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Math Help - Plotting a constant y-axis plane

  1. #1
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    Question Plotting a constant y-axis plane

    I am trying to plot a plane such that the y-axis is constant at 3 on a 3-D graph. So the plane is going to be stick upwards parrellel to x-axis while staying at y=3 throughout. The idea is to actually have this plane to cut another surface. I try to get the function in the form of z=f(x,y) and here's what I did:

    Since the plane is a constant y=3 plane, I assume (0, 3, 0) is a point on the plane. Then I move 5 in the z-axis and then minus them to get a vector direction.
    \begin{pmatrix}<br />
0\\ <br />
3\\ <br />
0<br />
\end{pmatrix}<br />
-<br />
\begin{pmatrix}<br />
5\\ <br />
3\\ <br />
0<br />
\end{pmatrix}<br />
=<br />
\begin{pmatrix}<br />
-5\\ <br />
0\\ <br />
0<br />
\end{pmatrix}<br />

    I then get another vector direction on this plane by moving 2steps up z-axis and 5 steps to x-axis; (2, 3, 5).
    <br />
\begin{pmatrix}<br />
0\\ <br />
3\\ <br />
0<br />
\end{pmatrix}<br />
-<br />
\begin{pmatrix}<br />
2\\ <br />
3\\ <br />
5<br />
\end{pmatrix}<br />
=<br />
\begin{pmatrix}<br />
-2\\ <br />
0\\ <br />
-5<br />
\end{pmatrix}<br />

    Then I cross these 2 direction vectors to get the normal line.
    <br />
\begin{pmatrix}<br />
-5\\ <br />
0\\ <br />
0<br />
\end{pmatrix}<br />
\times<br />
\begin{pmatrix}<br />
-2\\ <br />
0\\ <br />
-5<br />
\end{pmatrix}<br />
=<br />
\begin{pmatrix}<br />
0\\ <br />
-25\\ <br />
0<br />
\end{pmatrix}<br />

    I then did a dot product of this normal vector with the point (0, 3, 0) to get the equation of the plane: 0x -25y +0z = -75

    But, from 0x -25y +0z = -75, how do I form it to z=f(x, y)? z is zero in this case and I can't make it the subject in terms of x and y.

    How should I carry on from here? Thanks.
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  2. #2
    A Plied Mathematician
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    CT, USA
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    I think you're making this a lot more work than is necessary. The equation for the y = 3 plane is (drum roll) y = 3! (That's an exclamation point, not a factorial.) If you look carefully, that's the equation you got from your (erroneous) calculations. I say erroneous because you have, throughout, switched your x and z components; for some reason, the fact that you're technically using a left-handed coordinate system did not, in the end, mess up your cross product.

    You cannot find the equation of this particular plane in the form z = f(x,y), because the plane is massively multi-valued in z. The analogy is that the equation x = 3, a vertical line in the xy plane, cannot be solved for y.

    What you're really asking is how to plot that plane on whatever software you're using. What software are you using?
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  3. #3
    Member
    Joined
    Oct 2010
    Posts
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    oh...so y=3 cannot be plotted on a 3D graph...
    I am using Mathcad and thought I could create a y=3 plane to visualise a level curve or something of that sort related to interception. I saw some graphs having like a surface and y or x constant plane on the same graph showing the intersections and thought I could do the same.
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  4. #4
    A Plied Mathematician
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    No, I didn't say you couldn't plot y = 3 on a 3D graph. Of course you can. You just can't write it as z = f(x,y), like you can with many surfaces.

    Unfortunately, I don't know the first thing about Mathcad. I would look into implicit plots or maybe parametric plots. In Mathematica, for example, the command would be

    ParametricPlot3D[{x, 3, z}, {x, -5, 5, 0.1}, {z, -5, 5, 0.1}]

    So the {x,3,z} is the vector, then you have the arguments {x,xmin,xmax,xgridsize}, and then {y,ymin,ymax,ygridsize}.

    Try looking up the help in Mathcad and see if there isn't a command like either one of these.
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