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Prove a vector formula for the volume of a tetrahedron

  1. Consider a tetrahedron with vertices A,B,C,D. Prove that the volume of the tetrahedron is given by the formula

        \[ V = \frac{1}{6} |(B-A) \cdot (C -A) \times (D-A)|. \]

  2. Compute the volume in the case that

        \[ A = (1,1,1), \quad B = (0,0,2), \quad C = (0,3,0), \quad D = (4,0,0). \]


  1. Proof. We know the volume of a tetrahedron is given by \frac{1}{3} A_0 h (where A_0 denotes the altitude of the tetrahedron). We know (page 490 of Apostol) that the volume of the parallelepiped with base formed by vector A,B and height formed by vector C is given by A \times B \cdot C. In this case we have that the base of the tetrahedron is formed by the vectors (B-A) and (C-A), and the height is formed by the vector (D-A). Further, we know that the area of the base described by the vectors (B-A) and (C-A) is one half that of the parallelepiped whose base is given by vectors A and B (since the base of the parallelepiped described by vectors A and B is a rectangle, and the base of the tetrahedron is the triangle formed by cutting this rectangle along the diagonal). Therefore we have

        \begin{align*}  V &= \frac{1}{3} \left( \frac{1}{2} \right) (B -A)\cdot (C-A) \times (D - A) \\  &= \frac{1}{6} \big( (B-A) \cdot (C-A) \times (D-A) \big). \qquad \blacksquare \end{align*}

  2. Using the formula in part (a) with the given values of A,B,C,D we have

        \begin{align*}  V &= \frac{1}{6} \big( (-1,-1,1) \cdot (-1,2,-1) \times (3,-1,-1) \big) \\   &= \frac{1}{6} \big( (-1,-1,1) \cdot (-2-1, -3-1, 1-6) \big) \\  &= \frac{1}{6} \big( (-1,-1,1) \cdot (-3,-4,-5) \big) \\  &= \frac{1}{6} (3 + 4 - 5) \\  &= \frac{1}{3}. \end{align*}

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