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प्रश्न
Vectors \[\vec{a} \text{ and } \vec{b}\] are inclined at angle θ = 120°. If \[\left| \vec{a} \right| = 1, \left| \vec{b} \right| = 2,\] then \[\left[ \left( \vec{a} + 3 \vec{b} \right) \times \left( 3 \vec{a} - \vec{b} \right) \right]^2\] is equal to
पर्याय
300
325
275
225
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उत्तर
\[\left( \vec{a} + 3 \vec{b} \right) \times \left( 3 \vec{a} - \vec{b} \right)\]
\[ = 3 \left( \vec{a} \times \vec{a} \right) - \vec{a} \times \vec{b} + 9 \left( \vec{b} \times \vec{a} \right) - 3 \left( \vec{b} \times \vec{b} \right)\]
\[ = 3 \left( 0 \right) - \vec{a} \times \vec{b} - 9 \left( \vec{a} \times \vec{b} \right) - 3 \left( 0 \right)\]
\[ = - 10 \left( \vec{a} \times \vec{b} \right)\]
\[\text{ Now } ,\]
\[\left| \left( \vec{a} \times 3 \vec{b} \right) \times \left( 3 \vec{a} - \vec{b} \right) \right| {}^2 \]
\[ = \left| - 10 \left( \vec{a} \times \vec{b} \right) \right|^2 \]
\[ = 100 \left| \left( \vec{a} \times \vec{b} \right) \right|^2 \]
\[ = 100 \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \sin^2 120\]
\[ = 100 \left( 1 \right)^2 \left( 2 \right)^2 \left( \frac{\sqrt{3}}{2} \right)^2 \]
\[ = 400 \times \frac{3}{4}\]
\[ = 300\]
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