Question

If $A,B$ and $C$ are the vertices of a triangle whose position vectors are $a,b$ and $c$ respectively and $G$ is the centroid of the $∆ABC$, then $GA+GB+GC$ is?

• A. $0$
• B. $a+b+c$
• C. $\frac{(a+b+c)}{3}$
• D. $\frac{(a-b-c)}{3}$

Answer 1

The correct option is A

$0$

Find the value of $\mathit{G}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{G}\mathit{B}\mathbf{}\mathbf{+}\mathbf{}\mathit{G}\mathit{C}$:

Given that $A,B$ and $C$ are the vertices of a triangle whose position vectors are $a,b$ and $c$ .

and $G$ is the centroid of the $∆ABC$

$$\begin{gathered} G=\frac{(a+b+c)}{3} \\ GA=\frac{a-(a+b+c)}{3} \\ GB=\frac{b-(a+b+c)}{3} \\ GC=\frac{c-(a+b+c)}{3} \end{gathered}$$

Therefore,

$$\begin{gathered} GA+GB+GC=(a+b+c)-\frac{3(a+b+c)}{3} \\ \Rightarrow GA+GB+GC=(a+b+c)-(a+b+c) \\ \Rightarrow GA+GB+GC=0 \end{gathered}$$

Hence, the correct option is (A).