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If area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is 20 square units, then the area of the parallelogram having $3 \bar{a}$ $+\overline{\mathrm{b}}$ and $2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}$ as two adjacent sides in square units is
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Verified Answer
The correct answer is:
140
We have $|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=20$ and we have to find value of
$$
\begin{aligned}
& |(3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}})| \\
& (3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}) \\
& =6(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+3(\overline{\mathrm{b}} \times \overline{\mathrm{b}}) \\
& =0+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})-2(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+0=7(\overline{\mathrm{a}} \times \overline{\mathrm{b}})
\end{aligned}
$$
Hence area of the required parallelogram $=7 \times 20=140$
$$
\begin{aligned}
& |(3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}})| \\
& (3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}) \\
& =6(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+3(\overline{\mathrm{b}} \times \overline{\mathrm{b}}) \\
& =0+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})-2(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+0=7(\overline{\mathrm{a}} \times \overline{\mathrm{b}})
\end{aligned}
$$
Hence area of the required parallelogram $=7 \times 20=140$
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