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  • Provide solution for RD Sharma maths class 12 chapter 25 Scalar triple product exercise multiple choice question 14

Provide solution for RD Sharma maths class 12 chapter 25 Scalar triple product exercise multiple choice question 14

Answers (1)

Answer:

 \begin{aligned} &=-\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \end{aligned}

Hint:

 Simplify the expression

Given:

\begin{aligned} &\left ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \right ).\left [ \left ( (\overrightarrow{a} + \overrightarrow{b})\times (\overrightarrow{a} + \overrightarrow{c}) \right ) \right ]\\ &=\left ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \right ).\left [ (\overrightarrow{a}\times \overrightarrow{a})+(\overrightarrow{a}\times\overrightarrow{c})+(\overrightarrow{b}\times\overrightarrow{a})+(\overrightarrow{b}\times\overrightarrow{c}) \right ]\\ &=\left ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \right ).\left [ 0+(\overrightarrow{a}\times\overrightarrow{c})+(\overrightarrow{b}\times\overrightarrow{a})+(\overrightarrow{b}\times\overrightarrow{c}) \right ]\\ \end{aligned}

\begin{aligned} &=\overrightarrow{a}.(\overrightarrow{a}\times \overrightarrow{c})+ \overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{a})+ \overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})+ \overrightarrow{b}.(\overrightarrow{a}\times \overrightarrow{c})+ \overrightarrow{b}.(\overrightarrow{b}\times \overrightarrow{a})+ \overrightarrow{b}.(\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{c}.(\overrightarrow{a}\times \overrightarrow{c})+ \overrightarrow{c}.(\overrightarrow{b}\times \overrightarrow{a})+ \overrightarrow{c}.(\overrightarrow{b}\times \overrightarrow{c}) \end{aligned}

\begin{aligned} &=\left [ \overrightarrow{a} \quad \overrightarrow{a} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{a} \right ]+\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{b} \quad \overrightarrow{a} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{b} \quad \overrightarrow{b} \quad \overrightarrow{a} \right ]+\left [ \overrightarrow{b} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{c} \quad \overrightarrow{a} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{c} \quad \overrightarrow{b} \quad \overrightarrow{a} \right ]+\left [ \overrightarrow{c} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \end{aligned}

\begin{aligned} &=0+0+\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{b} \quad \overrightarrow{a} \quad \overrightarrow{c} \right ]+0+0+0+\left [ \overrightarrow{c} \quad \overrightarrow{b} \quad \overrightarrow{a} \right ]+0\\ &=\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]+\left ( -\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \right )+\left ( -\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \right )\\ &=-\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \end{aligned}

Posted by

Gurleen Kaur

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