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Math Help - Vector Problem

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by r2d2 View Post
    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.
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  3. #3
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    Quote Originally Posted by r2d2 View Post
    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.
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  4. #4
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    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?
    Last edited by mr fantastic; November 13th 2010 at 01:14 PM. Reason: Moved from thread created by duplicate post.
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  5. #5
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    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.
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