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MHT CET · Maths · Vector Algebra

If \(\bar{a}, \bar{b}\) and \(\bar{c}\) are any three non-zero vectors, then \((\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]=\)

  1. A \(\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]\)
  2. B \(2\left[\begin{array}{lll}-\bar{a} & \bar{b} & \bar{c}\end{array}\right]\)
  3. C \(3\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]\)
  4. D \(4\left[\begin{array}{lll}\vec{a} & \bar{b} & \bar{c}\end{array}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(3\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]\)

Step-by-step Solution

Detailed explanation

\((\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})] \)
\( =(\bar{a}+2 \bar{b}+\bar{c}) \cdot(\bar{a} \times \bar{a}-\bar{a} \times \bar{b}-\bar{a} \times \bar{c}-\bar{b} \times \bar{a}+\bar{b}\) \(\times~ \bar{b}+\bar{b} \times \bar{c}) \)
\( =(\bar{a}+2 \bar{b}+\bar{c}) \cdot(\overline{0}-\bar{a} \times \bar{b}-\bar{a} \times \bar{c}+\bar{a} \times \bar{b}+\overline{0}+\bar{b} \times \bar{c}) \)
\( =(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}}+\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \)
\( =\bar{a} \cdot(\bar{c} \times \bar{a})+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \bar{b} \cdot(\bar{b} \times \bar{c}) \) \( +\overline{\mathrm{c}} \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}})+\overline{\mathrm{c}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \)
\( =\overline{0}+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \times 0-0-0 \)
\(=\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]+2\left[\begin{array}{lll}\overline{ b } & \overline{ c } & \overline{ a }\end{array}\right]\)
\(=\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]+2\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]\)
\(=3\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]\)