Find the volume of a tetrahedron whose vertices are A $(- 1, 2, 3)$ ,B $(3, - 2, 1)$ , C $(2, 1, 3)$ and D $(- 1, - 2, 4)$ .

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Solution

solution-
$\to a = -^i + 2^j + 3^k$
$\to b = 3^i - 2^j +^k$
$\to c = 2^i +^j + 3^k$
$\to d =^i - 2^j + 4^k$

$- - \to A B = \to b - \to a = 4^i - 4^j - 2^k$

$- - \to A C = \to c - \to a = 3^i -^j$

$- - \to A D = \to d - \to a = - 4^j +^k$

Volume of Tetrahedral $= \frac{1}{6} [- - \to A B - - \to A C - - \to A D]$
$\frac{1}{6}$ $∣ ∣ ∣ ∣ \begin{matrix} 4 & - 4 & - 2 3 & - 1 & 0 0 & - 4 & 1 \end{matrix} ∣ ∣ ∣ ∣$

Volume $= \frac{1}{6} [4 (- 1 - 0) + 4 (3 - 0) - 2 (- 12 - 0)]$

$= \frac{32}{6} = \frac{16}{3}$ cubic units.

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