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Scalar Equations of a plane in space

  1. Feb 13, 2006 #1
    One question is bothering me..

    Find the scalar equation of the plane that contains the intersecting lines


    and (x-2)/-3=y\4=(z+3)\2

    What i've tried is doing the cross product of (1,2,3)and(-3,4,2)
    I get a Normal and then put it into scalar form...

    Substitute (2,0,-3) into it to find D and I get a wrong answer....

    Can anyone help, by the way it has to be done geometricaly because that's the course,.,,,thanks
  2. jcsd
  3. Feb 13, 2006 #2


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    It's hard to help without knowing exactly what you did. Yes, taking the cross product of (1, 2, 3) and (-3, 4, 2) will give a vector perpendicular to the plane. What result did you get?
    I'm guessing that you used the normal vector together with a point in the plane to find the equation of the plane. The obvious choice for that point, I would think, would be (2, 0, -3). That is, if <A, B, C> is your normal vector then A(x- 2)+ By+ C(z+ 3)= 0 is the equation of the plane.
    I take it your D is -2A+ 3C. How do you know you got a wrong answer? Were you told what the answer should be?
  4. Feb 13, 2006 #3
    Ok I did (1,2,3)x(-3,4,2)=(-8,-11,10)

    tHEN I wrote -8x-11y+10z+D=0

    Substituting the P(2,0,-3) into that -8(2)+10(-3)+D=0


    But the answer in the book is 10x+11y-10z-50=0
  5. Feb 14, 2006 #4


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    Try (1, 2, 3)x(-3, 4, 2) again! It's not just your D that is wrong.
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