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( → a + 2 → B − → C ) ⋅ { ( → a − → B ) × ( → a − → B − → C ) } is Equal to - Mathematics

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Question

\[\left( \vec{a} + 2 \vec{b} - \vec{c} \right) \cdot \left\{ \left( \vec{a} - \vec{b} \right) \times \left( \vec{a} - \vec{b} - \vec{c} \right) \right\}\] is equal to

Options

  • \[\left[ \vec{a} \vec{b} \vec{c} \right]\]

  • \[2\left[ \vec{a} \vec{b} \vec{c} \right]\]

  • \[3\left[ \vec{a} \vec{b} \vec{c} \right]\]

  • 0

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Solution

\[ 3 \left[ \vec{a} \vec{b} \vec{c} \right]\]

We have

\[\left( \vec{a} + 2 \vec{b} - \vec{c} \right) . \left\{ \left( \vec{a} - \vec{b} \right) \times \left( \vec{a} - \vec{b} - \vec{c} \right) \right\}\]

\[ = \left( \vec{a} + 2 \vec{b} - \vec{c} \right) . \left\{ \left( \vec{a} - \vec{b} \right) \times \vec{a} - \left( \vec{a} - \vec{b} \right) \times \vec{b} - \left( \vec{a} - \vec{b} \right) \times \vec{c} \right\}\]

\[ = \left( \vec{a} + 2 \vec{b} - \vec{c} \right) . \left\{ \vec{a} \times \vec{a} - \vec{b} \times \vec{a} - \vec{a} \times \vec{b} + \vec{b} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{c} \right\}\]

\[ = \left( \vec{a} + 2 \vec{b} - \vec{c} \right) . \left\{ 0 - \vec{b} \times \vec{a} - \vec{a} \times \vec{b} + 0 - \vec{a} \times \vec{c} + \vec{b} \times \vec{c} \right\}\]

\[ = \left( \vec{a} + 2 \vec{b} - \vec{c} \right) . \left\{ - \vec{a} \times \vec{c} + \vec{b} \times \vec{c} \right\} ( \because \vec{a} \times \vec{b} = - \vec{b} \times \vec{a} )\]

\[ = - \vec{a} . \left( \vec{a} \times \vec{c} \right) + \vec{a} . \left( \vec{b} \times \vec{c} \right) - 2 \vec{b} . \left( \vec{a} \times \vec{c} \right) + 2 \vec{b} . \left( \vec{b} \times \vec{c} \right) + \vec{c} . \left( \vec{a} \times \vec{c} \right) - \vec{c} . \left( \vec{b} \times \vec{c} \right)\]

\[ = 0 + \left[ \vec{a} \vec{b} \vec{c} \right] - 2 \left[ \vec{b} \vec{a} \vec{c} \right] + 0 + 0 - 0 ( \because \left[ \lambda \vec{a} \vec{b} \vec{c} \right] = \lambda\left[ \vec{a} \vec{b} \vec{c} \right] \text { for any scalar  }\lambda) \]

\[ = 3 \left[ \vec{a} \vec{b} \vec{c} \right] ( \because - \left[ \vec{b} \vec{a} \vec{c} \right] = \left[ \vec{a} \vec{b} \vec{c} \right]) \]

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Chapter 26: Scalar Triple Product - MCQ [Page 20]

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RD Sharma Mathematics [English] Class 12
Chapter 26 Scalar Triple Product
MCQ | Q 15 | Page 20

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