1. ## Vector Problem

Find all the points on the surface x^2+y^2-z^2=-1 where the tangent plane is perpendicular to the vector <1,0,2>.

I'm confused!

2. Originally Posted by r2d2
Find all the points on the surface x^2+y^2-z^2=-1 where the tangent plane is perpendicular to the vector <1,0,2>.

I'm confused!
Get the vector normal to the surface. Equate this vector to <1, 0, 2>. Solve the resulting equations simultaneously for x, y and z.

3. Originally Posted by r2d2
Find all the points on the surface x^2+y^2-z^2=-1 where the tangent plane is perpendicular to the vector <1,0,2>.

I'm confused!
What exactly are you confused about? Do you know that if f(x, y, z)= constant, then $\displaystyle \nabla f$ is normal to the surface (and so to its tangent plane). Do you know that two vectors are parallel if and only if one is a multiple of the other?

I disagree slightly with mr fantastic- equating the normal vector to the given vector will give a point that is NOT on the surface. Since you were asked to find points on the surface where the tangent plane is perpendicular to the vector <1, 0, 2>, set the normal vector equal to a multiple of that vector, say <a, 0, 2a>. That will give a "curve" of points, two of which are on the surface.

4. ## Attempt at solution.

The attempt at a solution:

So, the gradient would be: <2x,2y,-2z>, but what do I do with the gradient vector to find these points?

5. The gradient vector at each point is perpendicular to the tangent plane at that point. Since you want the tangent plane to be perpendicular to <1, 0, 2>, you want a point (x, y, z) on $\displaystyle x^2+ y^2- z^2= -1$ such that <2x, 2y. -2z> is a multiple of that- such that < 2x, 2y, 2z>= <a, 0, 2a> for some number a.