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Question
Find the area of the parallelogram whose diagonals are \[4 \hat{ i } - \hat{ j } - 3 \hat{ k } \text{ and } - 2 \hat{ j } + \hat{ j } - 2 \hat{ k } \]
Solution
\[\text{ Let } : \]
\[ \vec{a} = 4 \hat{ i } - \hat{ j } - 3 \hat{ k } \]
\[ \vec{b} = - 2 \hat{ i } + \hat{ j } - 2 \hat{ k } \]
\[ \therefore \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 4 & - 1 & - 3 \\ - 2 & 1 & - 2\end{vmatrix}\]
\[ = \left( 2 + 3 \right) \hat{ i } - \left( - 8 - 6 \right) \hat{ j } + \left( 4 - 2 \right) \hat{ k } \]
\[ = 5 \hat{ i } + 14 \hat{ j } + 2 \hat{ k } \]
\[ \Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{25 + 196 + 4}\]
\[ = \sqrt{225}\]
\[ = 15\]
\[\text{ Area of the parallelogram } =\frac{1}{2}\left| \vec{a} \times \vec{b} \right|\]
\[ =\frac{15}{2} \text{ sq. units. } \]
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