 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?

2. Originally Posted by Soldier
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 $\frac{1}{z_0}= 1$.