Finding points on surface whose tangent plane has a given normal vector

Problem: Find the points on the graph of z = 3x^{2} - 4y^{2} at which the vector **n** = < 3, 2, 2 > is normal to the tangent plane

I tried finding L(x,y) at point (a,b) and solving for a and b before plugging these values back into the equation for the surface and I got 3a^{2} - 4b^{2} = 6ax - 8by. I knew that the coefficients of the x, y, and z terms of the tangent plane equation would be the x, y, and z values of the normal vector, so I set 6a = 3 and -8b = 2 to get a = 1/2 and b = -1/4 however the answer is (-1/4, 1/8, 1/8).

I know I must be doing something wrong D:

Halp

Re: Finding points on surface whose tangent plane has a given normal vector

I assume you've found the equation of the tangent plane correctly, although you didn't explicitly write it down (I don't know the convention you used when you wrote "finding L(x, y) at point (a,b)").

The equation of the tangent plane at (a,b) is:

, or

From that, you can read off that the plane is normal to the vector (6a, -8b, -1). So far, so good, if I understood you.

Now you tried to set (6a, -8b, -1) = (3, 2, 2) and solve for a and b. That was your mistake. Look at the z-component for instance. There's no solution on your reasoning.

Here's what's happening: If a plane is normal to a vector , then it's also normal to , and , and .

That's because to be normal to a plane means that the vector defined by the difference between any two points in the plane is perpendicular to . And if that's perpendicularity holds with , then it holds with , and , and . It holds with any (non-zero) multiple of !!!

So it's not that your normal vector, (6a, -8b, -1) at (a, b), and the problems normal vector, (3, 2, 2), should be "equal", but rather they should represent the same direction, so that one should be a (real number) multiple of the other.

So what you should've done is this:

Find (a, b) such that for some .

Then , so

, so

the 3rd equation gives , so .