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Another vector proof

  1. Sep 5, 2005 #1
    hey guys, i am stuck on one more problem. Can anyone guide me onto the right path on how to start this?

    Use this formula for the perpendicular distance between a point and a plane
    D= |ax+by+cz-d| all over the square root of (a^2+b^2+c^2)

    to show that the perpendicular distance D between the two parallel planes ax+by+cz+d1=0 and ax+by+cz+d2=0 is

    D= |d1-d2| all over the square root of (a^2+b^2+c^2)
     
  2. jcsd
  3. Sep 5, 2005 #2

    AKG

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    For a plane {(x, y, z) : ax + by + cz + d = 0} and a point (x', y', z'), you have that the distance from the point to the plane is:

    [tex]D = \frac{|ax' + by' + cz' + d|}{\sqrt{a^2 + b^2 + c^2}}[/tex]

    The distance from one plane to another parallel plane is obviously the same as the distance from that plane to any single point on the other plane. Suppose (x1, y1, z1) is on plane 1. That means that this point satisfies the equation:

    ax1 + by1 + cz1 + d1 = 0

    If we isolate d1, we get:

    d1 = -(ax1 + by1 + cz1)

    -d1 = ax1 + by1 + cz1

    Now the distance between plane 2 and plane 1 is the same as the distance between plane 2 and (x1, y1, z1) since (x1, y1, z1) is on plane 1. Using the given formula:

    [tex]D = \frac{|ax_1 + by_1 + cz_1 + d_2|}{\sqrt{a^2 + b^2 + c^2}}[/tex]

    [tex]D = \frac{|-d_1 + d_2|}{\sqrt{a^2 + b^2 + c^2}}[/tex]

    [tex]D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}[/tex]
     
  4. Sep 5, 2005 #3
    i get that, but i dont get your last algebraic step. how did the -d1+d2 turn into d1-d2
     
  5. Sep 5, 2005 #4

    robphy

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    The question should really be
    "how did |-d1+d2| turn into |d1-d2|?"
    The answer is, of course, "because |-A|=|A|".
     
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