# Vector Problem

• Nov 12th 2010, 07:50 PM
r2d2
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!
• Nov 12th 2010, 08:55 PM
mr fantastic
Quote:

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.
• Nov 13th 2010, 03:44 AM
HallsofIvy
Quote:

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 $\nabla f$ 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.
• Nov 13th 2010, 07:37 AM
r2d2
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?
• Nov 13th 2010, 12:03 PM
HallsofIvy
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 $x^2+ y^2- z^2= -1$ such that <2x, 2y. -2z> is a multiple of that- such that < 2x, 2y, 2z>= <a, 0, 2a> for some number a.