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  • Please solve RD Sharma class 12 chapter 25 Scalar triple product exercise multiple choice question 5 maths textbook solution

Please solve RD Sharma class 12 chapter 25 Scalar triple product exercise multiple choice question 5 maths textbook solution

Answers (1)

Answer:

 0

Hint:

 Simplify the expression

Given:

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

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

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

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

Posted by

Gurleen Kaur

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