Question

Given position vectors $\overline{a},\overline{b},\overline{c}$ of points A, B, C for a triangle. The centroid can be given by

• A. $\frac{\overline{a}+\overline{b}+\overline{c}}{3}$
• B. $\frac{2}{3}(\overline{a}+\overline{b}+\overline{c})$
• C. $\overline{a}+\overline{b}+\overline{c}$
• D. $\frac{(\overline{a}.\overline{b})\times \overline{c}}{3}$

Answer 1

The correct option is A $\frac{\overline{a}+\overline{b}+\overline{c}}{3}$

Let’s construct the triangle BC. Let $P,M,N$ be the mid points of sides AB, BC and CA. By section formula P, M and N can be given by $\frac{\overline{a}+\overline{b}}{2},\frac{\overline{b}+\overline{a}}{2}and\frac{\overline{c}+\overline{a}}{2}$. Apart from that we also know the fact that centroid divides the line joining a vertex and midpoint of opposite side in the ratio 2: 3. $\therefore \overline{G}=\overline{a}+\frac{2}{3}\left(\overline{AM}\right)$ $\overline{AM}=\left(\frac{\overline{b}+\overline{c}}{2}\right)-\overline{a}\Rightarrow \overline{AM}=\frac{\overline{b}+\overline{c}-2\overline{a}}{2}\Rightarrow \overline{G}=\overline{a}+\frac{2}{3}\left(\frac{\overline{b}+\overline{c}-2\overline{a}}{2}\right)=\frac{\overline{a}+\overline{b}+\overline{c}}{3}$
$\therefore$ Centroid can be given by $\frac{\overline{a}+\overline{b}+\overline{c}}{3}$