A) \[5\pi \] sq units
B) \[10\pi \]sq units
C) \[25\pi \] sq units
D) \[35\pi \]sq units
Correct Answer: C
Solution :
We know that, \[|\overrightarrow{a}-\overrightarrow{b}{{|}^{2}}=|\overrightarrow{a}{{|}^{2}}+|\overrightarrow{b}{{|}^{2}}-2\overrightarrow{a}.\overrightarrow{b}\] \[=|\overrightarrow{a}|+|\overrightarrow{b}{{|}^{2}}-2|\overrightarrow{a}||\overrightarrow{b}|\cos 2\theta \] \[(\because |\overrightarrow{a}|=|\overrightarrow{b}|=1)\] \[\Rightarrow \] \[|\overrightarrow{a}-\overrightarrow{b}{{|}^{2}}=2-2\cos 2\theta \] \[\Rightarrow \] \[|\overrightarrow{a}-\overrightarrow{b}{{|}^{2}}=4{{\cos }^{2}}\theta \] \[\Rightarrow \] \[|\overrightarrow{a}-\overrightarrow{b}|=2|\sin \theta |\] But it is given that \[|\overrightarrow{a}-\overrightarrow{b}|<1\] \[\therefore \] \[2|\sin \theta |<1\] \[\Rightarrow \] \[|\sin \theta |<\frac{1}{2}\] \[\Rightarrow \] \[\theta \in \left[ 0,\frac{\pi }{6} \right)\]for\[\theta \in \left( \frac{5\pi }{6},\pi \right]\]
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