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  • Need solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 11.

Need solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 11.

Answers (1)

Answer:

-10

Hint:

Use scalar triple product formula.

Given:

Vectors are \vec{a}=2 \hat{\imath}+\hat{\jmath}+3 \hat{k}, \vec{b}=-\hat{\imath}+2 \hat{\jmath}+\hat{k} \text { and } \vec{c}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}

Solution:

We have

\left.\begin{array}{l} \vec{a}=2 \hat{\imath}+\hat{\jmath}+3 \hat{k} \\ \vec{b}=-\hat{\imath}+2 \hat{\jmath}+\hat{k} \\ \vec{c}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} \end{array}\right\}                          Eq.(i)

\begin{aligned} \text { Now }(\vec{b} \times \vec{c}) &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & k \\ -1 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right| \\ &=\hat{i}(4-1)-\hat{j}(-2-3)+\hat{k}(-1-6) \end{aligned}

=3 \hat{i}-(-5) \hat{j}+(-7) \hat{k}=3 \hat{i}+5 \hat{j}-7 \hat{k}

\therefore \overrightarrow{\boldsymbol{a}} \cdot(\overrightarrow{\boldsymbol{b}} \times \overrightarrow{\boldsymbol{c}})=(2 \hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+3 \hat{\boldsymbol{k}}) \cdot(3 \hat{\boldsymbol{\imath}}+5 \hat{\boldsymbol{\jmath}}-7 \widehat{\boldsymbol{k}})

=6(\hat{\imath} . \hat{\imath})+10(\hat{\imath} . \hat{\jmath})-14(\hat{\imath} . \hat{k})+3(\hat{\jmath} \cdot \hat{\imath})+5(\hat{\jmath} \cdot \hat{\jmath})-7(\hat{\jmath} \cdot \hat{k})+9(\hat{k} \cdot \hat{\imath})+15(\hat{k} \cdot \hat{\jmath})-21(\widehat{k}.\widehat{k})

\begin{aligned} &=6.1+0-0+0+5-0+0+0-21.1\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \hat{\imath} . \hat{\imath}=\hat{\jmath} \cdot \hat{\jmath}=\hat{k} \cdot \hat{k}=1, \hat{\imath} . \hat{\jmath}=\hat{\jmath} . \hat{\imath}=0 \\ \hat{\imath} \cdot \hat{k}=\hat{k} \cdot \hat{\imath}=0, \hat{\jmath} \cdot \hat{k}=\hat{k} \cdot \hat{\jmath}=0 \end{array}\right] \\ &=6+5-21 \\ &=11-21=-10 \end{aligned}

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