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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 10 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 25 Scalar Triple Product  Exercise 25.1 Question 10 Maths Textbook Solution.

Answers (1)

Answer:\left ( \vec{a}-\vec{b} \right ).\left \{ \left ( \vec{b}-\vec{c} \right )\times \left ( \vec{c}-\vec{a} \right ) \right \}=0

Hint :- Use cross-product to prove

Given:\left ( \vec{a}-\vec{b} \right ).\left \{ \left ( \vec{b}-\vec{c} \right )\times \left ( \vec{c}-\vec{a} \right ) \right \}=0

\begin{aligned} \text { L.H.S } &=\left(\vec{a}_{-} \vec{b}\right) \cdot\{(\vec{b}-\vec{c}) \times(\vec{c}-\vec{a})\} \\ &=(\vec{a}-\vec{b}) \cdot\{\vec{b} \times(\vec{c}-\vec{a})-\vec{c} \times(\vec{c}-\vec{a})\} \\ \end{aligned}

              \begin{aligned} &=(\vec{a}-\vec{b}) \cdot((\vec{b} \times \vec{c})-(\vec{b} \times \vec{a})-(\vec{c} \times \vec{c})+(\vec{c} \times \vec{a})) \: \: \: \quad(\because \text { Distributive law }) \\ \end{aligned}

              \begin{aligned} &=(\vec{a}-\vec{b}) \cdot((\vec{b} \times \vec{c})-(\vec{b} \times \vec{a})-0+(\vec{c} \times \vec{a})) \quad(\because \vec{c} \times \vec{c}=0) \\\\ &=(\vec{a}-\vec{b}) \cdot((\vec{b} \times \vec{c})+(\vec{a} \times \vec{b})+(\vec{c} \times \vec{a})) \quad(\vec{a}) \\\\ &=\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot(\vec{a} \times \vec{b})+\vec{a} \cdot(\vec{c} \times \vec{a})-\vec{b} \cdot(\vec{b} \times \vec{c})-\vec{b} \cdot(\vec{a} \times \vec{b})-\vec{b} \cdot(\vec{c} \times \vec{a}) \\\\ &=\left[\begin{array}{ll} \vec{a} \vec{b} & \vec{c}]+0+0-0-0-[\vec{a} \vec{b} \vec{c}] \end{array}\right.\\\\ &=0 \\ &=\mathrm{R} . \mathrm{H} . \mathrm{S} \text { (Hence proved) } \end{aligned}

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