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Homework Help: Finding the scalar equation for a plane

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the scalar equation for the plane containing L1 and L2.
    Consider the lines:
    L1 : x = t + 1, y = 2t, z = 3t - 1
    L2 : x = s - 1, y = 2s + 1, z = 3s - 1

    2. Relevant equations
    Scalar equation for a plane:
    a(x - x0) + b(y - y0) + c(z - z0) = 0

    3. The attempt at a solution
    These two lines have the same vector <1, 2, 3>, so they are parallel. I let t = 0 and s = 0 so that I found a point (1, 0, -1) for L1 and a point (-1, 1, -1) for L2. Where do I go from here? Thanks and I appreciate it.
     
  2. jcsd
  3. Nov 7, 2011 #2

    ehild

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    Find the vector which connects those points. It will not be parallel with the original lines.

    ehild
     
  4. Nov 7, 2011 #3
    Ok so I found the displacement vector between the two points to be (-1 - 1, 1 - 0, -1 -(-1)), which comes up to (-2, 1, 0). Now that I have the vector, does it matter which point I use to plug into the formula to find the scalar equation?
     
  5. Nov 7, 2011 #4

    ehild

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    You have two vectors lying in the plane, but you need a normal vector of the plane to get the scalar equation. How do you get a vector that is normal to the vectors of the plane?

    What does the scalar equation of the plane mean? What are the parameters a,b,c?

    ehild
     
  6. Nov 7, 2011 #5
    Okay, I took the cross product with the displacement vector <-2, 1, 0> and the other vector <1, 2, 3> which ended up as <-3, -6, 5>. So now I would use that vector and pick one of those points to come up with the scalar equation. I end up with -3x - 6y + 5z = -8
     
  7. Nov 7, 2011 #6

    HallsofIvy

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    Now, it is easy to check if that is correct, isn't it? The line L1 is given by x = t + 1, y = 2t, z = 3t - 1. Putting that into your equation for the plane, -3(t+1)- 6(2t)+ 5(3t-1)= -3t- 3- 12t+ 15t- 5= (-3- 12+ 15)t+ (-3- 5)= -8 which is correct for all t. Every point of L1 is in that plane. The line L2 is given by x = s - 1, y = 2s + 1, z = 3s - 1. Putting that into your equation, -3(s- 1)- 6(2s+ 1)+ 5(3x- 1)= -3s+ 3- 12s- 6+ 15x- 5= (-3-12+15)s+ (3- 6- 5)= -8 which is correct for all s. Every point of L2 is in that plane.
     
  8. Nov 7, 2011 #7
    Thanks a lot for the help! To follow up on this problem. If I wanted to find the distance between the lines, would I just use the distance formula between the two points?
     
  9. Nov 7, 2011 #8

    ehild

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    The distance between two parallel lines is the length of the segment of line, perpendicular to them. You know already a vector which connects the two lines: find its normal component.

    ehild
     
  10. Nov 8, 2011 #9
    Using the displacement vector <-2, 1, 0>, the component would be ( (-2)^2 + 1^2 + 0^2 ) ^ (1/2), which ends up as square root of 5. That would then be the distance correct?
     
  11. Nov 8, 2011 #10

    ehild

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    As the displacement vector happens to be perpendicular to the lines, the distance is correct.


    ehild
     
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