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If the area of the parallelogram with $\mathbf{a}$ and $\mathbf{b}$ as two adjacent sides is 15 sq units, then the area of the parallelogram having $3 \mathbf{a}+2 \mathbf{b}$ and $\mathbf{a}+3 \mathbf{b}$ as two adjacent sides in sq units is
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The correct answer is:
105
We know that, if $\mathbf{a}$ and $\mathbf{b}$ are two adjacent sides of a parallelogram, then
Area $=|a \times b|=15$ (given) ...(i)
If the sides are $(3 \mathbf{a}+2 \mathbf{b})$ and $(\mathbf{a}+3 \mathbf{b})$
Then, area of parallelogram
$\begin{aligned}
&=3|(3 \mathbf{a}+2 \mathbf{b}) \times(\mathbf{a}+3 \mathbf{b})| \\
&=|3 \mathbf{a} \times \mathbf{a}+9 \mathbf{a} \times \mathbf{b}+2 \mathbf{b} \times \mathbf{a}+6 \mathbf{b} \times \mathbf{a}| \\
&=|0+9 \mathbf{a} \times \mathbf{b}-2 \mathbf{a} \times \mathbf{b}+0| \\
&=|7(\mathbf{a} \times \mathbf{b})| \\
&=7|\mathbf{a} \times \mathbf{b}| \\
&=7 \times 15=105 \text { sq units }
\end{aligned}$
Area $=|a \times b|=15$ (given) ...(i)
If the sides are $(3 \mathbf{a}+2 \mathbf{b})$ and $(\mathbf{a}+3 \mathbf{b})$
Then, area of parallelogram
$\begin{aligned}
&=3|(3 \mathbf{a}+2 \mathbf{b}) \times(\mathbf{a}+3 \mathbf{b})| \\
&=|3 \mathbf{a} \times \mathbf{a}+9 \mathbf{a} \times \mathbf{b}+2 \mathbf{b} \times \mathbf{a}+6 \mathbf{b} \times \mathbf{a}| \\
&=|0+9 \mathbf{a} \times \mathbf{b}-2 \mathbf{a} \times \mathbf{b}+0| \\
&=|7(\mathbf{a} \times \mathbf{b})| \\
&=7|\mathbf{a} \times \mathbf{b}| \\
&=7 \times 15=105 \text { sq units }
\end{aligned}$
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