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प्रश्न
Find the area of the parallelogram determined by the vector \[2 \hat{ i } + \hat{ j } + 3 \hat{ k } \text{ and } \hat{ i } - \hat{ j } \] .
योग
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उत्तर
\[\text{ Let } : \]
\[ a^\to = 2 \hat{ i }+ \hat{ j } + 3\hat{ k } \]
\[ \vec{b} = \hat{ i } - \hat{ j } + 0 \hat{ k } \]
\[ \therefore \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i }& \hat{ j } & \hat{ k } \\ 2 & 1 & 3 \\ 1 & - 1 & 0\end{vmatrix}\]
\[ = \left( 0 + 3 \right) \hat{ i } - \left( 0 - 3 \right) \hat{ j } + \left( - 2 - 1 \right) \hat{ k } \]
\[ = 3 \hat{ i } + 3 \hat{ j } - 3 \hat{ k } \]
\[\text{ Area of the parallelogram }
=\left| \vec{a} \times \vec{b} \right|\]
\[ = \sqrt{3^2 + 3^2 + 3^2}\]
\[ = \sqrt{27}\]
\[ = 3\sqrt{3} \text{ sq. units } \]
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