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Distances between planes (intro lin alg class)

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Im working currently with vectors. The question asks for the distance between two planes given by the two following equations:
    x + y -2z = 0
    3x + 3y -6z = 1

    2. Relevant equations

    I know the planes, H1, and H2 are parallel, so I can pick any random point on either, call the point Q, use the other plane as my H, and apply the following formula:

    distance(H,Q) = (q-p) dot (n/|n|)

    3. The attempt at a solution

    Im assuming that someone familiar with vectors and lin. alg is going to answer this so Im not writing the whole essay. But yea if more explanation is needed then let me know.

    For H I chose the first plane, so that n from plane 1 is (1, 1, -2). Correct me if Im wrong but I understand that you do need to pick the appropriate plane/line to use the right n? Or can it be either?
    For Q on plane 2 I did - (1/3, 0, 0)
    For a point P on plane 1 I chose (0,0,0)

    After using the corresponding position vectors for the two points, I plug them in the formula and get the result:

    (1)/(3√6), but my answers say it is (√6)/(9)

    Thanks all.
  2. jcsd
  3. Feb 8, 2012 #2


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    You mean (1/3, 0,0,) for Q, do you?

    The correct answer is 1/(3√6)

  4. Feb 8, 2012 #3
    Yea. I use (1/3, 0, 0) as my Q, and purposefully taking it from plane 2.

    Would you mind following up on the other question I snuck in there as well? About picking correct "n"?
  5. Feb 9, 2012 #4


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    Yes, of course. Show the problem.

  6. Feb 9, 2012 #5
    Ok so that is where the question originated but I will rephrase it....

    When picking "n", if the lines are parallel, I can pick either plane 1 or plane 2 to extract "n" right? Or is there any sort of decision I need to make?

    Furthermore, if I were to pick the second plane 3x + 3y -6z = 1, my "n" from this plane could be any multiple of that expression, which is why I could say n = (1,1,-2) or n = (3, 3, -6).

  7. Feb 9, 2012 #6


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    The equation of a plane comes from that any vector in the plane is perpendicular to the normal vector, so their dot product is zero: (r-p)n=0. In components: xnx+yny+znz=pxnx+pyny+pznz=constant

    The magnitude of the normal vector does not influence the zero dot product. If it is (1,1,-2) , you equally can choose any vector which is a multiple of this. So (3,3,-6) is also a normal vector, or (-1,-1,2)...Often we choose a unit normal vector, with the components divided by the magnitude.

    In the problem, you had two planes. And you found that the normal vectors were multiples of each other. If the normal vectors are of identical direction, the planes are parallel.

  8. Feb 9, 2012 #7
    Thanks so much!
  9. Feb 9, 2012 #8


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    You are welcome:smile:

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