How can I visualize the plane y=z?

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The discussion focuses on visualizing the plane defined by the equation y=z. To understand this plane, one should first draw the y-z coordinate plane and visualize a line at a 45-degree angle representing y=z. This line extends into the x-axis, forming a plane that includes points of the form (x, a, a) for any values of x and a. The plane can be conceptualized as spanning vectors (1,0,0) and (0,1,1), and its orientation changes based on the observer's position relative to the x-axis. Overall, the plane can be imagined as a tilted surface that intersects the x-axis and the line y=z.
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Hello

I am trying to solve a problem that involves "the plane y=z". How exactly should I think of this plane? It's obviously one of three planes that can be formed between the three diffirent axes, but which one? What is the intuition behind "y=z"?

A really basic question to some of you, I bet. Any help would be appreciated. The understanding behind it all is what I really want to achieve (not just the right answer).

Thanks!
 
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In the y-z plane draw a line of 45 degrees. This line should be the cross section of the plane you want to imagine.
 
As Anshuman said, first draw a yz-coordinate plane and draw the line y= z (through the origin, at 45 degrees to the axes. Now imagine the x-axis coming out of the paper toward you and the plane being that line coming straight out.
For example, the plane contains the points (x, a, a) for any x or a.
 
another way to think of this plane, is that it consists of all vectors of the form:

x(1,0,0) + y(0,1,1), for any real numbers x,y.

that is, P = span({(1,0,0), (0,1,1)}).

this plane intersects the "x-axis" (the line x(1,0,0)) and the line y = z. (if we are looking perpendicular to the x-axis (from the "negative part" where x < 0, so the positive numbers are ahead of us) at the yz-plane, we would see our plane "tilted" 45 degrees to the left. if we were on the "positive side" of the x-axis, perpendicluar to it and looking towards the origin, we would see our plane as titled 45 degrees to the right).

driving up our plane parallel to the y-axis it's a rather steep grade, but if we make a left or right turn, and go parallel to the x-axis, it's perfectly flat (unless it's icy, in which case we'll slip down it sideways).
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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