Question

Let $\overrightarrow{OA} = \vec{a}, \overrightarrow{OB} = 12\vec{a} + 4\vec{b}$ and $\overrightarrow{OC} = \vec{b}$, where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $OA$ and $OC$, then$\frac{\text{area of the quadrilateral OABC}}{\text{area of } S}$is equal to ___.

• A. 6
• B. 10
• C. 7
• D. 8

Answer 1

Answer: (D)

8

Answer 2

Step 1. Area of parallelogram $S$ with adjacent sides $OA$ and $OC$:
$S = |\vec{a} \times \vec{b}|$
Step 2. Area of quadrilateral $OABC$:

$\text{Area of } OABC = \text{Area of } \triangle OAB + \text{Area of } \triangle OBC$
$= \frac{1}{2} \left| \vec{a} \times (12\vec{a} + 4\vec{b}) \right| + \frac{1}{2} \left| \vec{b} \times (12\vec{a} + 4\vec{b}) \right|$
$= \frac{1}{2} |4\vec{a} \times \vec{b}| + \frac{1}{2} |12\vec{a} \times \vec{b}|$
$= 8|\vec{a} \times \vec{b}|$

Step 3. Ratio:

$\text{Ratio} = \frac{\text{Area of quadrilateral } OABC}{\text{Area of parallelogram } S} = \frac{8|\vec{a} \times \vec{b}|}{|\vec{a} \times \vec{b}|} = 8$

The Correct Answer is: 8