One side and one diagonal of a parallelogram are represented by $3 \hat{i}+\hat{j}-\hat{k}$ and $2 \hat{i}+\hat{j}-2 \hat{k}$ respectively, then the area of parallelogram in square units is

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Aicharya · Accepted Answer

$\overline{\mathrm{a}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \text { and } \overline{\mathrm{c}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
In $\triangle \mathrm{ABC}$, $\begin{aligned} & \overline{\mathrm{a}+\overline{\mathrm{b}}}=\overline{\overline{\mathrm{c}}} \ldots \ldots[\text { Using triangle law of addition] } \\ & \begin{aligned} \Rightarrow \overline{\mathrm{b}} & =\overline{\mathrm{c}}-\overline{\mathrm{a}} \\ & =2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}-(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=-\hat{\mathrm{i}}-\hat{\mathrm{k}} \\ \overline{\mathrm{a}} \times \overline{\mathrm{b}} & =\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 1 & -1 \\ -1 & 0 & -1 \end{array}\right|=-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned} \end{aligned}$ $\begin{aligned} \therefore \quad \text { Area of parallelogram } & =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \\ & =\sqrt{1+16+1} \\ & =\sqrt{18}=3 \sqrt{2} \text { sq. units } \end{aligned}$

One side and one diagonal of a parallelogram are represented by $3 \hat{i}+\hat{j}-\hat{k}$ and $2…

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