A) \[60{}^\circ \]
B) \[45{}^\circ \]
C) \[30{}^\circ \]
D) \[15{}^\circ \]
Correct Answer: A
Solution :
[a] Let \[a=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\hat{k}\] and \[b=\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}+\hat{k}\] \[\therefore \,\cos \theta =\frac{a.b}{|a||b|}\] \[=\frac{\left( \frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\hat{k} \right).\left( \frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}+\hat{k} \right)}{\sqrt{\frac{1}{2}+\frac{1}{2}+1}\sqrt{\frac{1}{2}+\frac{1}{2}+1}}\] \[=\frac{1}{2}\left[ \frac{1}{2}-\frac{1}{2}+1 \right]=\frac{1}{2}=\cos \,\,60{}^\circ \] \[\therefore \,\,\theta =60{}^\circ \]
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