Question

If $G$ and $G^{\prime }$ be the centroids of the triangles $ABC$ and $A^{\prime }{B}^{\prime }{C}^{\prime }$ respectively, then $\overrightarrow{A{A}^{\prime }}+\overrightarrow{B{B}^{\prime }}+\overrightarrow{C{C}^{\prime }}=$

• A. $3\overrightarrow{G^{\prime }G}$
• B. $3\overrightarrow{G{G}^{\prime }}$
• C. $2\overrightarrow{G{G}^{\prime }}$
• D. $\frac{3}{2}\overrightarrow{G{G}^{\prime }}$

Answer 1

The correct option is B $3\overrightarrow{G{G}^{\prime }}$

Let position vector of $A\left(\overrightarrow{a}\right),B\left(\overrightarrow{b}\right),C\left(\overrightarrow{c}\right),A^{\prime }\left(\overrightarrow{a^{\prime }}\right),B^{\prime }\left(\overrightarrow{b^{\prime }}\right)$ and $C^{\prime }\left(\overrightarrow{c^{\prime }}\right)$
Now position vector of $G=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}$
Position vector of $G^{\prime }=\frac{\overrightarrow{a^{\prime }}+\overrightarrow{b^{\prime }}+\overrightarrow{c^{\prime }}}{3}$
Now $\overrightarrow{A{A}^{\prime }}=\overrightarrow{a^{\prime }}-\overrightarrow{a},\overrightarrow{B{B}^{\prime }}=\overrightarrow{b^{\prime }}-\overrightarrow{b},\overrightarrow{C{C}^{\prime }}=\overrightarrow{c^{\prime }}-\overrightarrow{c}$
So $\overrightarrow{A{A}^{\prime }}+\overrightarrow{B{B}^{\prime }}+\overrightarrow{C{C}^{\prime }}=(\overrightarrow{a^{\prime }}+\overrightarrow{b^{\prime }}+\overrightarrow{c^{\prime }})-(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$
$\Rightarrow \overrightarrow{A{A}^{\prime }}+\overrightarrow{B{B}^{\prime }}+\overrightarrow{C{C}^{\prime }}=3(-\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}+\frac{\overrightarrow{a^{\prime }}+\overrightarrow{b^{\prime }}+\overrightarrow{c^{\prime }}}{3})$
$\Rightarrow \overrightarrow{A{A}^{\prime }}+\overrightarrow{B{B}^{\prime }}+\overrightarrow{C{C}^{\prime }}=3\overrightarrow{G{G}^{\prime }}$