A) \[{{\sin }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]
B) \[{{\sin }^{-1}}\left( \sqrt{\frac{2}{3}} \right)\]
C) \[{{\cos }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]
D) \[{{90}^{o}}\]
Correct Answer: A
Solution :
Let \[\vec{A}=\hat{i}+\hat{j}+\hat{k}\] and \[\vec{B}=\hat{i}+\hat{j}\] and angle between \[\vec{A}\] and \[\vec{B}\] be \[\theta \] As \[\vec{A}\,.\,\vec{B}=AB\cos \theta \] \[\Rightarrow \] \[\cos \theta =\frac{\vec{A}\,.\,\vec{B}}{AB}\] ?.. (i) Now, \[\left| {\vec{A}} \right|=\sqrt{{{(1)}^{2}}+{{(1)}^{2}}+{{(1)}^{2}}}=\sqrt{3}\] and \[\vec{B}=\sqrt{{{(1)}^{2}}+{{(1)}^{2}}}=\sqrt{2}\] \[\therefore \] \[\cos \theta =\frac{(\hat{i}+\hat{j}+\hat{k})\,.\,(\hat{i}+\hat{j})}{\sqrt{3}\,.\,\sqrt{2}}=\frac{1+1}{\sqrt{6}}=\sqrt{\frac{2}{3}}\] and \[\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }\] or \[\sin \theta =\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}}\]
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