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Vectors → a and → b are inclined at angle θ = 120°. If | → a | = 1 , ∣ ∣ → b ∣ ∣ = 2 , then [ ( → a + 3 → b ) × ( 3 → a − → b ) ] 2 is equal to - Mathematics

MCQ

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 

 
  

Options

  • 300

  •  325

  •  275

  •  225 

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Solution

\[\left( \vec{a} + 3 \vec{b} \right) \times \left( 3 \vec{a} - \vec{b} \right)\]

\[ = 3 \left( \vec{a} \times \vec{a} \right) - \vec{a} \times \vec{b} + 9 \left( \vec{b} \times \vec{a} \right) - 3 \left( \vec{b} \times \vec{b} \right)\]

\[ = 3 \left( 0 \right) - \vec{a} \times \vec{b} - 9 \left( \vec{a} \times \vec{b} \right) - 3 \left( 0 \right)\]

\[ = - 10 \left( \vec{a} \times \vec{b} \right)\]

\[\text{ Now } ,\]

\[\left| \left( \vec{a} \times 3 \vec{b} \right) \times \left( 3 \vec{a} - \vec{b} \right) \right| {}^2 \]

\[ = \left| - 10 \left( \vec{a} \times \vec{b} \right) \right|^2 \]

\[ = 100 \left| \left( \vec{a} \times \vec{b} \right) \right|^2 \]

\[ = 100 \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \sin^2 120\]

\[ = 100 \left( 1 \right)^2 \left( 2 \right)^2 \left( \frac{\sqrt{3}}{2} \right)^2 \]

\[ = 400 \times \frac{3}{4}\]

\[ = 300\]

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APPEARS IN

RD Sharma Class 12 Maths
Chapter 25 Vector or Cross Product
MCQ | Q 6 | Page 35
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