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!

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- Nov 12th 2010, 07:50 PMr2d2Vector 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 PMmr fantastic
- Nov 13th 2010, 03:44 AMHallsofIvy
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. - Nov 13th 2010, 07:37 AMr2d2Attempt 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 PMHallsofIvy
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.