question_answer

In a parallelogram \[ABCD,\left| \overrightarrow{AB} \right|=a,\left| \overrightarrow{AD} \right|=b\]and \[\left| \overrightarrow{AC} \right|=c,\] then \[\overrightarrow{AB}.\overrightarrow{AB}\] has the value : [JEE Main Online Paper (Held On 11 April 2015)]  

A) \[\frac{1}{2}\left( {{a}^{2}}-{{b}^{2}}+{{c}^{2}} \right)\]

B) \[\frac{1}{4}\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)\]

C) \[\frac{1}{3}\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)\]

D) \[\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\]

E) None of these

Correct Answer: E

Solution :

\[\overline{a}.\overline{b}=\left| a \right|\left| b \right|\cos \theta \]22 ...(1) \[\overline{a}+\overline{b}=\overline{c}\]\[\Rightarrow \]\[{{\left[ a+b \right]}^{2}}={{\left| c \right|}^{2}}\] \[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}+2a\cdot b={{c}^{2}}\] \[\Rightarrow \]\[a\cdot b=\frac{{{c}^{2}}-{{a}^{2}}-{{b}^{2}}}{2}\] ?.(2) Now\[\overrightarrow{DB}\cdot \overrightarrow{AB}\] \[=\left( \vec{a}-\vec{b} \right)\cdot \vec{a}={{a}^{2}}-a.b\] \[={{a}^{2}}-\frac{{{c}^{2}}-{{a}^{2}}-{{b}^{2}}}{2}=\frac{1}{2}\left( 3{{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)\] Hence none of the answers is correct.