Let overrightarrow{OA}=overrightarrow{a},over | vedclass

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(D) The area of the parallelogram $S$ formed by vectors $\overrightarrow{OA}=\overrightarrow{a}$ and $\overrightarrow{OC}=\overrightarrow{b}$ is given

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The area of the parallelogram $S$ formed by vectors $\overrightarrow{OA}=\overrightarrow{a}$ and $\overrightarrow{OC}=\overrightarrow{b}$ is given by $|\overrightarrow{a} 	imes \overrightarrow{b}|$.The area of the quadrilateral $OABC$ can be calculated as the sum of the areas of $	riangle OAB$ and $	riangle OBC$.Area of $	riangle OAB = \frac{1}{2} |\overrightarrow{OA} 	imes \overrightarrow{OB}| = \frac{1}{2} |\overrightarrow{a} 	imes (12 \overrightarrow{a} + 4 \overrightarrow{b})| = \frac{1}{2} |12(\overrightarrow{a} 	imes \overrightarrow{a}) + 4(\overrightarrow{a} 	imes \overrightarrow{b})| = \frac{1}{2} |0 + 4(\overrightarrow{a} 	imes \overrightarrow{b})| = 2 |\overrightarrow{a} 	imes \overrightarrow{b}|$.Area of $	riangle OBC = \frac{1}{2} |\overrightarrow{OC} 	imes \overrightarrow{OB}| = \frac{1}{2} |\overrightarrow{b} 	imes (12 \overrightarrow{a} + 4 \overrightarrow{b})| = \frac{1}{2} |12(\overrightarrow{b} 	imes \overrightarrow{a}) + 4(\overrightarrow{b} 	imes \overrightarrow{b})| = \frac{1}{2} |12(\overrightarrow{b} 	imes \overrightarrow{a}) + 0| = 6 |\overrightarrow{b} 	imes \overrightarrow{a}| = 6 |\overrightarrow{a} 	imes \overrightarrow{b}|$.Total area of quadrilateral $OABC = 2 |\overrightarrow{a} 	imes \overrightarrow{b}| + 6 |\overrightarrow{a} 	imes \overrightarrow{b}| = 8 |\overrightarrow{a} 	imes \overrightarrow{b}|$.The ratio of the area of the quadrilateral $OABC$ to the area of $S$ is $\frac{8 |\overrightarrow{a} 	imes \overrightarrow{b}|}{|\overrightarrow{a} 	imes \overrightarrow{b}|} = 8$.

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