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¯.b¯=0a¯×b¯=a¯⇒a¯b¯ sin 90°=b¯⇒a¯=1a¯+b¯+a¯×b¯2=a¯2+b¯2+a¯×b¯2+0+0+0 =1+1+1=3a¯+b¯+a¯×b¯=3cos θ=(a¯+b¯+a¯×b¯).a¯a¯+b¯+a¯×b¯a¯=1+0+03×1=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¯.b¯=0a¯×b¯=a¯⇒a¯b¯ sin 90°=b¯⇒a¯=1a¯+b¯+a¯×b¯2=a¯2+b¯2+a¯×b¯2+0+0+0 =1+1+1=3a¯+b¯+a¯×b¯=3cos θ=(a¯+b¯+a¯×b¯).a¯a¯+b¯+a¯×b¯a¯=1+0+03×1=13

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

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

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

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

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

answer is B.

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Detailed Solution

$\begin{array}{rcl} |\bar{a}| & = & |\bar{b}|, \bar{a} . \bar{b} = 0 \\ |\bar{a} \times \bar{b}| & = & |\bar{a}| \Rightarrow |\bar{a}| |\bar{b}| \sin 90^{°} = |\bar{b}| \Rightarrow |\bar{a}| = 1 \\ {|\bar{a} + \bar{b} + \bar{a} \times \bar{b}|}^{2} & = & {|\bar{a}|}^{2} + {|\bar{b}|}^{2} + {|\bar{a} \times \bar{b}|}^{2} + 0 + 0 + 0 \\ = & 1 + 1 + 1 = 3 \\ |\bar{a} + \bar{b} + \bar{a} \times \bar{b}| & = & \sqrt{3} \\ \cos θ & = & \frac{(\bar{a} + \bar{b} + \bar{a} \times \bar{b}) . \bar{a}}{|\bar{a} + \bar{b} + \bar{a} \times \bar{b}| |\bar{a}|} = \frac{1 + 0 + 0}{\sqrt{3} \times 1} = \frac{1}{\sqrt{3}} \end{array}$