A) \[{{\left( {{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+}\frac{\text{AB}}{\sqrt{\text{3}}} \right)}^{1/2}}\]
B) \[\text{A+B}\]
C) \[{{\left( {{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{AB} \right)}^{1/2}}\]
D) \[{{\left( {{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+AB} \right)}^{1/2}}\]
Correct Answer: D
Solution :
[d] \[\text{ }\!\!|\!\!\text{ \vec{A} }\!\!\times\!\!\text{ }\,\text{\vec{B} }\!\!|\!\!\text{ =}\,\sqrt{\text{3}}\text{. \vec{A} }\text{. \vec{B}}\] \[\text{or AB sin}\theta \,\,\text{=}\,\sqrt{3}\text{ AB cos}\theta \] \[\therefore \,\,\,\,\text{tan}\theta \,\,\text{=}\,\sqrt{3}\text{ , or }\theta \,\text{=}\,\,\text{60}{}^\circ \] Thus \[\text{ }\!\!|\!\!\text{ \vec{A}+\vec{B} }\!\!|\!\!\text{ =}\sqrt{{{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+2}\,\text{AB}\,\text{cos}\,\text{60 }\!\!{}^\circ\!\!\text{ }}\] \[=\sqrt{{{\text{A}}^{\text{2}}}\text{+}{{\text{B}}^{\text{2}}}\text{+2AB}}.\]
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