A) 12
B) 3
C) 8
D) 4
Correct Answer: B
Solution :
Using the formula: \[{{\left( \vec{a}\times \vec{b} \right)}^{2}}+{{\left( \vec{a}\,.\,\vec{b} \right)}^{2}}={{\left| {\vec{a}} \right|}^{2}}\,{{\left| {\vec{b}} \right|}^{2}}\] Given, \[{{\left| \vec{a}\times \vec{b} \right|}^{2}}+{{\left| \vec{a}\,.\,\vec{b} \right|}^{2}}=\,\,144\] and \[{{\left| \vec{a}\times \vec{b} \right|}^{2}}+{{\left| \vec{a}\,.\,\vec{b} \right|}^{2}}=\,\,144\] \[{{\left| {\vec{a}} \right|}^{2}}+\,\,4\,\,\Rightarrow \,\,144=16{{\left| {\vec{b}} \right|}^{2}}\Rightarrow {{\left| {\vec{b}} \right|}^{2}}=9\Rightarrow \left| {\vec{b}} \right|=3\]
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