Question
Two vectors $$\overrightarrow a = \overrightarrow i + \frac{{\overrightarrow j }}{{\sqrt 3 }}$$ and $$\overrightarrow b = \frac{{\overrightarrow i }}{{\sqrt 3 }} + \overrightarrow j $$ are :
A.
perpendicular to each other
B.
parallel to each other
C.
inclined to each other at an angle $$\frac{\pi }{3}$$
D.
inclined to each other at an angle $$\frac{\pi }{6}$$
Answer :
inclined to each other at an angle $$\frac{\pi }{6}$$
Solution :
$$\eqalign{
& \overrightarrow a .\overrightarrow b = \sqrt {1 + \frac{1}{3}} .\sqrt {1 + \frac{1}{3}} \cos \,\theta \cr
& {\text{or }}\frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }} = \frac{2}{{\sqrt 3 }}.\frac{2}{{\sqrt 3 }}\cos \,\theta \cr
& {\text{or }}\cos \,\theta = \frac{{\sqrt 3 }}{2} \cr
& \therefore \,\cos \,\theta = {30^ \circ } \cr} $$