Question

Find the area of the parallelogram determined by the vectors:
(i) $2\hat{i}\mathrm{and}3\hat{j}$

(ii) $2\hat{i}+\hat{j}+3\hat{k}\mathrm{and}\hat{i}-\hat{j}$

(iii) $3\hat{i}+\hat{j}-2\hat{k}\mathrm{and}\hat{i}-3\hat{j}+4\hat{k}$

(iv) $\hat{i}-3\hat{j}+\hat{k}\mathrm{and}\hat{i}+\hat{j}+\hat{k}.$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}: \\ a=2\hat{i}+0\hat{j}+0\hat{k} \\ \overrightarrow{b}=0\hat{i}+3\hat{j}+0\hat{k} \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& 0\\ 0& 3& 0\end{array}\right| \\ =\left(0-0\right)\hat{i}-\left(0-0\right)\hat{j}+\left(6-0\right)\hat{k} \\ =0\hat{i}+0\hat{j}+6\hat{k} \\ \text{Area of the parallelogram=}\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ =\sqrt{0+0+6^2} \\ =6\text{sq. units} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let}: \\ \overrightarrow{a}=2\hat{i}+\hat{j}+3\hat{k} \\ \overrightarrow{b}=\hat{i}-\hat{j}+0\hat{k} \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 3\\ 1& -1& 0\end{array}\right| \\ =\left(0+3\right)\hat{i}-\left(0-3\right)\hat{j}+\left(-2-1\right)\hat{k} \\ =3\hat{i}+3\hat{j}-3\hat{k} \\ \text{Area of the parallelogram =}\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ =\sqrt{3^2+3^2+3^2} \\ =\sqrt{27} \\ =3\sqrt{3}\text{sq. units} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Let}: \\ \overrightarrow{a}=3\hat{i}+\hat{j}-2\hat{k} \\ \overrightarrow{b}=1\hat{i}-3\hat{j}+4\hat{k} \\ \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& -2\\ 1& -3& 4\end{array}\right| \\ =\hat{i}\left(4-6\right)-\hat{j}\left(12+2\right)+\hat{k}\left(-9-1\right) \\ =-2\hat{i}-14\hat{j}-10\hat{k} \\ \text{Area of the parallelogram=}\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ =\sqrt{{\left(-2\right)}^2+{\left(-14\right)}^2+{\left(-10\right)}^2} \\ =\sqrt{300} \\ =10\sqrt{3}\text{sq. units} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{Let}: \\ \overrightarrow{a}=\hat{i}-3\hat{j}+\hat{k} \\ \overrightarrow{b}=\hat{i}+\hat{j}+\hat{k} \\ \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 1\\ 1& 1& 1\end{array}\right| \\ =\left(-3-1\right)\hat{i}-\left(1-1\right)\hat{j}+\left(1+3\right)\hat{k} \\ =-4\hat{i}+0\hat{j}+4\hat{k} \\ \text{Area of the parallelogram=}\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ =\sqrt{{\left(-4\right)}^2+0+4^2} \\ =\sqrt{32} \\ =4\sqrt{2}\text{sq. units.} \end{gathered}$$