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Math Help - Tangent Vector to a Space Circle

  1. #1
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    Tangent Vector to a Space Circle

    Find a vector tangent to the space circle:

    x^2+y^2+z^2=1
    x+y+z=0

    At the point (1/\sqrt{14}, 2/\sqrt{14}, -3/\sqrt{14})

    I should know how to do this, but it's been 4 years since I had multivariable calc and I don't remember a darn thing from that class. :-(
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  2. #2
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    Since the circle belongs to the plane x + y + z = 0, the tangent vector belongs to it as well. Also, the vector is perpendicular to the radius-vector (1/\sqrt{14}, 2/\sqrt{14}, -3/\sqrt{14}). So you have two equations: x + y + z = 0 and x + 2y - 3z = 0. I get (-5, 4, 1) up to proportionality.
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  3. #3
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    <br />
r_0=(1/\sqrt{14}, 2/\sqrt{14}, -3/\sqrt{14})<br />

    The normal vector to the tangent plane to the sphere is r_0.

    The normal vector of the given plane is p=(1,1,1).

    The vector perpendicular to r_0 and p is (cross product)

    <br />
n=p\; \times \; r_0.<br />
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