Let overrightarrow{OA}=2 overrightarrow{a},ov | vedclass

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(D) The area of a parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is given by $|\overrightarrow{OA} 	imes \overrigh

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(D) The area of a parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is given by $|\overrightarrow{OA} 	imes \overrightarrow{OC}|$.Given,$|\overrightarrow{OA} 	imes \overrightarrow{OC}| = |2 \overrightarrow{a} 	imes 3 \overrightarrow{b}| = 15$.$6 |\overrightarrow{a} 	imes \overrightarrow{b}| = 15 \implies |\overrightarrow{a} 	imes \overrightarrow{b}| = \frac{15}{6} = \frac{5}{2} \dots (1)$.The area of a quadrilateral $OABC$ with diagonals $\overrightarrow{OB}$ and $\overrightarrow{AC}$ is given by $\frac{1}{2} |\overrightarrow{OB} 	imes \overrightarrow{AC}|$.We have $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = 3 \overrightarrow{b} - 2 \overrightarrow{a}$.Area $= \frac{1}{2} |(6 \overrightarrow{a} + 5 \overrightarrow{b}) 	imes (3 \overrightarrow{b} - 2 \overrightarrow{a})|$.$= \frac{1}{2} |18 (\overrightarrow{a} 	imes \overrightarrow{b}) - 12 (\overrightarrow{a} 	imes \overrightarrow{a}) + 15 (\overrightarrow{b} 	imes \overrightarrow{b}) - 10 (\overrightarrow{b} 	imes \overrightarrow{a})|$.Since $\overrightarrow{a} 	imes \overrightarrow{a} = 0$,$\overrightarrow{b} 	imes \overrightarrow{b} = 0$,and $\overrightarrow{b} 	imes \overrightarrow{a} = -(\overrightarrow{a} 	imes \overrightarrow{b})$:Area $= \frac{1}{2} |18 (\overrightarrow{a} 	imes \overrightarrow{b}) + 10 (\overrightarrow{a} 	imes \overrightarrow{b})| = \frac{1}{2} |28 (\overrightarrow{a} 	imes \overrightarrow{b})| = 14 |\overrightarrow{a} 	imes \overrightarrow{b}|$.Substituting from $(1)$:Area $= 14 	imes \frac{5}{2} = 35$ sq. units.

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