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Question
Find the area of the parallelogram determined by the vector \[3 \hat{ i } + \hat{ j } - 2 \hat{ k } \text{ and } \hat{ i } - 3 \hat{ j } + 4 \hat{ k } \] .
Solution
\[ \text{ Let } : \]
\[ \vec{a} = 3 \hat{ i } + \hat{ j } - 2 \hat{ k } \]
\[ \vec{b} = 1 \hat{ i } - 3 \hat{ j } + 4 \hat{ k } \]
\[ \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 3 & 1 & - 2 \\ 1 & - 3 & 4\end{vmatrix}\]
\[ = \hat{ i } \left( 4 - 6 \right) - \hat{ j } \left( 12 + 2 \right) + \hat{ k } \left( - 9 - 1 \right)\]
\[ = - 2 \hat{ i } - 14 \hat{ j } - 10 \hat{ k } \]
\[\text{ Area of the parallelogram } =\left| \vec{a} \times \vec{b} \right|\]
\[ = \sqrt{\left( - 2 \right)^2 + \left( - 14 \right)^2 + \left( - 10 \right)^2}\]
\[ = \sqrt{300}\]
\[ = 10\sqrt{3} \text{ sq. units } \]
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