Question

If $A , B$ and $C$ are the vertices of a triangle whose position vectors are $\vec{a}, \vec{b}$ and $\vec{c}$ respectively and $G$ is the centroid of the $\Delta ABC$, then $\overrightarrow{ GA }+\overrightarrow{ GB }+\overrightarrow{ GC }$ is

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

Answer 1

Answer: (A)

Let $A, B, C$ are the vertices of a $\Delta$ whose position vectors are $\vec{a}, \vec{b}$ and $\vec{c}$ respectivety. Let $G$ be the centroid
of $\Delta ABC $
$ \therefore $ Centroid of triangle $( G )=\frac{\vec{a}|\vec{b}| \vec{c}}{3}$
Consider, $\overrightarrow{ GA }+\overrightarrow{ GB }+\overrightarrow{ GC }$
$=\left(\vec{a}-\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)+\left(\vec{b}-\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)+\left(\vec{c}-\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)$
$=\frac{1}{3}[2 \vec{a}-\vec{b}-\vec{c}+2 \vec{b}-\vec{a}-\vec{c}+2 \vec{c}-\vec{a}-\vec{b}]=0 $