### Related Formula textArea of Parallelogram S = |veca times vecb|$\text{Area of Parallelogram } S = |\vec{a} \times \vec{b}|$ textArea of Quadrilateral OABC = textArea(Delta OAB) + textArea(Delta OBC)$\text{Area of Quadrilateral OABC} = \text{Area}(\Delta OAB) + \text{Area}(\Delta OBC)$ ### Core Logic Let us compute the component areas using vector cross products: textArea(Delta OAB) = frac12 |overrightarrowOA times overrightarrowOB| = frac12 |veca times (12veca + 4vecb)| = frac12 |4(veca times vecb)| = 2|veca times vecb|$\text{Area}(\Delta OAB) = \frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}| = \frac{1}{2} |\vec{a} \times (12\vec{a} + 4\vec{b})| = \frac{1}{2} |4(\vec{a} \times \vec{b})| = 2|\vec{a} \times \vec{b}|$ textArea(Delta OBC) = frac12 |overrightarrowOB times overrightarrowOC| = frac12 |(12veca + 4vecb) times vecb| = frac12 |12(veca times vecb)| = 6|veca times vecb|$\text{Area}(\Delta OBC) = \frac{1}{2} |\overrightarrow{OB} \times \overrightarrow{OC}| = \frac{1}{2} |(12\vec{a} + 4\vec{b}) \times \vec{b}| = \frac{1}{2} |12(\vec{a} \times \vec{b})| = 6|\vec{a} \times \vec{b}|$ ### Step 1: Finding Total Ratio Adding both triangles to find the total area of the quadrilateral OABC: textArea(OABC) = 2|veca times vecb| + 6|veca times vecb| = 8|veca times vecb|$\text{Area(OABC)} = 2|\vec{a} \times \vec{b}| + 6|\vec{a} \times \vec{b}| = 8|\vec{a} \times \vec{b}|$

Dividing this total by the area of the baseline parallelogram S = |veca times vecb|$S = |\vec{a} \times \vec{b}|$: textRatio = frac8|veca times vecb||veca times vecb| = 8$\text{Ratio} = \frac{8|\vec{a} \times \vec{b}|}{|\vec{a} \times \vec{b}|} = 8$ ### Pattern Recognition Since veca times veca = 0$\vec{a} \times \vec{a} = 0$ and vecb times vecb = 0$\vec{b} \times \vec{b} = 0$, cross product distributions yield terms with purely veca times vecb$\vec{a} \times \vec{b}$, ensuring absolute scalability independent of vectors chosen. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra