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Question
Find the area of the parallelogram whose diagonals are \[2 \hat{ i }+ \hat{ k } \text{ and } \hat{ i } + \hat{ j } + \hat{ k } \]
Solution
\[\text{ Let } : \]
\[ \vec{a} = 2 \hat{ i } + 0 \hat{ j } + \hat{ k } \]
\[ \vec{b} = \hat{ i } + \hat{ j } + \hat{ k } \]
\[ \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 0 & 1 \\ 1 & 1 & 1\end{vmatrix}\]
\[ = \left( 0 - 1 \right) \hat{ i } - \left( 2 - 1 \right) \hat{ j } + \left( 2 - 0 \right) \hat{ k } \]
\[ = - \hat{ i } - \hat{ j } + 2 \hat{ k } \]
\[ \Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{\left( - 1 \right)^2 + \left( - 1 \right)^2 + \left( 2 \right)}\]
\[ = \sqrt{6}\]
\[\text{ Area of the parallelogram } =\frac{1}{2}\left| \vec{a} \times \vec{b} \right|\]
\[ =\frac{\sqrt{6}}{2}\text{ sq. units } \]
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