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Math Help - Tangent plane

  1. #1
    Junior Member
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    Tangent plane

    Find a point on the surface where the tangent plane is perpendicular to the following tangent plane.
    <br /> <br />
x^2+2y^2+3z^2=66<br /> <br />
    at the point(6,3,2)

    so i got the gradient
    <2x,4y,6z>
    grad of(6,3,2)=<1,1,1>(<12,12,12>)

    now how to go about finding a point on the surface where the tangent plane is perpendicular to the original plane?
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  2. #2
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    Quote Originally Posted by statman101 View Post
    Find a point on the surface where the tangent plane is perpendicular to the following tangent plane.
    <br /> <br />
x^2+2y^2+3z^2=66<br /> <br />
    at the point(6,3,2)

    so i got the gradient
    <2x,4y,6z>
    grad of(6,3,2)=<1,1,1>(<12,12,12>)

    now how to go about finding a point on the surface where the tangent plane is perpendicular to the original plane?
    Do you mean find another point, say (x, y, z), on that same surface so that the tangent plane at (x, y, z) is perpendicular to the tangent plane at (6, 3, 2)?
    Okay, \nabla f= <2x, 4y, 6z> as you say. At (6, 3,2 ) that is <2(6), 4(3), 6(2)= <12, 12, 12> which is, as I think you mean, parallel to <1, 1, 1>. And that is perpendicular to the tangent plane at (6, 3, 2). Two planes are perpendicular if and only if their normal vectors are perpendicular. You want to find (x, y, z) such that <2x, 4y, 6z>\cdot<1, 1, 1>= 0 and, of course, such that x^2+ 2y^2+ 3z^2= 66.
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