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

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Ocata
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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.
 
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Ocata said:
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,
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.
Ocata said:
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.
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.
 
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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.
 
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.