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Cross Product of Two Planes

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Find a plane perpendicular to the two planes, X+Y=3 and X+2y-z=4

    I know i take the cross product of both so i get

    <1,1,0> and <1,2,-1>

    But when i do the cross product i get x-y+z

    book tells me x-y-z

    what am i doing wrong? Not sure why the z is negative. The cross product formula tells me. that the only the J component is negative. (Using Rogawskis Multivariable book. Page. 698, 13.4.)
     
  2. jcsd
  3. Sep 20, 2011 #2

    Dick

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    The book is right. Your cross product must be wrong. Can you show how you got it?
     
  4. Sep 20, 2011 #3
    This is what i did.

    |i j k |
    |1 1 0|
    |1 2 -1| =

    *simplified*

    |-1-0|i-|-1-0|j+|2-1|k

    which gives me

    i - j +k
     
    Last edited: Sep 20, 2011
  5. Sep 20, 2011 #4

    Dick

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    There's some mistakes in there alright. Take the i component. I get the coefficient to be 1*(-1)-0*2=(-1). Check your examples again.
     
  6. Sep 20, 2011 #5

    dynamicsolo

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    Are those really absolute value operations you're doing there? If so, they don't belong in the calculation.
     
  7. Sep 20, 2011 #6
    I thought that there were supposed to be absolute values in the equation.

    If there are not, the way i would calculate it would be then:

    -i + j + k
     
  8. Sep 20, 2011 #7

    Dick

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    Now that's right! Thanks for picking up on the problem, dynamicsolo.
     
  9. Sep 20, 2011 #8

    dynamicsolo

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    No, they shouldn't be there.

    Much better. You now have the direction vector for a line mutually perpendicular to the two given planes and thus the normal vector to the plane you seek. You next need to find a point that lies in that plane in order to write its equation.
     
  10. Sep 20, 2011 #9
    Okay so i just used (2, 1, 0) final equation -x+y+z=-1

    but the book is saying x-y-z=f

    I guess they just wanted the normal but why -y-z???
     
  11. Sep 20, 2011 #10
    Is it something as simple as multiplying by -1? if so, i feel stupid. but thanks on the absolute values, that would have been a face palm if i hadn't known that on the test day.
     
  12. Sep 20, 2011 #11

    dynamicsolo

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    There is no single way to write the normal vector for a plane. Any non-zero scalar (numerical) multiple of the vector will serve; multiplying by a negative number changes the signs of all the components, but that just gives you a vector which points in exactly the opposite direction along the same line.

    So you may use < -1, 1, 1 > or < 1, -1, -1 > (or, heck, < 2011, -2011, -2011 > , were you so inclined) to represent a normal vector to the desired plane. Flipping the signs on the equation for the plane (or multiplying the equation by any non-zero real number) just gives another name for the same plane, since the set of points that satisfy the algebraic equation will remain unchanged.
     
  13. Sep 20, 2011 #12
    Yea i figured that. Thanks so much Dick and Dynamicsolo! Don't worry ill be back. lol
     
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