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Visualizing the plane, 1x+1y+1z = 0

  1. Jun 7, 2015 #1
    Hello,

    I made traces for the equation, x+y+z=0, but they don't seem to connect in an intuitive way as other equations do. For instance, even with x+y+z=1, I can make traces where the 3 lines connect to make a triangle in the first/positive quadrant. But x+y+z=0 has traces that all run through the origin. Not sure how to draw/connect traces for a plane whose traces seem to only intersect at the origin.
     
  2. jcsd
  3. Jun 7, 2015 #2
  4. Jun 7, 2015 #3

    Mark44

    Staff: Mentor

    x + y + z = 1 is a different plane than x + y + z = 0. The two planes are parallel, though, but don't share any points.
    The origin is a point on your plane. It might help to draw traces in the three coordinate planes. For example, in the x-y plane (where z = 0), the trace is the line x + y = 0, or equivalently, the line y = -x.

    To get a three-dimensional view of this plane, calculate two points other than the origin (which is on the plane). Those three points should give you some idea of how the plane looks.
     
  5. Jun 7, 2015 #4
    Thank you Spinner,

    I did graph it on paper but it still didn't make sense because I was tracing through the intercepts and I wasn't able to get a clear idea of the plane. In the equation, d = 0. Now I see that d = 0 implied the plane goes through the origin. Thank you.
     
  6. Jun 7, 2015 #5
    Thank you Mark44.

    I have a clear visualization of x + y + z = 0. Your first statement sealed the deal visually for me. Your suggestion to calculate points other than the origin allowed me to realize I can find intercepts other than 0 + y = 0, x + 0 = 0, and z + 0 = 0.

    Thank you.
     
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