Hello math help forums! This is my first post on these forums and I hope to come back here often. Go calc!

Okay, so I just finished a problem on my homework, but for some reason the logic I used in completing this problem seems odd. I don't know why though.

Here is the problem:

Find a point on the surface of $\displaystyle z=8-3x^2-2y^2$ at which the tangent plane is perpendicular to the line $\displaystyle x=2-3t, y=7+8t, z=5-t$. Find the equation of the tangent plane at that point.

First, I know that a plane in space is defined by normal vector to the plane and a point on that plane. Because the given line is perpendicular to the tangent plane, I used the direction numbers of the line to form a scalar equation for the tangent plane:

$\displaystyle -3(x-x_{0}) + 8(y-y_{0}) -1(z-z_{0})$

Where $\displaystyle (x_{0}, y_{0}, z_{0})$ is the point on the surface at which the tangent plane is perpendicular to the line.

The equation of the tangent plane is also given by:

$\displaystyle z-z_{0} = f_{x}(x_{0}, y_{0})(x-x_{0}) + f_{y}(x_{0}, y_{0})(y-y_{0})$

$\displaystyle f_{x}(x_{0}, y_{0}) = -6x_{0}$

$\displaystyle f_{y}(x_{0}, y_{0}) = -4y_{0} $

Thus,

$\displaystyle -6x_{0}(x-x_{0}) + -4y_{0}(y-y_{0}) + z_{0} - z = 0$

Since both equations equal 0, and because the point $\displaystyle (x_{0}, y_{0}, z_{0})$ is the same for each equation, I set the equations equal to each other. For some reason, I feel like I am doing something wrong at this point... I don't know why.

$\displaystyle -6x_{0}(x-x_{0}) + -4y_{0}(y-y_{0}) + z_{0} - z = -3(x-x_{0}) + 8(y-y_{0}) -1(z-z_{0}) $

The z terms cancel out, and I am left with:

$\displaystyle -6x_{0}(x-x_{0}) + -4y_{0}(y-y_{0})= -3(x-x_{0}) + 8(y-y_{0}) $

Thus,

$\displaystyle -6x_{0}=-3 \Rightarrow x_{0} = \frac{1}{2}$

and

$\displaystyle -4y_{0}=8 \Rightarrow y_{0} = -2 $

I then plugged these points into the original equation for the surface to obtain $\displaystyle z_{0} = \frac{-3}{4}$

Does this logic seem right?

Thank you for your help! It is greatly appreciated.

Sincerely,

Opiaboy