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Question
Find the area of the parallelogram determined by the vector \[2 \hat{ i } \text{ and } 3 \hat{ j } \] .
Solution
\[\left( i \right) Let: \]
\[a = 2 \hat{ i } + 0 \hat{ j } + 0 \hat{ k } \]
\[ \vec{b} = 0 \hat{ i } + 3 \hat{ j } + 0 \hat{ k } \]
\[ \therefore \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } &\hat{ k } \\ 2 & 0 & 0 \\ 0 & 3 & 0\end{vmatrix}\]
\[ = \left( 0 - 0 \right) \hat{ i } - \left( 0 - 0 \right) \hat{ j } + \left( 6 - 0 \right) \hat{ k } \]
\[ = 0 \hat{ i } + 0 \hat{ j } + 6 \hat{ k } \]
\[\text{ Area of the parallelogram } =\left| \vec{a} \times \vec{b} \right| \]
\[ = \sqrt{0 + 0 + 6^2}\]
\[ = 6 \text{ sq. units } \]
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