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If the area of the parallelogram with $\vec{a}$ and $\vec{b}$ as two adjacent sides is 15 sq. units, then the area of the parallelogram having $3 \vec{a}+\vec{b}$ and $\vec{a}+3 \vec{b}$ as two adjacent sides, in square units, is
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120
Area of parallelogram $=|\vec{a} \times \vec{b}|=15$ [given]
Area of second parallelogram $=|(3 \vec{a}+\vec{b}) \times(\vec{a}+3 \vec{b})|$
$\begin{aligned} & =|3 \vec{a} \times \vec{a}+9 \vec{a} \times \vec{b}+\vec{b} \times \vec{a}+3 \vec{b} \times \vec{b}| \\ & =|0+9 \vec{a} \times \vec{b}-\vec{a} \times \vec{b}+0| \\ & =|8 \vec{a} \times \vec{b}|=8|\vec{a} \times \vec{b}|=8 \times 15=120\end{aligned}$
Area of second parallelogram $=|(3 \vec{a}+\vec{b}) \times(\vec{a}+3 \vec{b})|$
$\begin{aligned} & =|3 \vec{a} \times \vec{a}+9 \vec{a} \times \vec{b}+\vec{b} \times \vec{a}+3 \vec{b} \times \vec{b}| \\ & =|0+9 \vec{a} \times \vec{b}-\vec{a} \times \vec{b}+0| \\ & =|8 \vec{a} \times \vec{b}|=8|\vec{a} \times \vec{b}|=8 \times 15=120\end{aligned}$
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