Question

If in a $△ABC$, $A\equiv (0,0)$ , $B\equiv (3,3\sqrt{3})$ , $C\equiv (-3\sqrt{3},3)$, then the vector of magnitude $2\sqrt{2}$ units directed along where , is the circumcentre of $△ABC$ is :

• A. $(1-\sqrt{3})\hat{i}+(1+\sqrt{3})\hat{j}$
• B. $(1+\sqrt{3})\hat{i}+(1-\sqrt{3})\hat{j}$
• C. $(1+\sqrt{3})\hat{i}+(\sqrt{3}-1)\hat{j}$
• D. None of these

Answer 1

The correct option is A $(1-\sqrt{3})\hat{i}+(1+\sqrt{3})\hat{j}$

The given information can be represented using the figure shown

Slope of $AC=\frac{0-3}{0+3\sqrt{3}}=\frac{-1}{\sqrt{3}}$

Slope of $AB=\frac{0-3\sqrt{3}}{0-3}=\sqrt{3}$

Slope of $AC\times$Slope of $AB=\frac{-1}{\sqrt{3}}\times \sqrt{3}=-1$

$\therefore \angle CAB={90}^{\circ }$

Thus, the circumcentre$\left(O\right)$ of $△ABC$ is the mid-point of $BC$ and its co-ordinates are $(\frac{3-3\sqrt{3}}{2},\frac{3+3\sqrt{3}}{2})$

Now, $\overrightarrow{OA}=(\frac{3-3\sqrt{3}}{2}-0)\hat{i}+\left(\frac{3+3\sqrt{3}}{2}\right)\hat{j}$

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

Now, $\left|\overrightarrow{OA}\right|=\sqrt{{\left(\frac{3-3\sqrt{3}}{2}\right)}^2+{\left(\frac{3+3\sqrt{3}}{2}\right)}^2}$

$=\sqrt{\frac{9+27-18\sqrt{3}+9+27+18\sqrt{3}}{4}}$

$=\sqrt{18}=3\sqrt{2}$ on simplification

Unit vector along $\overrightarrow{OA}=\frac{\overrightarrow{OA}}{\left|\overrightarrow{OA}\right|}$

$=\frac{1}{3\sqrt{2}}(\frac{3-3\sqrt{3}}{2}\hat{i}+\frac{3+3\sqrt{3}}{2}\hat{j})$

Thus, the vector along $\overrightarrow{OA}$ whose magnitude is $2\sqrt{2}$

$=2\sqrt{2}\times \frac{1}{3\sqrt{2}}(\frac{3-3\sqrt{3}}{2}\hat{i}+\frac{3+3\sqrt{3}}{2}\hat{j})$

$=\frac{2}{3}\times \frac{3}{2}[(1-\sqrt{3})\hat{i}+(1+\sqrt{3})\hat{j}]$

$=(1-\sqrt{3})\hat{i}+(1+\sqrt{3})\hat{j}$