Vector Algebra
Area of Parallelogram
nta_pyq_2024_jan
Grade 12
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 $\dfrac{\text{Area of quadrilateral }OABC}{\text{Area of }S}$ is equal to:
Step-by-Step Solution
Key Concept: Area of $S=|\vec{a}\times\vec{b}|$. Area of quadrilateral $OABC=$ Area$(\triangle OAB)+$ Area$(\triangle OBC)=\frac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OB}|+\frac{1}{2}|\overrightarrow{OC}\times\overrightarrow{OB}|$.
Area of $S=|\vec{a}\times\vec{b}|$. Area$(OABC)=\frac{1}{2}|4(\vec{a}\times\vec{b})|+\frac{1}{2}|12(\vec{a}\times\vec{b})|=8|\vec{a}\times\vec{b}|$. Ratio $=8$.
Correct Answer: 4