Question

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

Answer 1

$$\begin{gathered} \mathrm{Suppose}\square ABCD\mathrm{is}\mathrm{the}\mathrm{given}\mathrm{parallelogram}\mathrm{and}AC\mathrm{is}\mathrm{its}\mathrm{diagonal}. \\ \mathrm{Let}: \\ \overrightarrow{AB}=2\hat{i}-4\hat{j}+5\hat{k} \\ \overrightarrow{BC}=\hat{i}-2\hat{j}-3\hat{k} \\ \therefore \text{Diagonal}\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC} \\ =3\hat{i}-6\hat{j}+2\hat{k} \\ \Rightarrow \left|\overrightarrow{AC}\right|=\sqrt{9+36+4} \\ =7 \\ \text{Unit vector parallel to}\overrightarrow{AC}\text{=}\frac{\overrightarrow{AC}}{\left|\overrightarrow{AC}\right|} \\ \text{=}\frac{3\hat{i}-6\hat{j}+2\hat{k}}{7} \\ \text{Now,} \\ \overrightarrow{AB}\times \overrightarrow{BC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -4& 5\\ 1& -2& -3\end{array}\right| \\ =22\hat{i}+11\hat{j}+0\hat{k} \\ \Rightarrow \left|\overrightarrow{AB}\times \overrightarrow{AC}\right|=\sqrt{484+121} \\ =\sqrt{605} \\ =11\sqrt{5} \\ \text{Area of triangle}\mathit{\text{ABC}}\text{=}\frac{1}{2}\left|\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}\right| \\ =\frac{11\sqrt{5}}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$