Question

The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}-5\hat{k}\mathrm{and}2\hat{i}+2\hat{j}+3\hat{k}$. Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.

Answer 1

The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}-5\hat{k}\mathrm{and}2\hat{i}+2\hat{j}+3\hat{k}$.
Suppose $\overrightarrow{a}=2\hat{i}-4\hat{j}-5\hat{k}\mathrm{and}\overrightarrow{b}=2\hat{i}+2\hat{j}+3\hat{k}$



Then any one diagonal of a parallelogram is $\overrightarrow{P}=\overrightarrow{a}+\overrightarrow{b}$.
$$\begin{gathered} \overrightarrow{P}=\overrightarrow{a}+\overrightarrow{b} \\ =2\hat{i}-4\hat{j}-5\hat{k}+2\hat{i}+2\hat{j}+3\hat{k} \\ =4\hat{i}-2\hat{j}-2\hat{k} \end{gathered}$$
Therefore, unit vector along the diagonal is $\frac{\overrightarrow{P}}{\left|\overrightarrow{P}\right|}=\frac{4\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{16+4+4}}=\frac{2\hat{i}-\hat{j}-\hat{k}}{\sqrt{6}}$.
Another diagonal of a parallelogram is $\overrightarrow{P\text{'}}=\overrightarrow{b}-\overrightarrow{a}$.
$$\begin{gathered} \overrightarrow{P}\text{'}=\overrightarrow{b}-\overrightarrow{a} \\ =2\hat{i}+2\hat{j}+3\hat{k}-2\hat{i}+4\hat{j}+5\hat{k} \\ =6\hat{j}+8\hat{k} \end{gathered}$$
Therefore, unit vector along the diagonal is $\frac{\overrightarrow{P\text{'}}}{\left|\overrightarrow{P\text{'}}\right|}=\frac{6\hat{j}+8\hat{k}}{\sqrt{36+64}}=\frac{6\hat{j}+8\hat{k}}{10}=\frac{3\hat{j}+4\hat{k}}{5}$.
Now,
$$\begin{gathered} \overrightarrow{P}\times \overrightarrow{P\text{'}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -2& -2\\ 0& 6& 8\end{array}\right| \\ =\hat{i}\left(-16+12\right)-\hat{j}\left(32-0\right)+\hat{k}\left(24-0\right) \\ =-4\hat{i}-32\hat{j}+24\hat{k} \end{gathered}$$
Area of parallelogram = $\frac{\left|\overrightarrow{P}\times \overrightarrow{P\text{'}}\right|}{2}=\frac{\sqrt{16+1024+576}}{2}=\frac{\sqrt{1616}}{2}=\frac{4\sqrt{101}}{2}=2\sqrt{101}$ square units