Let $\bar{O A} = \bar{a}$ , $\bar{O B} = 8 \bar{a} + 4 \bar{b}$ , $\bar{O C} = \bar{b}$ where O,A,C are non-collinear points. Let $Δ_{1}$ denote the area of the quadrilateral OABC and let $Δ_{2}$ denote the area of the parallelogram with $\bar{O A}$ and $\bar{O C}$ as adjacent sides then $\frac{Δ_{1}}{Δ_{2}} =$

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Detailed Solution

Given $\bar{O A} = \bar{a}, \bar{O B} = 8 \bar{a} + 4 \bar{b}, \bar{O C} = \bar{b}$ $Δ_{1} =$ area of quadrilateral OABC

$= \frac{1}{2} (\bar{O A} \times \bar{O B}) + \frac{1}{2} (\bar{O B} \times \bar{O C}) = \frac{1}{2} [\bar{a} \times (8 \bar{a} + 4 \bar{b})] + \frac{1}{2} [(8 \bar{a} + 4 \bar{b}) \times \bar{b}] = \frac{1}{2} [4 (\bar{a} \times \bar{b})] + \frac{1}{2} [8 (\bar{a} \times \bar{b})] = 6 (\bar{a} \times \bar{b}) = 6 (\bar{O A} \times \bar{O C}) = 6 Δ_{2} \frac{Δ_{1}}{Δ_{2}} = 6$