Question
If $$\overrightarrow a $$ and $$\overrightarrow b $$ are unit vectors and $$\alpha $$ is the angle between them then $$\cos \frac{\alpha }{2}$$ is equal to :
A.
$$\frac{1}{2}\left| {\overrightarrow a + \overrightarrow b } \right|$$
B.
$$\frac{1}{2}\left| {\overrightarrow a - \overrightarrow b } \right|$$
C.
$$\left| {\overrightarrow a + \overrightarrow b } \right|$$
D.
none of these
Answer :
$$\frac{1}{2}\left| {\overrightarrow a + \overrightarrow b } \right|$$
Solution :
$$\eqalign{
& {\text{Here,}} \cr
& \cos \,\alpha = \overrightarrow a .\overrightarrow b {\text{ or }}2{\cos ^2}\frac{\alpha }{2} - 1 = \overrightarrow a .\overrightarrow b \,\,\,{\text{or }}4{\cos ^2}\frac{\alpha }{2} = 2 + 2\overrightarrow a .\overrightarrow b \cr
& \therefore \,{\left( {2{{\cos }^2}\frac{\alpha }{2}} \right)^2} = {\overrightarrow a ^2} + {\overrightarrow b ^2} + 2\overrightarrow a .\overrightarrow b = {\left( {\overrightarrow a + \overrightarrow b } \right)^2} = {\left| {\overrightarrow a + \overrightarrow b } \right|^2} \cr
& {\text{or }}2\cos \frac{\alpha }{2} = \left| {\overrightarrow a + \overrightarrow b } \right| \cr
& \therefore \,\cos \frac{\alpha }{2} = \frac{1}{2}\left| {\overrightarrow a + \overrightarrow b } \right|. \cr} $$