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Equation of Parallel Plane Equidistant to a Point

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the plane parallel to 2x - y + 2z = 4 if the point P = (3,2,-1) is equidistant from both planes.

    2. Relevant equations

    a(x-x0) + b(y-y0) + c(z-z0) = 0

    3. The attempt at a solution

    I found the normal vector to the given plane first, as this is quite easy. [itex]\stackrel{\rightarrow}{n}[/itex] = <2,-1,2>, which is also the normal vector for the plane I am finding since they are parallel.

    After that, I found a point that is on the given plane. For simplicity, I picked Q = (2,0,0). I then found the equation of the vector from the given point and the point I found, which is [itex]\stackrel{\rightarrow}{PQ}[/itex] = <1,2,-1>. I used this vector and the normal vector to find the scalar projection of [itex]\stackrel{\rightarrow}{PQ}[/itex] along [itex]\stackrel{\rightarrow}{n}[/itex]. This came out to be [itex]\frac{-2}{3}[/itex].

    So, [itex]\frac{-2}{3}[/itex] is the distance from the point to both planes. This is where I got stuck.

    On my first attempt, I tried to find a point that, when I used it to find the scalar projection again, gave me the same distance.

    If the scalar projection is [itex]\frac{-2}{3}[/itex], that means that [itex]\stackrel{\rightarrow}{n}[/itex][itex]\bullet[/itex][itex]\stackrel{\rightarrow}{PQ}[/itex] = -2. So:

    2a - b + 2c = -2

    Now, I'm not sure if I can pick any point that makes that true and it gives me a valid equation. I assume that as long as the scalar projection is the same, and the normal vector is the same, the equation would be a valid, parallel plane?

    I used the point (-1,0,0), which gives me the equation 2(x+1)- y + 2z = 0

    I would like to know if this is correct. Since afterward, I looked at the work of my professor and he used a different method that I. Here's what he did.

    He started out the same, by picking a point on the given plane and finding [itex]\stackrel{\rightarrow}{PQ}[/itex] and finding the scalar projection. After that, he used this equation:

    point on parallel plane = (starting point) + distance<normal vector>

    So he got...

    (3,2,-1) - 2/3<2,-1,2> = (5/3, 8/3, -7/3)

    He then used that point to write his equation. His method makes sense to me intuitively, and seems more "concrete" than my method. The problem is, I decided to calculate the scalar projection with his calculated point and I something 2 instead of -2/3. This would imply to me that the second plane is not equidistant to the point.

    So what I would like to know here, which method is correct, if either? Was I wrong to assume that I should just guess a point and that point would work? Was his equation wrong? Are we both wrong?

    Thanks!
     
  2. jcsd
  3. Sep 9, 2012 #2

    vela

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    Your professor should have used the unit-length normal vector when calculating where the point on the other plane is.
     
  4. Sep 9, 2012 #3
    Oh yes, that makes sense. I was thinking that distance * normal vector would not give the proper distance, but I couldn't figure out why.

    So is my method correct or incorrect?

    And one last general question...is it correct to have a negative scalar projection, or should it be given as positive? Or does it not matter (so -2/3 and 2/3 is the same thing)?
     
  5. Sep 9, 2012 #4

    vela

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    With the vectors you used, the scalar projection is -2/3. I wouldn't say that -2/3 is the distance of the point from the plane, however, since distances are positive quantities.

    I don't think your answer is correct. I got 2x-y+2z=0 for the other plane. (But I may have messed up as well.)
     
  6. Sep 9, 2012 #5

    LCKurtz

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    Here's another approach you might try. You have a unit normal ##\hat n = \frac 1 3\langle 2,-1,2\rangle##, and using that in your formula gave ##-\frac 2 3##. This just means your normal points to the other side of the plane than where the given point is.

    You know the plane you seek must have the form ##2x-y+2z=k##. It will be on the other side of the given point so the same normal will give a positive scalar projection. An easy point on that plane is ##(0,-k,0)##. So do the same work and choose ##k## to get ##\frac 2 3##. I think you will discover Vela's answer is correct.
     
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