Question

Let $\mathrm{G}$ be the centroid of a triangle $\mathrm{ABC}$. If$\mathrm{AB}=\mathrm{a},\mathrm{AC}=\mathrm{b}$, then the bisector$\mathrm{AG}$, in terms of vectors $\mathrm{a}$ and $\mathrm{b}$ is?

• A. $\frac{1}{2}(\mathrm{a}+\mathrm{b})$
• B. $\frac{1}{3}(\mathrm{a}+\mathrm{b})$
• C. $\frac{2}{3}(\mathrm{a}+\mathrm{b})$
• D. $\frac{1}{6}(\mathrm{a}+\mathrm{b})$

Answer 1

The correct option is B

$\frac{1}{3}(\mathrm{a}+\mathrm{b})$

Determine the value of bisector $\mathrm{AG}$.

We have a triangle $\mathrm{ABC}$ in which $\mathrm{G}$ is its centroid and $\mathrm{AB}=\overrightarrow{\mathrm{a}}$ and $\mathrm{AC}=\overrightarrow{\mathrm{b}}$. Let us take $\mathrm{A}$ as an origin, hence position vectors of $\mathrm{A}$, $\mathrm{B}$ and $\mathrm{C}$ are $\overrightarrow{0}$, $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ respectively.

Now we know that if the position vectors of three vertices of a triangle are given then the position vector of a centroid is given by adding the three position vectors of vertices and dividing it by $3$. Here we have the position vector of the centroid $\mathrm{G}=\overrightarrow{\mathrm{AG}}$. Therefore:

$$\begin{gathered} \overrightarrow{\mathrm{AG}}=\frac{\overrightarrow{0}+\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}}{3} \\ \Rightarrow \overrightarrow{\mathrm{AG}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}}{3} \end{gathered}$$

Hence option B is the correct answer.