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    Okay, you have 4 points A B C and D with position vectors a, b,c, and d

    And you’re told to find the volume of the tetrahedron ABCD

    and you have volume tetrahedron = (1/6) a.bxc

    do you just take bxc and dot product it to a, or is it more complicated than that?

    and in finding the area of the triangle ABC is that = ½ |axb + bxc +cxa|
    ?

    Thanks
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    (Original post by lesser weevil)

    Okay, you have 4 points A B C and D with position vectors a, b,c, and d

    And you’re told to find the volume of the tetrahedron ABCD

    and you have volume tetrahedron = (1/6) a.bxc
    do you just take bxc and dot product it to a, or is it more complicated than that?

    and in finding the area of the triangle ABC is that = ½ |axb + bxc +cxa|
    ?

    Thanks
    To find the volume you need to work out

    1/6 (d-a).[(b-a)x(c-a)]

    It's the vectors along the edges from a vertex that go into the bold formula above, not the position vectors. Note that if your formula had been correct then every tetradhedron with a vertex at the origin would have had zero volume.

    And the tetrahedron has four faces. One of their areas can be found from the formula

    1/2 |(b-a) x (c-a)|

    I'll leave you to work out the other three faces.

    Note also to add the moduluses not take the modulus of the sums as these will give different answers.
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    thanks... but is slightly confused. where'd I get the equation (1/6) a.bxc then? it was in the heinneman book...
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    (Original post by lesser weevil)
    thanks... but is slightly confused. where'd I get the equation (1/6) a.bxc then? it was in the heinneman book...
    This is the correct formula but where a,b,c are the vectors along the edges of the tetrahedron from one vertex - NOT the position vectors of the four vertices

    In the same way 1/2 |axb| is the area of the triangle where two of the edges are a and b - not two of the position vectors
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    ah i see, I get it. although the new formula for the tetrahedron is horribly complicated IMO
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    (Original post by lesser weevil)
    Okay, you have 4 points A B C and D with position vectors a, b,c, and d

    And you’re told to find the volume of the tetrahedron ABCD

    and you have volume tetrahedron = (1/6) a.bxc

    do you just take bxc and dot product it to a, or is it more complicated than that?

    and in finding the area of the triangle ABC is that = ½ |axb + bxc +cxa|
    ?

    Thanks
    V=\frac{1}{6}(\vec{AB}.(\vec{AC}  )x(\vec{AD})
    That is, you triple cross the three vectors connecting one point with the other three points.
    You can only use the position vectors if you're working from the origin (as the position vectors in this case are the vectors connecting the origin and the other points).
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    (Original post by lesser weevil)
    ah i see, I get it. although the new formula for the tetrahedron is horribly complicated IMO
    Well I wouldn't recommend learning that one - use yours, but just remember what the a,b,c stand for.

    If you're asked though for the volume in terms of posn vectors a,b,c,d (as you posted) then the formula I wrote is the correct one. It will simplify a little as axa = 0 etc
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    imaho.... erm, how can i use mine if it's not correct? how do i change the abc into what it should be...
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    (Original post by lesser weevil)
    imaho.... erm, how can i use mine if it's not correct? how do i change the abc into what it should be...
    Yours is correct as long as you remember what a,b,c denote - namely three edge vectors emanating from a vertex.

    You can't let maths faze you just because someone goes and uses different notation. Your formula isn't the correct one for a tetrahedron with vertices at a,b,c,d - after all it doesn't even include d. I worked out the correct formula for this from the one you had - it wasn't one I remembered
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    In practice you're going to be given four vertices - by way of example

    (1,0,1), (2,1,0), (3,0,1), (1,2,3)

    so set

    a = (1,1,-1), b = (2,0,0), c = (0,2,2)

    and use your formula

    V = 1/6 |a.(bxc)|
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    just work out AB and AC then do vector product on them to calculate their normal vector. Then do scalar product of AD and your previously calculated vector. then take modulus of this vector and div by 6... that is the easy way
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    ah ok! thanks!
 
 
 
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