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Thread: Linear algebra - tetrahedron problem

  1. #1
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    Linear algebra - tetrahedron problem

    Exercise:
    For which values of a ∈ R are the points P = (-1, 1, 2), Q = (0, a, 1), R = (a, 4, -1) and S = (-11, -1, 0) corners of a tetrahedron? Cacluate the volume of the tetrahedron PQRS for these values of a.

    I really need some advice on how to execute this on, as I have no idea where to begin.

    Best regards.
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  2. #2
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    Re: Linear algebra - tetrahedron problem

    Four points are the four corners of a tetrahedron as long as they do NOT all lie in the same plane. Since three points will always determine a plane, the simplest way to deal with this is to use three of these points to find the plane that those three points form. If we take, for example, (-1, 1, 2), (0, a, 1), and (-11, -1, 0) the plane determined by those points is of the form Ax+ By+ Cz= D. Since we can always divide through by any non 0 number, we can write any plane that does not include the origin as Ax+ By+ Cz= 1. Taking x= -1, y= 1, z= 2 that becomes -A+ B+ 2C= 1. Taking x= 0, y= a,, z= 1, that becomes aB+ C= 1. Taking x= -11, y= -1, z= 0, that becomes -11A- B= 1. From the last equation, B= -11A- 1. From the second equation, C= 1- aB= 1+ 11aA+ a. So the first equation becomes -A+ (-11A- 1)+ 2(1+ 11aA+ a)= (-12+ 11a)A+ 1+ 2a= 1. That gives A= 2a/(12- 11a), B= -11A- 1= -22a/(12- 11a)- (12- 11a)/(12- 11a)= (-11a- 12)/(12- 11a), C= 1+ 22a^2/(12- 11a)= (12- 11a+ 22a^2/(12- 11a). Now, does put x= a, y= 4, z= -1 into Ax+ By+ Cz= 1. Any value of a that does NOT satisfy that equation will give a tetrahedron. The volume of a tetrahedron is 1/3 the area of the base (any of the four faces) times the distance from that base to the fourth vertex.
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  3. #3
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    Re: Linear algebra - tetrahedron problem

    using the determinant formula for the volume of a tetrahedron we find

    $\displaystyle V=\frac{1}{3} |(a-2) (a+18)|$

    $\displaystyle a\neq 2,-18$
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