Question

$ABC$ is an equilateral triangle of side $a$. The value of $\overrightarrow{AB}.\overrightarrow{BC}+\overrightarrow{BC}.\overrightarrow{CA}+\overrightarrow{CA}.\overrightarrow{AB}$ is equal to

• A. $\frac{3a^2}{2}$
• B. $3a^2$
• C. $-\frac{3a^2}{2}$
• D. None of these

Answer 1

The correct option is C $-\frac{3a^2}{2}$

Let $B$ be the origin.
$C=(a,0)$ and $A=(\frac{a}{2},\frac{a\sqrt{3}}{2})$

Thus, $\overrightarrow{AB}=-\frac{a}{2}\hat{i}-\frac{a\sqrt{3}}{2}\hat{j}$
$\overrightarrow{BC}=a\hat{i}$

$\overrightarrow{CA}=-\frac{a}{2}\hat{i}+\frac{a\sqrt{3}}{2}\hat{j}$

Hence, $\overrightarrow{AB}.\overrightarrow{BC}+\overrightarrow{BC}.\overrightarrow{CA}+\overrightarrow{CA}.\overrightarrow{AB}$

$=\frac{-a^2}{2}+\left(\frac{-a^2}{2}\right)+(\frac{a^2}{4}-\frac{3a^2}{4})$

$=\frac{-3a^2}{2}$