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Math Help - an equation of a plane

  1. #1
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    an equation of a plane

    Find an equation of the plane that contains the point (-1,0,0) , and is perpendicular to the plane 4x-8y+3z=-14.

    What about this question? How should I solve it? Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    Find an equation of the plane that contains the point (-1,0,0) , and is perpendicular to the plane 4x-8y+3z=-14.

    What about this question? How should I solve it? Thank you very much.
    One way of going about this (not necessarily the best but one that
    comes to mind) is the following.

    All the points p(lambda)=(0,7/4,0)+lambda(4,-8,3) are on a line perpendicular
    to the given plane. This is because the vector (4, -8, 3) is the gradient
    of 4x-8y+3z+14, and hence is normal to the given plane.

    Then a plane perpendicular to the given plane through the given point is:

    Ax+By+Cz=K,

    where:

    1.) -A = K (force the given point (-1,0,0) onto the plane))
    2.) (7/4) B = K (force the point p(0) onto the plane)
    3.) 4A + (7/4-8)B +3C=K (force the point p(1) onto the plane)

    Which defines all three coefficients in terms of K. So pick any
    non-zero value for K, and the equation of the plane through
    the three points follows.

    RonL
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