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Question
The two adjacent sides of a parallelogram are \[2 \hat{ i } - 4 \hat{ j } + 5 \hat{ k } \text{ and } \hat{ i } - 2 \hat{ j } - 3\hat{ k } .\]\ Find the unit vector parallel to one of its diagonals. Also, find its area.
Solution
\[\text{ Suppose } \square ABCD \text{ is the given parallelogram and AC is its diagonal } . \]
\[\text{ Let } : \]
\[ \vec{AB} = 2 \hat{ i } - 4 \hat{ j } + 5 \hat{ k } \]
\[ \vec{BC} = \hat{ i } - 2 \hat{ j } - 3 \hat{ k } \]
\[ \therefore \text{ Diagonal } \vec{AC} = \vec{AB} + \vec{BC} \]
\[ = 3 \hat{ i } - 6 \hat{ j } + 2 \hat{ k } \]
\[ \Rightarrow \left| \vec{AC} \right| = \sqrt{9 + 36 + 4}\]
\[ = 7\]
\[\text{ Unit vector parallel to } \vec{AC} =\frac{\vec{AC}}{\left| \vec{AC} \right|}\]
\[ =\frac{3 \hat{ i } - 6 \hat{ j } + 2 \hat{ k } }{7}\]
\[\text{ Now } ,\]
\[ \vec{AB} \times \vec{BC} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & - 4 & 5 \\ 1 & - 2 & - 3\end{vmatrix}\]
\[ = 22 \hat{ i } + 11 \hat{ j } + 0 \hat{ k } \]
\[ \Rightarrow \left| \vec{AB} \times \vec{AC} \right| = \sqrt{484 + 121}\]
\[ = \sqrt{605}\]
\[ = 11\sqrt{5}\]
\[Area of triangleABC=\frac{1}{2}\left| \vec{AB} \times \vec{AC} \right|\]
\[ = \frac{11\sqrt{5}}{2}\text{ sq . units } \]
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