A) 2
B) 3/2
C) 1/2
D) 1
Correct Answer: C
Solution :
Since, \[{{\overrightarrow{a}}_{_{1}}}\] and \[{{\overrightarrow{a}}_{_{2}}}\]are non-collinear \[\therefore \] \[{{a}_{1}}={{a}_{2}}=1\] Now, \[|\,\,{{\overrightarrow{a}}_{_{1}}}+{{\overrightarrow{a}}_{_{2}}}|\,\,=\sqrt{3}\] \[a_{1}^{2}+a_{2}^{2}+2{{a}_{1}}{{a}_{2}}\cos \theta ={{(\sqrt{3})}^{2}}\] or \[1+1+2\cos \theta =3\] or \[\cos \theta =\frac{1}{2}\] Now, \[({{\overrightarrow{a}}_{_{1}}}-{{\overrightarrow{a}}_{_{2}}})\cdot (2\,{{\overrightarrow{a}}_{_{1}}}+{{\overrightarrow{a}}_{_{2}}})\] \[=2a_{1}^{2}-a_{2}^{2}-{{a}_{1}}{{a}_{2}}\cos \theta \] \[=2-1-\frac{1}{2}=\frac{1}{2}\]You need to login to perform this action.
You will be redirected in
3 sec