MHT CET · Maths · Vector Algebra
If \([\bar{a} \bar{b} \bar{c}] \neq 0\), then \(\frac{[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]}{[\bar{b} \bar{c} \bar{a}]}=\)
- A 0
- B 1
- C 2
- D 4
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
(D)
\(\left[\begin{array}{lll}\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}\end{array}\right]\)
\(=(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]\)
\(=(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{a})]+[\bar{a} \cdot(\bar{c} \times \bar{a})]+[\bar{b} \cdot(\bar{b} \times \bar{c})]\) \(+~[\bar{b} \cdot(\bar{b} \times \bar{a})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]\)
\(=2[\bar{a} \cdot(\bar{b} \times \bar{c})] \quad=2\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]\)
Hence give expression \(=\frac{2[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]}=2\)
\(\left[\begin{array}{lll}\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}\end{array}\right]\)
\(=(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]\)
\(=(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{a})]+[\bar{a} \cdot(\bar{c} \times \bar{a})]+[\bar{b} \cdot(\bar{b} \times \bar{c})]\) \(+~[\bar{b} \cdot(\bar{b} \times \bar{a})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]\)
\(=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]\)
\(=2[\bar{a} \cdot(\bar{b} \times \bar{c})] \quad=2\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]\)
Hence give expression \(=\frac{2[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]}=2\)
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