Let a→ and b→ be two non-zero vectors perpendicular to each other and |a→|=|b→|. If |a→×b→|=|a→|, then the angle between the vectors a→+b→+(a→×b→)) and a→ is equal to:

|a→×b→|=|a→|⇒ |a→||b→|sinπ2|=|a→|⇒|a→|=|b→∣=1|a→+b→+a→×b→|2=|a|2+|b|2+|a→×b→|2=3⇒|a→+b→+a→×b→|=3cosθ=a→·(a→+b→+a→×b→)|a||a→+b→+a→×b→|=13

Let a→ and b→ be two non-zero vectors perpendicular to each other and |a→|=|b→|. If |a→×b→|=|a→|, then the angle between the vectors a→+b→+(a→×b→)) and a→ is equal to:

|a→×b→|=|a→|⇒ |a→||b→|sinπ2|=|a→|⇒|a→|=|b→∣=1|a→+b→+a→×b→|2=|a|2+|b|2+|a→×b→|2=3⇒|a→+b→+a→×b→|=3cosθ=a→·(a→+b→+a→×b→)|a||a→+b→+a→×b→|=13

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$Let \vec{a} and \vec{b} be two non-zero vectors perpendicular to each other and | \vec{a} | = | \vec{b} | . If | \vec{a} \times \vec{b} |$ $= | \vec{a} |, then the angle between the vectors \vec{a} + \vec{b} + (\vec{a} \times \vec{b})) and \vec{a} is equal to:$

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$\sin^{- 1} (\frac{1}{\sqrt{6}})$

$\cos^{- 1} (\frac{1}{\sqrt{2}})$

$\cos^{- 1} (\frac{1}{\sqrt{3}})$

$\sin^{- 1} (\frac{1}{\sqrt{3}})$

answer is B.

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Detailed Solution

$\begin{array}{l} | \vec{a} \times \vec{b} | = | \vec{a} | \Rightarrow | \vec{a} | | \vec{b} | \sin \frac{π}{2} | = | \vec{a} | \Rightarrow | \vec{a} | = | \vec{b} ∣ = 1 \\ | \vec{a} + \vec{b} + \vec{a} \times \vec{b} |^{2} = | a |^{2} + | b |^{2} + | \vec{a} \times \vec{b} |^{2} = 3 \Rightarrow | \vec{a} + \vec{b} + \vec{a} \times \vec{b} | = \sqrt{3} \\ \cos θ = \frac{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{a} \times \vec{b})}{| a | | \vec{a} + \vec{b} + \vec{a} \times \vec{b} |} = \frac{1}{\sqrt{3}} \end{array}$