Question

Let $\overrightarrow{OA} = 2\vec{a}$, $\overrightarrow{OB} = 6\vec{a} + 5\vec{b}$, and $\overrightarrow{OC} = 3\vec{b}$, where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $OABC$ is equal to:

• A. 38
• B. 40
• C. 32
• D. 35

Answer 1

Answer: (D)

35

Answer 2

Given the quadrilateral OABC, we calculate its area step by step.

Area of the parallelogram:

The area of the parallelogram formed by sides OA and OC is given by:

$\left| \overrightarrow{OA} \times \overrightarrow{OC} \right| = 2 \left| \overrightarrow{a} \times 3 \overrightarrow{b} \right| = 15$

Simplify:

$6 \left| \overrightarrow{a} \times \overrightarrow{b} \right| = 15$

$\therefore \left| \overrightarrow{a} \times \overrightarrow{b} \right| = \frac{5}{2} \quad \cdots (1)$

Area of the quadrilateral $OABC$:

The area of the quadrilateral is:

$\text{Area} = \frac{1}{2} \left| \overrightarrow{d_1} \times \overrightarrow{d_2} \right|$

Here:

$\overrightarrow{d_1} = \overrightarrow{AC}, \quad \overrightarrow{d_2} = \overrightarrow{OB}$

Substituting:

$\text{Area} = \frac{1}{2} \left| \overrightarrow{AC} \times \overrightarrow{OB} \right| = \frac{1}{2} \left| (3 \overrightarrow{b} - 2 \overrightarrow{a}) \times (6 \overrightarrow{a} + 5 \overrightarrow{b}) \right|$

Expanding the cross product:

$= \frac{1}{2} \left| 18 \overrightarrow{b} \times \overrightarrow{a} - 10 \overrightarrow{a} \times \overrightarrow{b} \right|$

Using $\left| \overrightarrow{a} \times \overrightarrow{b} \right| = \frac{5}{2}$ from (1):

$= \left( 14 \left| \overrightarrow{a} \times \overrightarrow{b} \right| \right)$

$= 14 \times \frac{5}{2} = 35$

Final Answer: The area of the quadrilateral $OABC$ is 35.