Let $\bar{a},\bar{b}$ be two vectors, and let P, Q and R be the points with position vectors $\bar{a},\bar{b}$ and $\bar{a}+\bar{b}$, respectively, with respect to the origin O. If $|\bar{a}+\bar{b}|=\sqrt{21}$, $|\bar{a}-\bar{b}|=3$, and $\bar{a}$ and $(\bar{a}-\bar{b})$ are perpendicular to each other, then the area of the triangle OPR is

$\text{Given: } |\bar{a}+\bar{b}|=\sqrt{21}$

$|\bar{a}-\bar{b}|=3$

$\bar{a} \perp (\bar{a}-\bar{b})$

To find: area of triangle OPR

$(1)\ \sqrt{3}\qquad (2)\ \frac{\sqrt{3}}{2}$

$(3)\ \frac{3\sqrt{3}}{2}\qquad (4)\ \frac{3}{2}$

$\vec{OR} = \bar{a}+\bar{b}$ $1.\ \bar{a}\cdot(\bar{a}-\bar{b})=0\ \Rightarrow\ \text{a relation between } |\bar{a}| \text{ and } \bar{a}\cdot\bar{b}$ $2.\ |\bar{a}+\bar{b}|^2,\ |\bar{a}-\bar{b}|^2\ \Rightarrow\ |\bar{a}|,\ |\bar{b}|$ $3.\ \text{area} = \tfrac12\,|\bar{a}\times\bar{b}|$ $\bar{a}\cdot(\bar{a}-\bar{b}) = 0$ $|\bar{a}|^2 - \bar{a}\cdot\bar{b} = 0 \Rightarrow |\bar{a}|^2 = \bar{a}\cdot\bar{b}$ $|\bar{a}+\bar{b}|^2 = (\sqrt{21})^2 = 21$ $|\bar{a}|^2 + |\bar{b}|^2 + 2\bar{a}\cdot\bar{b} = 21$ $|\bar{a}-\bar{b}|^2 = 3^2 = 9$ $|\bar{a}|^2 + |\bar{b}|^2 - 2\bar{a}\cdot\bar{b} = 9$ $2\!\left(|\bar{a}|^2+|\bar{b}|^2\right) = 21+9 = 30$ $|\bar{a}|^2 + |\bar{b}|^2 = 15$ $4\bar{a}\cdot\bar{b} = 21-9 = 12$ $\bar{a}\cdot\bar{b} = 3$ $|\bar{a}|^2 = \bar{a}\cdot\bar{b} = 3 \Rightarrow |\bar{a}| = \sqrt{3}$ $|\bar{b}|^2 = 15-3 = 12 \Rightarrow |\bar{b}| = 2\sqrt{3}$ $\bar{a} = \sqrt{3}\,\hat{i}$ $\cos\theta = \dfrac{\bar{a}\cdot\bar{b}}{|\bar{a}|\,|\bar{b}|} = \dfrac{3}{\sqrt{3}\cdot 2\sqrt{3}} = \dfrac12 \Rightarrow \theta = 60^\circ$ $\bar{b} = \sqrt{3}\,\hat{i} + 3\,\hat{j}$ $\text{Area}(\triangle OPR) = \tfrac12 \cdot \text{Area}(OPQR) = \tfrac12\,|\bar{a}\times\bar{b}|$ $|\bar{a}\times\bar{b}|^2 = |\bar{a}|^2|\bar{b}|^2 - (\bar{a}\cdot\bar{b})^2$ $= 3\cdot 12 - 3^2 = 36 - 9 = 27$ $|\bar{a}\times\bar{b}| = \sqrt{27} = 3\sqrt{3}$ $\text{Area}(\triangle OPR) = \tfrac12 \cdot 3\sqrt{3} = \dfrac{3\sqrt{3}}{2}$ $\boxed{\text{Answer: (3)}\quad \dfrac{3\sqrt{3}}{2}}$