A) \[\frac{2}{3}\overrightarrow{GG}'\]
B) \[\overrightarrow{GG}'\]
C) \[2\,\overrightarrow{GG}'\]
D) \[3\,\overrightarrow{GG}'\]
Correct Answer: D
Solution :
\[\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\mathbf{0}\] and \[\overrightarrow{{G}'{A}'}+\overrightarrow{{G}'{B}'}+\overrightarrow{{G}'{C}'}=\mathbf{0}\] \[\Rightarrow (\overrightarrow{GA}-\overrightarrow{{G}'{A}'})+(\overrightarrow{GB}-\overrightarrow{{G}'{B}'})+(\overrightarrow{GC}-\overrightarrow{{G}'{C}'})=\mathbf{0}\] \[\Rightarrow (\overrightarrow{GA}+\overrightarrow{{G}'G}-\overrightarrow{{G}'{A}'})+(\overrightarrow{GB}+\overrightarrow{{G}'G}-\overrightarrow{{G}'{B}'})\]\[+(\overrightarrow{GC}+\overrightarrow{{G}'G}-\overrightarrow{{G}'{C}'})=3\overrightarrow{{G}'G}\] \[\Rightarrow (\overrightarrow{GA}-\overrightarrow{G{A}'})+(\overrightarrow{GB}-\overrightarrow{G{B}'})+(\overrightarrow{GC}-\overrightarrow{G{C}'})=3\overrightarrow{{G}'G}\] \[\Rightarrow \overrightarrow{{A}'A}+\overrightarrow{{B}'B}+\overrightarrow{{C}'C}=3\overrightarrow{{G}'G}\] \[\Rightarrow \overrightarrow{A{A}'}+\overrightarrow{B{B}'}+\overrightarrow{C{C}'}=3\overrightarrow{G{G}'}\].You need to login to perform this action.
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