Correct Answer - Option 1 :
\(\frac{2}{3}\)
Concept:
Unit vector along x-axis = î + 0ĵ +0k̂ = î → (1)
Unit vector along y-axis = ĵ
Unit vector along z-axis = k̂
And angle between two vectors \(\vec a {\:}\& {\:}\vec b\)
\(\cos \theta = \frac{{\vec a.\vec b}}{{\left| {\vec a} \right|.\left| {\vec b} \right|}}\)
Where
\(\left| {\vec a} \right| = \sqrt {a_1^2 + b_1^2 + c_1^2} \)
\(\left| {\vec b} \right| = \sqrt {a_2^2 + b_2^2 + c_2^2} \)
Calculation:
Given,
Vector, \(\vec a = 2\widehat i + 2\widehat j + \widehat k\)
By equation (1), unit vector along x-axis, \(\vec b = \widehat i + 0\widehat j + 0\widehat k\)
Then, value of cosine,
\(\cos \theta = \frac{{\left( {2\widehat i + 2\widehat j + \widehat k} \right).\left( {\widehat i + 0\widehat j + 0\widehat k} \right)}}{{\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{1^2} + 0 + 0} }}\)
\( \Rightarrow \cos \theta = \frac{2}{{\sqrt 9 .\sqrt 1 }}\)
\(\Rightarrow \cos \theta = \frac{2}{3}Ans\)
