Let , and , where O is the origin. If the area of the parallelogram with adjacent sides and is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to: [2024]

Question

Let $\vec{O A} = 2 a$ and $\vec{O B} = 6 \vec{a} + 5 \vec{b}$ and $\vec{O C} = 3 \vec{b}$ where O is the origin. If the area of the parallelogram with adjacent sides $\vec{O A}$ and $\vec{O C}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to: [2024]

Smart Achievers · Accepted Answer

(1)

We have, $| \vec{O A} \times \vec{O C} |$ $= 15$

$\Rightarrow$ $| 2 \vec{a} \times 3 \vec{b} |$ $= 15 \Rightarrow$ $| \vec{a} \times \vec{b} |$ $= \frac{5}{2}$ ... (i)

Area of quadrilateral $O A B C = \frac{1}{2}$ $| \vec{O B} \times \vec{A C} |$

$= \frac{1}{2}$ $| (6 \vec{a} + 5 \vec{b}) \times (3 \vec{b} - 2 \vec{a}) |$ $= \frac{1}{2}$ $| 18 (\vec{a} \times \vec{b}) - 10 (\vec{b} \times \vec{a}) |$

$= \frac{1}{2}$ $| 18 (\vec{a} \times \vec{b}) + 10 (\vec{a} \times \vec{b}) |$ $= 14$ $| \vec{a} \times \vec{b} |$

$= 14 \times \frac{5}{2} [usin g (i)]$

= 35 sq. units.

Solution

Q.
Let $\vec{O A} = 2 a$ and $\vec{O B} = 6 \vec{a} + 5 \vec{b}$ and $\vec{O C} = 3 \vec{b}$ where O is the origin. If the area of the parallelogram with adjacent sides $\vec{O A}$ and $\vec{O C}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to: [2024]

1 35

2 40

3 38

4 32

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