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Volume of tetrahedra formed from coordinate and tangent planes

  1. Nov 26, 2009 #1
    I have that P is the tangent plane to the surface xyz=a[tex]^{3}[/tex] at the point (r,s,t). I need to show that the volume of the tetrahedron, T, formed by the coordinate planes and the tangent plane to P is indepedent of the point (r,s,t).

    I have found that P is;

    [tex]\frac{x}{r}[/tex] + [tex]\frac{y}{s}[/tex] + [tex]\frac{z}{t}[/tex] = 3

    A know that the volume of a tetrahedron is giving by 1/3(area of base [tex]\times[/tex] height)

    But I just can't picture what this looks like, as far as I can see the volume of T has to be dependent of the point (r,s,t).

    Any help anyone could give would be great! Thanks.
     
  2. jcsd
  3. Nov 26, 2009 #2

    LCKurtz

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    You would expect the volume to depend on r,s, and t. But if you work it out you will find that it doesn't. Just write the equation of the tangent plane, find its intercepts and the corresponding volume. Here is a picture with a = 1 of one of the tangent planes to help you visualize it:
    [​IMG]
     
  4. Nov 26, 2009 #3
    Thanks for that, think I might be starting to understand this a little bit more now.

    Right I have that

    [tex]\frac{x}{r}[/tex] + [tex]\frac{y}{s}[/tex] + [tex]\frac{z}{t}[/tex] = 3

    and so this plane intersects the coordinate planes at x=3r, y=3s and z=3t but all of these points you know that xyz=a[tex]^{3}[/tex] so is right to then say that the intersects occur at x=3a[tex]^{3}[/tex], y=3a[tex]^{3}[/tex] and z=3a[tex]^{3}[/tex].

    Hence the volume of T is given by [tex]\frac{9a^{9}}{2}[/tex]
     
  5. Nov 26, 2009 #4

    LCKurtz

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    Almost right. But check the plane intercepts again. For example, 3r doesn't equal 3a3.
     
  6. Nov 26, 2009 #5
    Right think I've got it this time.

    The intercepts are at x=3r, y=3s and z=3t.

    Now 3r=[tex]\frac{3a^{3}}{st}[/tex], 3s=[tex]\frac{3a^{3}}{rt}[/tex] and 3z=3r=[tex]\frac{3a^{3}}{rs}[/tex]

    Hence the volume of T is given by [tex]\frac{9a^{3}}{2}[/tex]
     
  7. Nov 26, 2009 #6
    Sorry that should say 3t=[tex]\frac{3a^{3}}{rs}[/tex]
     
  8. Nov 26, 2009 #7

    LCKurtz

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    Looks good.
     
  9. Nov 26, 2009 #8
    I get the feeling that this can be proved with only geometric considerations with eyes closed...
     
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