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  • Need solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 10.

Need solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 10.

Answers (1)

Answer:

0

Hint:

Use scalar triple product formula. 

Given:

a,b,c are non co planar vectors and\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}}+\frac{\vec{b} \cdot(\vec{a} \times \vec{c})}{\vec{c} \cdot(\vec{a} \times \vec{b})}

Solution:

\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}}+\frac{\vec{b} \cdot(\vec{a} \times \vec{c})}{\vec{c} \cdot(\vec{a} \times \vec{b})}
=\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{\vec{c} \cdot(\vec{a} \times \vec{b})}+\frac{(\vec{b} \times \vec{a}) \cdot \vec{c}}{\vec{c} \cdot(\vec{a} \times \vec{b})} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \text { The position of dot and cross can be interchanged } \\ \text { provided that the cyclic order of the vectors remain } \\ \text { same in scalar triple product. } \end{array}\right]

=\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{(\vec{a} \times \vec{b}) \cdot \vec{c}}+\left\{-\frac{(\vec{a} \times \vec{b}) \cdot \vec{c}}{(\vec{a} \times \vec{b}) \cdot \vec{c}}\right\}\; \; \; \; \; \; \; \; \quad[\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a} \text { and }(\vec{a} \times \vec{b})=-(\vec{b} \times \vec{a})]

                        \left[\because(\vec{a} \times \vec{b}) \cdot \vec{c}=\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\vec{a} \cdot(\vec{b} \times \vec{c})\right]

=1-1=0

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