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Distance between 2 planes

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the shortest distance between the 2 planes:

    2x+2y-z=1 and 4x+4y-2z=5


    How do we approach this problem?
    I used the approach of finding the point at which the normal of one plane intersects the other plane and then determining the length of this vector. The answer I've got is 5/6 which is not correct. The correct answer is 1/2.

    What is wrong with my approach? And how should we approach this problem?

    Thanks.
     
  2. jcsd
  3. Sep 28, 2011 #2

    dynamicsolo

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    You'll need to show what you calculated in order for someone to judge what you might have done wrong. Your verbal description sounds fine as far as it goes...
     
  4. Sep 28, 2011 #3
  5. Sep 28, 2011 #4

    dynamicsolo

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    I'm a bit unclear about how you are setting up what you call your "point of interesection" equations. But you can do this:

    You know that the common normal vector for the parallel planes is < 2, 2, -1 >. Pick a point in the plane 2x + 2y - z = 1 ; we can make this easy and use ( 0, 0, -1 ). Then this normal line passing through the chosen point has the vector equation

    < x, y, z > = < 2t , 2t, -t - 1 > .

    This line intersects the second plane for 4x + 4y - 2z = 5 , or

    4 · 2t + 4 · 2t - 2 ( -t - 1 ) = 5 --> 18t + 2 = 5 ---> t = 1/6 .

    The point of intersection in that plane is then ( 2 · 1/6 , 2 · 1/6 , -1/6 - 1 ) = ( 1/3 , 1/3 , -7/6 ). The distance between the two points along this line mutually perpendicular to the two planes (what is called the "perpendicular distance", the shortest distance between the planes) is given by

    [tex]\sqrt{ ( \frac{1}{3} - 0 )^{2} + ( \frac{1}{3} - 0 )^{2} + ( -\frac{7}{6} - [-1] )^{2} } = \sqrt{ ( \frac{1}{3} )^{2} + ( \frac{1}{3} )^{2} + ( -\frac{1}{6} )^{2} } = \sqrt{ \frac{1}{9} + \frac{1}{9} + \frac{1}{36} } [/tex]

    [tex]= \sqrt{ \frac{4 + 4 + 1}{36} } = \sqrt{ \frac{9}{36} } = \frac{1}{2} .[/tex]

    I think the problem you may have made for yourself is that you didn't actually choose a point of intersection in either plane to build a normal line from.

    A more general argument along these lines gives the perpendicular distance between two parallel planes ax + by + cz = d1 and ax + by + cz = d2 as [itex] D = \frac{\vert d_{1} - d_{2} \vert}{\sqrt{a^{2} + b^{2} + c^{2} } } . [/itex] For this problem, we would write

    2x + 2y - z = 1 and 2x + 2y - z = 5/2 ;

    The square root in the denominator gives 3 and the absolute value of the difference in the numerator is 3/2 , so the formula yields D = (3/2) / 3 = 1/2 .
     
    Last edited: Sep 28, 2011
  6. Sep 28, 2011 #5
    "Then this normal line passing through the chosen point has the vector equation
    < x, y, z > = < 2t , 2t, -t - 1 > "

    Is this derived from :

    r(t) = r0 + tv ,where r0 and v are vectors

    which is the vector equation for a straight line?
     
  7. Sep 28, 2011 #6

    dynamicsolo

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    Yes, that's right: this is just using vector notation, rather than writing three separate parametric equations for the three coordinates of the points on the line.
     
  8. Sep 29, 2011 #7
    You can use this "formula" but you can attempt to prove its derivative.

    Given a point [tex]P(x_{0},y_{0},z_{0})[/tex], and the equation of a plane is ax+by+cz+d=0,

    The distance from the point to the plane is:

    D=[tex]\mid \frac{ax_{0}+by_{0}+cz_{0}+d}{\sqrt{a^{2}+b^{2}+c^{2}}[/tex]
     
  9. Sep 29, 2011 #8
    Got it! Yes. I did not pick a point on the first plane through which the normal vector passes.

    Thanks.
     
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