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Question
Find the area of the parallelogram whose diagonals are \[2 \hat{ i } + 3 \hat{ j } + 6 \hat{ k } \text{ and } 3 \hat{ i } - 6 \hat{ j } + 2 \hat{ k } \]
Solution
\[ \text{ Let } : \]
\[ \vec{a} = 2 \hat{ i } + 3 \hat{ j } + 6 \hat{ k } \]
\[ \vec{b} = 3 \hat{ i } - 6 \hat{ j } + 2 \hat{ k } \]
\[ \therefore \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 3 & 6 \\ 3 & - 6 & 2\end{vmatrix}\]
\[ = \left( 6 + 36 \right) \hat{ i } - \left( 4 - 18 \right) \hat[{ j } + \left( - 12 - 9 \right) \hat{ k } \]
\[ = 42 \hat{ i } + 14 \hat{ j } - 21 \hat{ k } \]
\[ \Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{{42}^2 + {14}^2 + \left( - 21 \right)^2}\]
\[ = \sqrt{2401}\]
\[ = 49\]
\[\text{ Area of the parallelogram } =\frac{1}{2}\left| \vec{a} \times \vec{b} \right|\]
\[ = \frac{49}{2} \text{ sq. units } \]
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