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Volume of tetrahedra formed from coordinate and tangent planes 
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#1
Nov2609, 10:42 AM

P: 82

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
Nov2609, 12:46 PM

HW Helper
Thanks
PF Gold
P: 7,659




#3
Nov2609, 01:14 PM

P: 82

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] 


#4
Nov2609, 03:04 PM

HW Helper
Thanks
PF Gold
P: 7,659

Volume of tetrahedra formed from coordinate and tangent planes
Almost right. But check the plane intercepts again. For example, 3r doesn't equal 3a^{3}.



#5
Nov2609, 04:23 PM

P: 82

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] 


#6
Nov2609, 04:30 PM

P: 82

Sorry that should say 3t=[tex]\frac{3a^{3}}{rs}[/tex]



#8
Nov2609, 11:16 PM

P: 4,512

I get the feeling that this can be proved with only geometric considerations with eyes closed...



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