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Question
Find the area of the parallelogram whose diagonals are \[3 \hat{ i } + 4 \hat{ j } \text{ and } \hat{ i } + \hat{ j } + \hat{ k }\]
Solution
\[ \text{ Let: } \]
\[ \vec{a} = 3 \hat{ i } + 4 \hat{ j } + 0 \hat{ k } \]
\[ \vec{b} = \hat{ i } + \hat{ j } + \hat{ k } \]
\[ \therefore \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 3 & 4 & 0 \\ 1 & 1 & 1\end{vmatrix}\]
\[ = \left( 4 - 0 \right) \hat{ i } - \left( 3 - 0 \right) \hat{ j } + \left( 3 - 4 \right) \hat{ k} \]
\[ = 4 \hat{ i } - 3 \hat{ j } - \hat{ k } \]
\[ \Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{4^2 + \left( - 3 \right)^2 + \left( - 1 \right)^2}\]
\[ = \sqrt{26}\]
\[\text{ Area of the parallelogram } =\frac{1}{2}\left| \vec{a} \times \vec{b} \right|\]
\[ = \frac{\sqrt{26}}{2} \text{ sq. units } \]
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