If a→ and b→ are any two vectors of magnitudes 1 and 2 respectively, and (1−3a→⋅b→)2+|2a→+b→+3(a→×b→)|2=47, then the angle between a→ and b→ is

1+9(a→⋅b→)2−6(a→⋅b→)+4|a→|2+|b→|2+9|a→×b→|2+4a→⋅b→=47 or 1+4+4+36−4cos⁡θ=47 or cos⁡θ=−12 Hence, the angle between a→ and b→ is 2π3.

If a→ and b→ are any two vectors of magnitudes 1 and 2 respectively, and (1−3a→⋅b→)2+|2a→+b→+3(a→×b→)|2=47, then the angle between a→ and b→ is

1+9(a→⋅b→)2−6(a→⋅b→)+4|a→|2+|b→|2+9|a→×b→|2+4a→⋅b→=47 or 1+4+4+36−4cos⁡θ=47 or cos⁡θ=−12 Hence, the angle between a→ and b→ is 2π3.

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$If \vec{a} and \vec{b} are any two vectors of magnitudes 1 and 2 respectively, and (1 - 3 \vec{a} \cdot \vec{b})^{2} + | 2 \vec{a} + \vec{b} + 3 (\vec{a} \times \vec{b}) |^{2}$ $= 47,$
$then the angle between \vec{a} and \vec{b} is$

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$π / 3$

$π - \cos^{- 1} (1 / 4)$

$\frac{2 π}{3}$

$\cos^{- 1} (1 / 4)$

answer is C.

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Detailed Solution

$\begin{array}{l} 1 + 9 (\vec{a} \cdot \vec{b})^{2} - 6 (\vec{a} \cdot \vec{b}) + 4 | \vec{a} |^{2} + | \vec{b} |^{2} + 9 | \vec{a} \times \vec{b} |^{2} + 4 \vec{a} \cdot \vec{b} = 47 \\ or 1 + 4 + 4 + 36 - 4 \cos θ = 47 \\ or \cos θ = - \frac{1}{2} \end{array}$