If the volume of the parallelopiped whose conterminus edges are along the vectors $\bar{a},…

If the volume of the parallelopiped whose conterminus edges are along the vectors $\bar{a}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ is 12, then the volume of the tetrahedron whose conterminus edges are $\bar{a}+\overline{\mathrm{b}}, \overline{\mathrm{b}}+\overline{\mathrm{c}}$ and $\overline{\mathrm{c}}+\overline{\mathrm{a}}$ is
  1. 4 (units) $^{3}$
  2. 24 (units) $^{3}$
  3. 6 (units) $^{3}$
  4. 12 (units) $^{3}$

Solution

Volume of parallelepiped $=[\bar{a} \bar{b} \bar{c}]=12$ $\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=12 \ldots(1)$ volume of tetrahedrom $=\frac{1}{6}\left[\begin{array}{lll} \bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a} \end{array}\right]$ $=\frac{1}{6}(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]=\frac{1}{6}(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]$ $=\frac{1}{6}\{\bar{a} \cdot(\bar{b} \times \bar{c})+\bar{a}(\bar{b} \times \bar{a})+\bar{a}(\bar{c} \times \bar{a})+\bar{b} \cdot(\bar{b} \times \bar{c})+\bar{b} \cdot(\bar{b} \times \bar{a})+\bar{b}(\bar{c} \times \bar{a})\} \quad \ldots[\because \bar{c} \times \bar{c}=0]$ $=\frac{1}{6}[\bar{a} \cdot(\bar{b} \times \bar{c})+0+\bar{b} \cdot(\bar{c} \times \bar{a})]=\frac{1}{6}\{[\bar{a} \bar{b} \bar{c}]+[\bar{a} \bar{b} \bar{c}]\}$ $=\frac{2}{6}[\bar{a} \bar{b} \bar{c}]=\frac{1}{3}(12)=4$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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