The volume of a tetrahedron whose vertices are $\mathrm{A} \equiv(-1,2,3), \mathrm{B} \equiv(3,-2,1)$,…

The volume of a tetrahedron whose vertices are $\mathrm{A} \equiv(-1,2,3), \mathrm{B} \equiv(3,-2,1)$, $\mathrm{C} \equiv(2,1,3)$ and $\mathrm{D} \equiv(-1,-2,4)$ is
  1. $\frac{14}{3}$ cu. units
  2. $\frac{16}{3}$ cu. units
  3. $\frac{17}{3}$ cu. units
  4. $\frac{15}{3}$ cu. units

Solution

Here $\overline{\mathrm{AB}}=4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \overline{\mathrm{AC}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}$ and $\overline{\mathrm{AD}}=-4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ Volume of tetrahedron $\begin{array}{l} =\frac{1}{6} \overline{\mathrm{AB}} \cdot(\overline{\mathrm{AC}} \times \overline{\mathrm{AD}}) \\ =\frac{1}{6}\left|\begin{array}{ccc} 4 & -4 & -2 \\ 3 & -1 & 0 \\ 0 & -4 & 1 \end{array}\right|=\frac{1}{6}[4(-1)+4(3)-2(-12)] \\ \quad=\frac{1}{6}(32)=\frac{16}{3} \text { cu.units } \end{array}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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