Area of Triangle Formed by Vector Position Points

To find the area of the triangle $OPR$, we start by expressing the vector area of the triangle in terms of the given vectors.

The position vectors of the vertices with respect to the origin $O$ are:

• $O = \vec{0}$
• $P = \vec{a}$
• $R = \vec{a} + \vec{b}$

The vectors representing the sides $\vec{OP}$ and $\vec{OR}$ are: $\vec{OP} = \vec{a}$ $\vec{OR} = \vec{a} + \vec{b}$

The area of triangle $OPR$ is given by: $\text{Area}(\Delta OPR) = \frac{1}{2} |\vec{OP} \times \vec{OR}| = \frac{1}{2} |\vec{a} \times (\vec{a} + \vec{b})|$

Using the distributive property of the vector cross product and noting that $\vec{a} \times \vec{a} = \vec{0}$: $\vec{a} \times (\vec{a} + \vec{b}) = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} = \vec{a} \times \vec{b}$

Thus, $\text{Area}(\Delta OPR) = \frac{1}{2} |\vec{a} \times \vec{b}|$

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Now, we determine the magnitude $|\vec{a} \times \vec{b}|$ using the given conditions:

1. $|\vec{a} + \vec{b}| = \sqrt{21} \implies |\vec{a} + \vec{b}|^2 = 21$ $|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 21 \quad \text{--- (1)}$

2. $|\vec{a} - \vec{b}| = 3 \implies |\vec{a} - \vec{b}|^2 = 9$ $|\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 9 \quad \text{--- (2)}$

Subtracting equation (2) from equation (1): $4(\vec{a} \cdot \vec{b}) = 21 - 9 = 12 \implies \vec{a} \cdot \vec{b} = 3$

Adding equation (1) and equation (2): $2(|\vec{a}|^2 + |\vec{b}|^2) = 30 \implies |\vec{a}|^2 + |\vec{b}|^2 = 15 \quad \text{--- (3)}$

1. We are also given that $\vec{a}$ and $(\vec{a} - \vec{b})$ are perpendicular: $\vec{a} \cdot (\vec{a} - \vec{b}) = 0$ $|\vec{a}|^2 - \vec{a} \cdot \vec{b} = 0 \implies |\vec{a}|^2 = \vec{a} \cdot \vec{b}$

Since $\vec{a} \cdot \vec{b} = 3$, we have: $|\vec{a}|^2 = 3$

Substituting $|\vec{a}|^2 = 3$ into equation (3): $3 + |\vec{b}|^2 = 15 \implies |\vec{b}|^2 = 12$

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Using Lagrange's Identity to calculate $|\vec{a} \times \vec{b}|^2$: $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$ $|\vec{a} \times \vec{b}|^2 = (3)(12) - (3)^2 = 36 - 9 = 27$ $|\vec{a} \times \vec{b}| = \sqrt{27} = 3\sqrt{3}$

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Finally, substituting $|\vec{a} \times \vec{b}|$ back into the area formula: $\text{Area}(\Delta OPR) = \frac{1}{2} (3\sqrt{3}) = \frac{3\sqrt{3}}{2}$

Hence, the correct option is (C).