Key Concepts and Formulas

• Area of a Triangle: The area of a triangle with adjacent sides represented by vectors $\vec{u}$ and $\vec{v}$ is given by $\frac{1}{2} |\vec{u} \times \vec{v}|$.
• Area of a Parallelogram: The area of a parallelogram with adjacent sides represented by vectors $\vec{u}$ and $\vec{v}$ is given by $|\vec{u} \times \vec{v}|$.
• Properties of Cross Product:
  • Distributivity: $\vec{u} \times (\vec{v} + \vec{w}) = (\vec{u} \times \vec{v}) + (\vec{u} \times \vec{w})$
  • Scalar Multiplication: $(k\vec{u}) \times \vec{v} = k(\vec{u} \times \vec{v})$
  • Cross product of parallel vectors: $\vec{u} \times \vec{u} = \vec{0}$

Step-by-Step Solution

We are given the position vectors $\overrightarrow{OA} = 2\vec{a}$, $\overrightarrow{OB} = 6\vec{a} + 5\vec{b}$, and $\overrightarrow{OC} = 3\vec{b}$. The area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is 15 sq. units.

Step 1: Determine the value of $|\vec{a} \times \vec{b}|$.

• Why this step? The given area of the parallelogram formed by $\overrightarrow{OA}$ and $\overrightarrow{OC}$ involves the base vectors $\vec{a}$ and $\vec{b}$. By calculating this area in terms of $\vec{a}$ and $\vec{b}$, we can find a fundamental unit of area related to $|\vec{a} \times \vec{b}|$, which will be essential for calculating other areas.
• The area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is $|\overrightarrow{OA} \times \overrightarrow{OC}|$.
• Substituting the given vectors: $|\overrightarrow{OA} \times \overrightarrow{OC}| = |(2\vec{a}) \times (3\vec{b})|$
• Using the scalar multiplication property of the cross product: $|(2\vec{a}) \times (3\vec{b})| = |(2 \times 3) (\vec{a} \times \vec{b})| = |6 (\vec{a} \times \vec{b})|$
• Since 6 is a positive scalar: $6 |\vec{a} \times \vec{b}|$
• We are given that this area is 15 sq. units: $6 |\vec{a} \times \vec{b}| = 15$
• Solving for $|\vec{a} \times \vec{b}|$: $|\vec{a} \times \vec{b}| = \frac{15}{6} = \frac{5}{2}$

Step 2: Calculate the area of triangle $OAB$.

• Why this step? The quadrilateral $OABC$ can be divided into two triangles, $\triangle OAB$ and $\triangle OBC$. We need to find the area of each of these triangles to sum them up for the total area of the quadrilateral.
• The area of $\triangle OAB$ is given by $\frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}|$.
• Substitute the expressions for $\overrightarrow{OA}$ and $\overrightarrow{OB}$: $\text{Area}(\triangle OAB) = \frac{1}{2} |(2\vec{a}) \times (6\vec{a} + 5\vec{b})|$
• Apply the distributive property of the cross product: $= \frac{1}{2} |(2\vec{a} \times 6\vec{a}) + (2\vec{a} \times 5\vec{b})|$
• Apply the scalar multiplication property: $= \frac{1}{2} |12(\vec{a} \times \vec{a}) + 10(\vec{a} \times \vec{b})|$
• Since $\vec{a} \times \vec{a} = \vec{0}$: $= \frac{1}{2} |12(\vec{0}) + 10(\vec{a} \times \vec{b})| = \frac{1}{2} |10(\vec{a} \times \vec{b})|$
• This simplifies to: $= \frac{1}{2} \times 10 |\vec{a} \times \vec{b}| = 5 |\vec{a} \times \vec{b}|$
• Substitute the value of $|\vec{a} \times \vec{b}| = \frac{5}{2}$ from Step 1: $\text{Area}(\triangle OAB) = 5 \times \frac{5}{2} = \frac{25}{2} \text{ sq. units.}$

Step 3: Calculate the area of triangle $OBC$.

• Why this step? This is the second component triangle needed to find the total area of the quadrilateral $OABC$.
• The area of $\triangle OBC$ is given by $\frac{1}{2} |\overrightarrow{OB} \times \overrightarrow{OC}|$.
• Substitute the expressions for $\overrightarrow{OB}$ and $\overrightarrow{OC}$: $\text{Area}(\triangle OBC) = \frac{1}{2} |(6\vec{a} + 5\vec{b}) \times (3\vec{b})|$
• Apply the distributive property of the cross product: $= \frac{1}{2} |(6\vec{a} \times 3\vec{b}) + (5\vec{b} \times 3\vec{b})|$
• Apply the scalar multiplication property: $= \frac{1}{2} |18(\vec{a} \times \vec{b}) + 15(\vec{b} \times \vec{b})|$
• Since $\vec{b} \times \vec{b} = \vec{0}$: $= \frac{1}{2} |18(\vec{a} \times \vec{b}) + 15(\vec{0})| = \frac{1}{2} |18(\vec{a} \times \vec{b})|$
• This simplifies to: $= \frac{1}{2} \times 18 |\vec{a} \times \vec{b}| = 9 |\vec{a} \times \vec{b}|$
• Substitute the value of $|\vec{a} \times \vec{b}| = \frac{5}{2}$ from Step 1: $\text{Area}(\triangle OBC) = 9 \times \frac{5}{2} = \frac{45}{2} \text{ sq. units.}$

Step 4: Calculate the total area of the quadrilateral $OABC$.

• Why this step? This is the final step where we sum the areas of the two constituent triangles to obtain the area of the quadrilateral.
• The area of the quadrilateral $OABC$ is the sum of the areas of $\triangle OAB$ and $\triangle OBC$: $\text{Area}(OABC) = \text{Area}(\triangle OAB) + \text{Area}(\triangle OBC)$ $\text{Area}(OABC) = \frac{25}{2} + \frac{45}{2}$ $\text{Area}(OABC) = \frac{25 + 45}{2} = \frac{70}{2} = 35$

The area of the quadrilateral $OABC$ is 35 sq. units.

Common Mistakes & Tips

• Cross Product Properties: Incorrect application of the distributive law or forgetting that $\vec{u} \times \vec{u} = \vec{0}$ are common errors. Always simplify terms involving the cross product of a vector with itself.
• Area Formulas: Ensure you are using the correct formula for the area of a triangle ($\frac{1}{2} |\vec{u} \times \vec{v}|$) versus the area of a parallelogram ($|\vec{u} \times \vec{v}|$).
• Magnitude of Scalars: When factoring out scalars from a cross product magnitude, remember to take the absolute value of the scalar. For example, $|-5\vec{v}| = |-5||\vec{v}| = 5|\vec{v}|$. In this problem, all scalars were positive.

Summary

The problem was solved by decomposing the quadrilateral $OABC$ into two triangles, $\triangle OAB$ and $\triangle OBC$. We first used the given area of a parallelogram to find the fundamental quantity $|\vec{a} \times \vec{b}|$. Then, using the properties of the cross product and the area formulas for triangles, we calculated the areas of $\triangle OAB$ and $\triangle OBC$. Finally, summing these areas yielded the total area of the quadrilateral $OABC$.

The final answer is $\boxed{35}$.