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Posted by on Monday, July 7, 2008 at 12:39pm.

Find the scalar equation of the plane through the points M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z +1 = 0. I know that normal of the latter equation is (3,2,6) but now what do I do? Thanks.

  • Calculus - , Monday, July 7, 2008 at 5:23pm

    You are on the right track, so (3,2,6) must be a direction vector on your new plane.
    you also have two points M(1,2,3) and N(3,2,-1), so the vector MN or (2,0,-1) is another direction vector.

    So by taking the cross-product of vectors (3,2,6) and (2,0,-1) I got (2,-6,1)

    which must be the normal to the new plane.

    So the new plane has equation
    2x - 6y + z = k
    but M(1,2,3) lies on it, so 2 - 12 - 1 = k
    k = -7

    The new plane has equation 2x - 6y + x = -7

    Check:
    1. Both M and N satisfy the new equation
    2. Is the dot-product of their normals zero? (3,2,6)∙(2,-6,1) = 6-12+6 = 0

    There you go!

  • typo correction - , Monday, July 7, 2008 at 5:28pm

    my line
    <but M(1,2,3) lies on it, so 2 - 12 - 1 = k >

    should say :
    but M(1,2,3) lies on it, so 2 - 12 + 3 = k

  • Calculus - , Monday, July 7, 2008 at 9:16pm

    I appreciate the help, but you made another mistake. You got a wrong answer, it should be (2,0,-4) not (2,0,-1) so thus, you used the wrong information. But thanks, I know what to do now :)

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