Question

Find the area of the parallelogram whose diagonals are:
(i) $4\hat{i}-\hat{j}-3\hat{k}\mathrm{and}-2\hat{j}+\hat{j}-2\hat{k}$

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

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

(iv) $2\hat{i}+3\hat{j}+6\hat{k}\mathrm{and}3\hat{i}-6\hat{j}+2\hat{k}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}: \\ \overrightarrow{a}=4\hat{i}-\hat{j}-3\hat{k} \\ \overrightarrow{b}=-2\hat{i}+\hat{j}-2\hat{k} \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -1& -3\\ -2& 1& -2\end{array}\right| \\ =\left(2+3\right)\hat{i}-\left(-8-6\right)\hat{j}+\left(4-2\right)\hat{k} \\ =5\hat{i}+14\hat{j}+2\hat{k} \\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{25+196+4} \\ =\sqrt{225} \\ =15 \\ \text{Area of the parallelogram =}\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ \text{=}\frac{15}{2}\text{sq. units.} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let}: \\ \overrightarrow{a}=2\hat{i}+0\hat{j}+\hat{k} \\ \overrightarrow{b}=\hat{i}+\hat{j}+\hat{k} \\ \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& 1\\ 1& 1& 1\end{array}\right| \\ =\left(0-1\right)\hat{i}-\left(2-1\right)\hat{j}+\left(2-0\right)\hat{k} \\ =-\hat{i}-\hat{j}+2\hat{k} \\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2+\left(2\right)} \\ =\sqrt{6} \\ \text{Area of the parallelogram =}\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ \text{=}\frac{\sqrt{6}}{2}\text{sq. units} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Let}: \\ \overrightarrow{a}=3\hat{i}+4\hat{j}+0\hat{k} \\ \overrightarrow{b}=\hat{i}+\hat{j}+\hat{k} \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 0\\ 1& 1& 1\end{array}\right| \\ =\left(4-0\right)\hat{i}-\left(3-0\right)\hat{j}+\left(3-4\right)\hat{k} \\ =4\hat{i}-3\hat{j}-\hat{k} \\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{4^2+{\left(-3\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{26} \\ \text{Area of the parallelogram =}\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ =\frac{\sqrt{26}}{2}\text{sq. units} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{Let}: \\ \overrightarrow{a}=2\hat{i}+3\hat{j}+6\hat{k} \\ \overrightarrow{b}=3\hat{i}-6\hat{j}+2\hat{k} \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 6\\ 3& -6& 2\end{array}\right| \\ =\left(6+36\right)\hat{i}-\left(4-18\right)\hat{j}+\left(-12-9\right)\hat{k} \\ =42\hat{i}+14\hat{j}-21\hat{k} \\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{{42}^2+{14}^2+{\left(-21\right)}^2} \\ =\sqrt{2401} \\ =49 \\ \text{Area of the parallelogram =}\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ =\frac{49}{2}\text{sq. units} \end{gathered}$$

Disclaimer: The answer given for (iii) and (iv) in the textbook is incorrect.