Key Concepts and Formulas

1. Area of a Parallelogram: The area of a parallelogram with adjacent sides represented by vectors $\vec{u}$ and $\vec{v}$ is given by the magnitude of their cross product: $\text{Area} = |\vec{u} \times \vec{v}|$.
2. Area of a Triangle: The area of a triangle with two sides originating from a common vertex represented by vectors $\vec{u}$ and $\vec{v}$ is half the magnitude of their cross product: $\text{Area} = \frac{1}{2} |\vec{u} \times \vec{v}|$.
3. Properties of the Cross Product:
   • $\vec{x} \times \vec{x} = \vec{0}$
   • $\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})$
   • $k(\vec{x} \times \vec{y}) = (k\vec{x}) \times \vec{y} = \vec{x} \times (k\vec{y})$
   • $\vec{x} \times (\vec{y} + \vec{z}) = (\vec{x} \times \vec{y}) + (\vec{x} \times \vec{z})$

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Step-by-Step Solution

We are given the position vectors of points A, B, and C with respect to the origin O: $\overrightarrow{OA} = \vec{a}$ $\overrightarrow{OB} = 12\vec{a} + 4\vec{b}$ $\overrightarrow{OC} = \vec{b}$

Step 1: Calculate the Area of Parallelogram S

Parallelogram S has adjacent sides OA and OC. The area of S is determined by the magnitude of the cross product of these two vectors. $\text{Area}(S) = |\overrightarrow{OA} \times \overrightarrow{OC}|$ Substituting the given vectors: $\text{Area}(S) = |\vec{a} \times \vec{b}|$ We will use this as our reference area.

Step 2: Calculate the Area of Quadrilateral OABC

The quadrilateral OABC can be divided into two triangles with a common vertex at the origin O: $\triangle OAB$ and $\triangle OBC$. The total area of the quadrilateral is the sum of the areas of these two triangles. $\text{Area(OABC)} = \text{Area}(\triangle OAB) + \text{Area}(\triangle OBC)$

Step 2a: Calculate the Area of $\triangle OAB$

The sides of $\triangle OAB$ originating from O are $\overrightarrow{OA}$ and $\overrightarrow{OB}$. $\text{Area}(\triangle OAB) = \frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}|$ Substitute the given vectors: $\text{Area}(\triangle OAB) = \frac{1}{2} |\vec{a} \times (12\vec{a} + 4\vec{b})|$ Using the distributive property of the cross product: $\vec{a} \times (12\vec{a} + 4\vec{b}) = (\vec{a} \times 12\vec{a}) + (\vec{a} \times 4\vec{b})$ Using the scalar multiplication and the property $\vec{x} \times \vec{x} = \vec{0}$: $= 12(\vec{a} \times \vec{a}) + 4(\vec{a} \times \vec{b}) = 12(\vec{0}) + 4(\vec{a} \times \vec{b}) = 4(\vec{a} \times \vec{b})$ Therefore, the area of $\triangle OAB$ is: $\text{Area}(\triangle OAB) = \frac{1}{2} |4(\vec{a} \times \vec{b})| = \frac{1}{2} \cdot 4 |\vec{a} \times \vec{b}| = 2 |\vec{a} \times \vec{b}|$

Step 2b: Calculate the Area of $\triangle OBC$

The sides of $\triangle OBC$ originating from O are $\overrightarrow{OC}$ and $\overrightarrow{OB}$. $\text{Area}(\triangle OBC) = \frac{1}{2} |\overrightarrow{OC} \times \overrightarrow{OB}|$ Substitute the given vectors: $\text{Area}(\triangle OBC) = \frac{1}{2} |\vec{b} \times (12\vec{a} + 4\vec{b})|$ Using the distributive property of the cross product: $\vec{b} \times (12\vec{a} + 4\vec{b}) = (\vec{b} \times 12\vec{a}) + (\vec{b} \times 4\vec{b})$ Using the scalar multiplication, the property $\vec{x} \times \vec{x} = \vec{0}$, and the anti-commutative property $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$: $= 12(\vec{b} \times \vec{a}) + 4(\vec{b} \times \vec{b}) = 12(-(\vec{a} \times \vec{b})) + 4(\vec{0}) = -12(\vec{a} \times \vec{b})$ Therefore, the area of $\triangle OBC$ is: $\text{Area}(\triangle OBC) = \frac{1}{2} |-12(\vec{a} \times \vec{b})| = \frac{1}{2} \cdot |-12| |\vec{a} \times \vec{b}| = \frac{1}{2} \cdot 12 |\vec{a} \times \vec{b}| = 6 |\vec{a} \times \vec{b}|$

Step 2c: Sum the Areas to find Area(OABC)

Now, add the areas of the two triangles: $\text{Area(OABC)} = \text{Area}(\triangle OAB) + \text{Area}(\triangle OBC)$ $\text{Area(OABC)} = 2 |\vec{a} \times \vec{b}| + 6 |\vec{a} \times \vec{b}|$ $\text{Area(OABC)} = 8 |\vec{a} \times \vec{b}|$

Step 3: Calculate the Ratio of Areas

We need to find the ratio of the area of quadrilateral OABC to the area of parallelogram S. $\frac{\text{Area of quadrilateral OABC}}{\text{Area of S}} = \frac{8 |\vec{a} \times \vec{b}|}{|\vec{a} \times \vec{b}|}$ Since $|\vec{a} \times \vec{b}|$ is a non-zero magnitude (assuming $\vec{a}$ and $\vec{b}$ are not collinear), we can cancel it out: $\frac{\text{Area of quadrilateral OABC}}{\text{Area of S}} = 8$

Common Mistakes & Tips

• Sign Errors with Cross Products: Be careful with the anti-commutative property ($\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})$). When calculating the area of $\triangle OBC$, the order of vectors in the cross product matters for the intermediate calculation, but the final magnitude of the area will be positive.
• Forgetting the 1/2 for Triangle Area: Always remember that the area of a triangle is half the magnitude of the cross product of its two adjacent sides.
• Misinterpreting Quadrilateral Decomposition: Ensure that the quadrilateral is divided into triangles that share a common diagonal and whose areas can be calculated using the given vectors from the origin.

Summary

The problem requires calculating the areas of a parallelogram and a quadrilateral using vector cross products. The area of parallelogram S, with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$, is $|\vec{a} \times \vec{b}|$. The quadrilateral OABC is decomposed into two triangles, $\triangle OAB$ and $\triangle OBC$. The area of $\triangle OAB$ is found to be $2|\vec{a} \times \vec{b}|$, and the area of $\triangle OBC$ is $6|\vec{a} \times \vec{b}|$. Summing these gives the area of quadrilateral OABC as $8|\vec{a} \times \vec{b}|$. The ratio of the area of OABC to the area of S is then $8|\vec{a} \times \vec{b}| / |\vec{a} \times \vec{b}| = 8$.

The final answer is \boxed{7}.