MHT CET · Maths · Vector Algebra
If \(\bar{a}, \bar{b}, \bar{c}\) are nonzero vectors along the coterminus edges of a parallelopiped with volume 7 cubic units, then the volume of a parallelopiped with \(\overline{\mathrm{a}}+\overline{\mathrm{b}}, \overline{\mathrm{b}}+\overline{\mathrm{c}}, \overline{\mathrm{c}}+\overline{\mathrm{a}}\) as the coterminus edges is
- A 49 cubic units
- B 2 cubic units
- C 14 cubic units
- D 7 cubic units
Answer & Solution
Correct Answer
(C) 14 cubic units
Step-by-step Solution
Detailed explanation
(A)
We have \([\bar{a} \cdot(\bar{b} \times \bar{c})]=7\)
\([\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]\)
\(=(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]\)
\(=(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{a})]+[\bar{a}(\bar{c} \times \bar{a})]+[\bar{b} \cdot(\bar{b} \times \bar{c})]\) \(+~[\bar{b} \cdot(\bar{b} \times \bar{a})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+0+[\bar{b} \cdot(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{c})]=2[\bar{a} \cdot(\bar{b} \times \bar{c})]\)
\(=2(7)=14\)
We have \([\bar{a} \cdot(\bar{b} \times \bar{c})]=7\)
\([\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]\)
\(=(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]\)
\(=(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{a})]+[\bar{a}(\bar{c} \times \bar{a})]+[\bar{b} \cdot(\bar{b} \times \bar{c})]\) \(+~[\bar{b} \cdot(\bar{b} \times \bar{a})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+0+[\bar{b} \cdot(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{c})]=2[\bar{a} \cdot(\bar{b} \times \bar{c})]\)
\(=2(7)=14\)
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