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  • Need solution for RD Sharma maths class 12 chapter 25 Scalar triple product exercise multiple choice question 7

Need solution for RD Sharma maths class 12 chapter 25 Scalar triple product exercise multiple choice question 7

Answers (1)

Answer:

 =\frac{\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2}{4}

Hint:

 Use scalar triple product &

sin \frac{\pi }{6}=\frac{1}{2}

Given:

\overrightarrow{a},\overrightarrow{b}are two vectors &

\overrightarrow{c} is unit perpendicular to both

\therefore \left |\overrightarrow{c} \right |=1

For

\begin{aligned} &\begin{vmatrix} a_1 &a_2 &a_3 \\ b_1 &b_2 &b_3 \\ c_1 &c_2 &c_3 \end{vmatrix}^2\\ &=\left ( \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right )^2\\ &=\left ( \left | \overrightarrow{a}\times \overrightarrow{b} \right |.\overrightarrow{c} \right )^2\\ &=\left | \overrightarrow{a}\times \overrightarrow{b} \right |^2 \end{aligned}

\begin{aligned} &\because \overrightarrow{c} \text { is unit vector }\\ &=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2sin^2\frac{\pi }{6}\\ &=\frac{\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2}{4} \end{aligned}

Posted by

Gurleen Kaur

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