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Finding the scalar equation for a plane

  • Thread starter silver1x
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  • #1
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Homework Statement


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

Homework Equations


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

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.
 

Answers and Replies

  • #2
ehild
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Find the vector which connects those points. It will not be parallel with the original lines.

ehild
 
  • #3
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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?
 
  • #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
 
  • #5
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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
 
  • #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.
 
  • #7
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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?
 
  • #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
 
  • #9
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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?
 
  • #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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