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If the area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is 16 sq. units, then the area of the parallelogram having $3 \bar{a}+2 \bar{b}$ and $\overline{\mathrm{a}}+3 \overline{\mathrm{b}}$ as two adjacent sides (in sq. units) is
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The correct answer is:
$112$
Area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|$
$\therefore \quad|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=16$
$\therefore$ Area of the required parallelogram
$\begin{aligned}
& =|(3 \overline{\mathrm{a}}+2 \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+3 \overline{\mathrm{b}})| \\
& =|3(\overline{\mathrm{a}} \times \overline{\mathrm{a}})+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+(\overline{\mathrm{b}} \times \overline{\mathrm{b}})| \\
& =0+9|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|-2|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|+0 \\
& =7|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \\
& =7 \times 16 \\
& =112
\end{aligned}$
$\therefore \quad|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=16$
$\therefore$ Area of the required parallelogram
$\begin{aligned}
& =|(3 \overline{\mathrm{a}}+2 \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+3 \overline{\mathrm{b}})| \\
& =|3(\overline{\mathrm{a}} \times \overline{\mathrm{a}})+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+(\overline{\mathrm{b}} \times \overline{\mathrm{b}})| \\
& =0+9|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|-2|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|+0 \\
& =7|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \\
& =7 \times 16 \\
& =112
\end{aligned}$
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