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Math Help - Vector calculus - planes

  1. #1
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    Exclamation Vector calculus - planes

    A plane has equation x- 2y +z = 4. Show that the square, say D^2, of the distance from the point P = (-1,3,2) to a point (x,y,z) on the plane is given by
    D^2 = 2x^2 - 2x + 14 + 5y^2 + 2y - 4xy.
    By finding the minimum of D^2 determine the shortest distance from P to the plane. (You should verify that the distance is indeed a minimum.) Let Q be the point on the plane clasest to P. Show that the line QP is perpendicular to the plane at Q.
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  2. #2
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    Quote Originally Posted by njr008 View Post
    A plane has equation x- 2y +z = 4. Show that the square, say D^2, of the distance from the point P = (-1,3,2) to a point (x,y,z) on the plane is given by
    D^2 = 2x^2 - 2x + 14 + 5y^2 + 2y - 4xy.
    By finding the minimum of D^2 determine the shortest distance from P to the plane. (You should verify that the distance is indeed a minimum.) Let Q be the point on the plane clasest to P. Show that the line QP is perpendicular to the plane at Q.
    D^2 = (x - [-1])^2 + (y - 3)^2 + (z - 2)^2.

    Now substitute z = 4 - x + 2y. Expand. Simplify. Spadework for you to do.

    ------------------------------------------------------------------------

    Let f(x, y) = 2x^2 - 2x + 14 + 5y^2 + 2y - 4xy and find extrema in the usual way:

    Solve simultaneously:

    \frac{\partial f}{\partial x} = 0 .... (1)

    \frac{\partial f}{\partial y} = 0 .... (2)

    Spadework for you.

    --------------------------------------------------------------------------

    Get the vector QP and show that QP is parallel to the normal to the plane (which is seen from the Cartesian equation of the plane to be i - 2j + k).
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