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Math Help - Volume of a tetrahedron

  1. #1
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    Volume of a tetrahedron

    Given a tetrahedron ABCD with A = (1,1,1), B = (3,2,1), C = (0,1,2) , D = (3,2,2)

    a) Find volume of ABCD:

    I have  V= \frac {|(A-D) \cdot (B-D) \times (C-D)|}{6}

     = \frac {|(-2,-1,-1) \cdot (0,0,-1) \times (-3,-1,0)|} {6} = \frac {|-1|}{6} = \frac {1}{6}

    Is that correct?

    b) Find distance between A and face BCD:

    Face BCD is  ( \bar {x} - D) \cdot (B-D) \times (C-D) = -x-3y+9=0

    But I can't remember the distance formula, any help?

    I think is this one:

    dist(A,BCD) = | \frac {-(1)-3(1)+9}{||(-1,-3,0)||} | = \frac {4}{10^{1/2}}

    Sorry, we don't have a textbook for this course, and the professor didn't have the formula clearly, I tried my best to take the notes but I just can't follow him.

    c) Find the distance between the lines AC and BD:

    AC = (1,1,1)+t(-1,0,1) , BD = (3,2,1)+s(0,0,1)

    dist(AC,BD) =  | \frac {(B-A) \cdot (0,0,1) \times (-1,0,1)}{(0,0,1) \times (-1,0,1)}| = \frac {|(2,1,0) \cdot (0,-1,0)|}{|(0,-1,0)|} = 1

    Is that right?
    Last edited by tttcomrader; January 27th 2008 at 10:41 AM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Given a tetrahedron ABCD with A = (1,1,1), B = (3,2,1), C = (0,1,2) , D = (3,2,2)

    a) Find volume of ABCD:

    I have  V= \frac {|(A-D) \cdot (B-D) \times (C-D)|}{6}

     = \frac {|(-2,-1,-1) \cdot (0,0,-1) \times (-3,-1,0)|} {6} = \frac {|-1|}{6} = \frac {1}{6}

    Is that correct?

    b) Find distance between A and face BCD:

    Face BCD is  ( \bar {x} - D) \cdot (B-D) \times (C-D) = -x-3y+9=0

    But I can't remember the distance formula, any help?

    I think is this one:

    dist(A,BCD) = | \frac {-(1)-3(1)+9}{||(-1,-3,0)||} | = \frac {4}{10^{1/2}}

    Sorry, we don't have a textbook for this course, and the professor didn't have the formula clearly, I tried my best to take the notes but I just can't follow him.

    c) Find the distance between the lines AC and BD:

    AC = (1,1,1)+t(-1,0,1) , BD = (3,2,1)+s(0,0,1)

    dist(AC,BD) =  | \frac {(B-A) \cdot (0,0,1) \times (-1,0,1)}{(0,0,1) \times (-1,0,1)}| = \frac {|(2,1,0) \cdot (0,-1,0)|}{|(0,-1,0)|} = 1

    Is that right?
    Hello,

    to #a) and #c): In my opinion your results are OK.

    to #b) I have got p_{BCD} : -x+3y+3 = 0 as the equation of the plane BCD.
    The distance formula you use is correct. But with my equation I got a distance of d(A, p_{BCD})=\frac12 \cdot \sqrt{10}
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