Question
If $$\overrightarrow p $$ and $$\overrightarrow q $$ are two unit vectors inclined at an angle $$\alpha $$ to each other then $$\left| {\overrightarrow p + \overrightarrow q } \right| < 1$$ if :
A.
$$\frac{{2\pi }}{3} < \alpha < \frac{{4\pi }}{3}$$
B.
$$\frac{{4\pi }}{3} < \alpha < 2\pi $$
C.
$$0 < \alpha < \frac{\pi }{3}$$
D.
$$\alpha = \frac{\pi }{2}$$
Answer :
$$\frac{{2\pi }}{3} < \alpha < \frac{{4\pi }}{3}$$
Solution :
$$\eqalign{
& \left| {\overrightarrow p + \overrightarrow q } \right| = \left( {\overrightarrow p + \overrightarrow q } \right).\left( {\overrightarrow p + \overrightarrow q } \right) \cr
& = {\left| {\overrightarrow p } \right|^2} + {\left| {\overrightarrow q } \right|^2} + 2\overrightarrow p .\overrightarrow q \cr
& = 2 + 2\,\cos \,\alpha , \cr
& {\text{where }}\alpha {\text{ is the angle between}}\overrightarrow p {\text{ and }}\overrightarrow q \cr
& = 2\left( {1 + \cos \,\alpha } \right) \cr
& = 4\,{\cos ^2}\left( {\frac{\alpha }{2}} \right) \cr
& {\left| {\overrightarrow p + \overrightarrow q } \right|^2} < 1 \cr
& \Rightarrow \left( {4\,{{\cos }^2}\frac{\alpha }{2} - 1} \right) < 0 \cr
& \Rightarrow \left( {2\,\cos \frac{\alpha }{2} - 1} \right)\left( {2\,\cos \frac{\alpha }{2} + 1} \right) < 0,\, - \frac{1}{2} < \cos \frac{\alpha }{2} < \frac{1}{2} \cr
& \Rightarrow \frac{\pi }{3} < \frac{\alpha }{2} < \frac{{2\pi }}{3} \cr
& \Rightarrow \frac{{2\pi }}{3} < \alpha < \frac{{4\pi }}{3} \cr} $$