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प्रश्न
\[\left( \vec{a} + 2 \vec{b} - \vec{c} \right) \cdot \left\{ \left( \vec{a} - \vec{b} \right) \times \left( \vec{a} - \vec{b} - \vec{c} \right) \right\}\] is equal to
विकल्प
\[\left[ \vec{a} \vec{b} \vec{c} \right]\]
\[2\left[ \vec{a} \vec{b} \vec{c} \right]\]
\[3\left[ \vec{a} \vec{b} \vec{c} \right]\]
0
उत्तर
\[ 3 \left[ \vec{a} \vec{b} \vec{c} \right]\]
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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