A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[p\mathbf{i}+q\mathbf{j}+r\mathbf{k}\]
D) \[\frac{2\pi }{3}\]
Correct Answer: B
Solution :
\[{{(\mathbf{a}-\sqrt{2}\mathbf{b})}^{2}}=1\Rightarrow 1+2-2\sqrt{2}\mathbf{a}\,.\,\mathbf{b}=1\] \[\Rightarrow \mathbf{a}\,.\,\mathbf{b}=\frac{1}{\sqrt{2}}\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4}.\]
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