A) \[30{}^\circ \]
B) \[45{}^\circ \]
C) \[60{}^\circ \]
D) \[90{}^\circ \]
Correct Answer: C
Solution :
Given,\[\overrightarrow{a}=\hat{i}+2\hat{j}-\hat{k}\]and \[\overrightarrow{b}=2\hat{i}+\hat{j}+\hat{k}\] \[\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|.|\overrightarrow{b}|}\] \[=\frac{(\hat{i}+2\hat{j}-\hat{k}).(2\hat{i}+\hat{j}+\hat{k})}{\sqrt{1+{{2}^{2}}+{{1}^{2}}}\sqrt{{{2}^{2}}+{{1}^{2}}+1}}\] \[=\frac{2+2-1}{\sqrt{1+4+1}\sqrt{4+1+1}}\] \[=\frac{3}{\sqrt{6}\sqrt{6}}\] \[=\frac{3}{6}\] \[\Rightarrow \] \[\cos \theta =\frac{1}{2}\cos 60{}^\circ \] \[\Rightarrow \] \[\theta =60{}^\circ \]
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