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Let $\overrightarrow{O A}=2 \vec{a}, \overrightarrow{O B}=6 \vec{a}+5 \vec{b}$ and $\overrightarrow{O C}=3 \vec{b}$, where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OC}}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $\mathrm{OABC}$ is equal to :
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The correct answer is:
35
Area of parallelogram having sides
$\begin{aligned}
& \overrightarrow{\mathrm{OA}} \& \overrightarrow{\mathrm{OC}}=|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OC}}|=|2 \overrightarrow{\mathrm{a}} \times 3 \overrightarrow{\mathrm{b}}|=15 \\
& 6|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=15 \\
& \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{5}{2} \ldots \ldots . .(1)
\end{aligned}$
Area of quadrilateral
$\begin{aligned}
& \mathrm{OABC}=\frac{1}{2}\left|\overrightarrow{\mathrm{d}}_1 \times \overrightarrow{\mathrm{d}}_2\right| \\
& =\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{OB}}|=\frac{1}{2}|(3 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}) \times(6 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}})| \\
& =\frac{1}{2}|18 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-10 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=14|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \\
& =14 \times \frac{5}{2}=35
\end{aligned}$
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