A) \[2\]
B) \[\frac{3}{2}\]
C) \[\frac{1}{2}\]
D) \[1\]
Correct Answer: C
Solution :
Since,\[{{\mathbf{\vec{a}}}_{1}}\]and\[{{\mathbf{\vec{a}}}_{2}}\]are non-collinear \[\therefore \] \[{{a}_{1}}={{a}_{2}}=1\] and \[|{{\mathbf{\vec{a}}}_{1}}+{{\mathbf{\vec{a}}}_{2}}|=\sqrt{3}\] \[a_{1}^{2}+a_{2}^{2}+2{{a}_{1}}{{a}_{2}}=\cos \theta {{(\sqrt{3})}^{2}}\] or \[1+1+2\cos \theta =3\] or \[\cos \theta =\frac{1}{2}\] Now, \[({{\mathbf{\vec{a}}}_{1}}-{{\mathbf{\vec{a}}}_{2}})\cdot (2{{\mathbf{\vec{a}}}_{1}}+{{\mathbf{\vec{a}}}_{2}})\] \[=2a_{1}^{2}-a_{2}^{2}-{{a}_{1}}{{a}_{2}}\cos \theta \] \[=2-1-\frac{1}{2}=\frac{1}{2}\]
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