A) \[\overset{\to }{\mathop{A}}\,.(\overset{\to }{\mathop{A}}\,+\overset{\to }{\mathop{B}}\,)={{A}^{2}}+AB\]
B) \[\overset{\to }{\mathop{A}}\,\times (\overset{\to }{\mathop{A}}\,+\overset{\to }{\mathop{B}}\,)=AB\]_
C) \[\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\,=0\]
D) \[\overset{\to }{\mathop{A}}\,\times (\overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\,)=9\]
Correct Answer: A
Solution :
\[\overset{\to }{\mathop{A}}\,.\left( \overset{\to }{\mathop{A}}\,+\overset{\to }{\mathop{B}}\, \right)={{A}^{2}}+\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\,\] \[={{A}^{2}}+AB\,\cos {{0}^{{}^\circ }}\] \[={{A}^{2}}+AB\]
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