A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{4}\]
Correct Answer: C
Solution :
| [c] Let \[\vec{c}=\hat{a}+2\hat{b}\] and \[\vec{d}=5\hat{a}-4\hat{b}\] |
| Since \[\text{\vec{c}}\] and \[\text{\vec{d}}\]are perpendicular to each other |
| \[\therefore \,\,\text{\vec{c}}\,\text{.}\,\text{\vec{d}=0}\Rightarrow \left( \text{\hat{a}+2}\,\text{\hat{b}} \right).\left( \text{5\hat{a}}-4\,\text{\hat{b}} \right)=0\] |
| \[\Rightarrow 5+\text{6}\,\text{\hat{a}}\,\text{.\hat{b}}-8=0\text{ }\left( \therefore \text{\vec{a}}\,\text{.}\,\text{\vec{a}}\,\text{=1} \right)\] |
| \[\Rightarrow \text{\hat{a}}\,\text{.\hat{b}=}\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}\] |
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