Question

Q ​Find the area of a parallelogram whose adjacent sides are $\hat{i}+2\hat{j}+3\hat{k}and-3\hat{i}-2\hat{j}+\hat{k}$.

Answer 1

Dear student
$$\begin{gathered} \mathrm{Let}\overrightarrow{\mathrm{a}}=\mathrm{i}+2\mathrm{j}+3\mathrm{k} \\ \mathrm{and}\overrightarrow{\mathrm{b}}=-3\mathrm{i}-2\mathrm{j}+\mathrm{k} \\ \mathrm{Area}\mathrm{of}\mathrm{parallelogram}=\left|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right| \\ \overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& 2& 3\\ -3& -2& 1\end{array}\right| \\ =\mathrm{i}(2+6)-\mathrm{j}(1+9)+\mathrm{k}(-2+6) \\ =8\mathrm{i}-10\mathrm{j}+4\mathrm{k} \\ \mathrm{So},\left|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right|=\sqrt{8^2+(-10{)}^2+4^2} \\ =\sqrt{64+100+16} \\ =\sqrt{180} \\ =6\sqrt{5} \\ \mathrm{Therefore},\mathrm{area}\mathrm{of}\mathrm{parallelogram}=6\sqrt{5} \end{gathered}$$
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