A) \[{{\cos }^{-1}}\left( \frac{1}{10} \right)\]
B) \[{{\cos }^{-1}}\left( \frac{9}{11} \right)\]
C) \[{{\cos }^{-1}}\left( \frac{9}{91} \right)\]
D) \[{{\cos }^{-1}}\left( \frac{1}{9} \right)\]
Correct Answer: C
Solution :
Let\[\overset{\to }{\mathop{\mathbf{a}}}\,=2\widehat{\mathbf{i}}+6\widehat{\mathbf{j}}+3\widehat{\mathbf{k}}\]and\[\overset{\to }{\mathop{\mathbf{b}}}\,=12\widehat{\mathbf{i}}-4\widehat{\mathbf{j}}+3\widehat{\mathbf{k}}\] \[\therefore \] \[\cos \theta =\frac{\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,}{|\overset{\to }{\mathop{\mathbf{a}}}\,||\overset{\to }{\mathop{\mathbf{b}}}\,|}\] \[=\frac{(2\widehat{\mathbf{i}}+6\widehat{\mathbf{j}}+3\widehat{\mathbf{k}})\cdot (12\widehat{\mathbf{i}}-4\widehat{\mathbf{j}}+3\widehat{\mathbf{k}})}{\sqrt{4+36+9}\sqrt{144+16+9}}\] \[=\frac{24-24+9}{7\cdot 13}=\frac{9}{91}\] \[\Rightarrow \] \[\theta ={{\cos }^{-1}}\left( \frac{9}{91} \right)\]
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