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I am very new to mathematica and was wondering how I could 3Dplot -5(x-2)-7y+3(z+4)=0 I've tried 3DPlot and Contour but neither have seemed to work. PLease help.

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    $\begingroup$ Look up ContourPlot3D[]. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Jun 3, 2016 at 7:41
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    $\begingroup$ @J.M. it is 2D case but I think it is good enough to be a duplicate, what do you think? How can I plot implicit equations? $\endgroup$
    – Kuba ♦
    Jun 3, 2016 at 7:51
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    $\begingroup$ @Kuba, let's leave it open for now until somebody finds an actual dupe. I have the feeling closing a 3D question as a dupe of a 2D problem is a slippery slope. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Jun 3, 2016 at 8:04
  • $\begingroup$ @J.M. mathematica.stackexchange.com/q/10710/5478? $\endgroup$
    – Kuba ♦
    Jun 3, 2016 at 8:11
  • $\begingroup$ for some reason the ContourPlot3D doesn't seem to work on my mathematica $\endgroup$
    – Lorianna Esparza
    Jun 3, 2016 at 8:54

2 Answers 2

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Try the following: first solve with respect to z:

sl = Solve[-5 (x - 2) - 7 y + 3 (z + 4) == 0, z]

(*  {{z -> 1/3 (-22 + 5 x + 7 y)}}  *)

Then plot it in 3D:

Plot3D[sl[[1, 1, 2]], {x, -1, 1}, {y, -1, 1}]

enter image description here

Have fun!

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I'm on 10.0 for Mac OS X x86 (64-bit) (December 4, 2014)

Plotting a Function see Function Visualization,

ContourPlot3D[-5 (x - 2) - 7 y + 3 (z + 4) == 0, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]

enter image description here

Adding a plan z an y,

cp1 = ContourPlot3D[{-5 (x - 2) - 7 y + 3 (z + 4) == 0, z, y}
, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}
, ContourStyle -> {Red, Blue, Green}]

enter image description here

Solve for Intersections,

Solve[{-5 (x - 2) - 7 y + 3 (z + 4) == 0 && z == 0 && y == 0, z == 0}, {x, y, z}]

{{x->22/5,y->0,z->0}}

And Visualise the whole gang:

Show[cp1, Graphics3D[{White, Sphere[{22/5, 0, 0}]}]]

enter image description here

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