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

The Correct Option is D

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.