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Linear Algebra: Finding A k-value

  1. Jul 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the value of k such that no solutions are shared between the planes

    x + 2y + kz = 6

    and

    3x + 6y + 8z = 4


    2. Relevant equations



    3. The attempt at a solution

    I figured that if I calculated [itex]\vec{n_1} \times \vec{n_2} = \vec{0}[/itex], and then take the magnitude of this vector equation, I could procure the correct k-value.

    Here are some parts of my work:

    [itex]\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}[/itex]

    [itex]0=320 -240k + 39k^2[/itex]

    [itex]k = \frac{120 \pm 8 \sqrt{30}}{39}[/itex]

    When I graphed the two planes, it was quite conspicuous that the two planes were not parallel.

    What is wrong with my work and reasoning?
     
  2. jcsd
  3. Jul 2, 2013 #2

    Mark44

    Staff: Mentor

    You're making this much harder than it needs to be. The only way the planes won't share any points is if they are parallel and distinct (not coplanar). This problem can be solved by inspection; you don't need cross products or much else for this problem.
     
  4. Jul 2, 2013 #3

    LCKurtz

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    You want ##\vec n_1 \times \vec n_2## to be the zero vector, so just set its components equal to 0.

    Wouldn't it be easier to note that the planes are parallel if the normal vectors are proportional?
     
    Last edited: Jul 2, 2013
  5. Jul 2, 2013 #4

    LCKurtz

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    Also, that would have worked except you have 39 where you should have 45.
     
  6. Jul 5, 2013 #5
    Thank you both for your help.
     
  7. Jul 5, 2013 #6

    mfb

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    2016 Award

    Staff: Mentor

    k (scalar parameter) and ##\hat{k}## (direction in space) are completely unrelated, they just share the same letter here. Both brackets have to be zero.

    I agree with Mark44, this can be solved just by looking at it.
     
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