Discussion

Answer: C

The area of triangle \(OPR\) is \(\frac{3\sqrt{3}}{2}\).

Perpendicularity gives \(\vec{a}\cdot(\vec{a} - \vec{b}) = 0\), so \(|\vec{a}|^{2} = \vec{a}\cdot\vec{b}\). Expand the two given lengths:

\[ \begin{aligned} |\vec{a}|^{2} + |\vec{b}|^{2} + 2\,\vec{a}\cdot\vec{b} &= 21 \\ |\vec{a}|^{2} + |\vec{b}|^{2} - 2\,\vec{a}\cdot\vec{b} &= 9 \end{aligned} \]

Subtracting, \(4\,\vec{a}\cdot\vec{b} = 12\), so \(\vec{a}\cdot\vec{b} = 3\) and \(|\vec{a}|^{2} = 3\). Adding, \(|\vec{a}|^{2} + |\vec{b}|^{2} = 15\), so \(|\vec{b}|^{2} = 12\).

With \(P\) at \(\vec{a}\) and \(R\) at \(\vec{a} + \vec{b}\), the area is half the cross product, and \(\vec{a} \times \vec{a} = \vec{0}\):

\[ \begin{aligned} \text{Area} &= \tfrac{1}{2}|\vec{a} \times (\vec{a} + \vec{b})| \\ &= \tfrac{1}{2}|\vec{a} \times \vec{b}| \end{aligned} \]

\[ \begin{aligned} |\vec{a} \times \vec{b}|^{2} &= |\vec{a}|^{2}|\vec{b}|^{2} \\ &\quad - (\vec{a}\cdot\vec{b})^{2} \\ &= 36 - 9 = 27 \\ \text{Area} &= \tfrac{1}{2}\sqrt{27} = \tfrac{3\sqrt{3}}{2} \end{aligned} \]

\(\sqrt{3}\) is \(|\vec{a}|\) itself, and \(\frac{3}{2}\) is half the dot product rather than half the cross product; \(\frac{\sqrt{3}}{2}\) is half of \(|\vec{a}|\).