A) 1
B) \[{{\cos }^{-1}}\frac{1}{6}\]
C) \[\frac{1}{\sqrt{6}}\]
D) none of these
Correct Answer: B
Solution :
Key Idea: The angle between the vectors \[\vec{a}\]and\[\vec{b}\]is \[{{\cos }^{-1}}\left( \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|} \right).\] We have \[\vec{a}=2\hat{i}-\hat{j}+\hat{k}\]and \[\vec{b}=\hat{i}+2\hat{j}+\hat{k}\] The angle between them is \[\theta ={{\cos }^{-1}}\left( \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|} \right)\]\[={{\cos }^{-1}}\left( \frac{(2\hat{i}-\hat{j}+\hat{k}).(\hat{i}+2\hat{j}+\hat{k})}{\sqrt{4+1+1}\sqrt{1+4+1}} \right)\] \[={{\cos }^{-1}}\left( \frac{2-2+1}{\sqrt{6}+\sqrt{6}} \right)\] \[={{\cos }^{-1}}\frac{1}{6}.\]
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