Question

Find the area of the parallelogram whose diagonals are  \[4 \hat{ i } - \hat{ j }  - 3 \hat{ k }  \text{ and }  - 2 \hat{ j }  + \hat{ j }  - 2 \hat{ k } \]

 

Answer 1

 \[\text{ Let } : \]
\[ \vec{a} = 4 \hat{ i }  - \hat{ j }  - 3 \hat{ k } \]
\[ \vec{b} = - 2 \hat{ i } + \hat{ j } - 2 \hat{ k } \]
\[ \therefore \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i }  & \hat{ j }  & \hat{ k } \\ 4 & - 1 & - 3 \\ - 2 & 1 & - 2\end{vmatrix}\]
\[ = \left( 2 + 3 \right) \hat{ i }  - \left( - 8 - 6 \right) \hat{ j }  + \left( 4 - 2 \right) \hat{ k }  \]
\[ = 5 \hat{ i }  + 14 \hat{ j }  + 2 \hat{ k }   \]
\[ \Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{25 + 196 + 4}\]
\[ = \sqrt{225}\]
\[ = 15\]
\[\text{ Area of the parallelogram  } =\frac{1}{2}\left| \vec{a} \times \vec{b} \right|\]
\[ =\frac{15}{2} \text{ sq. units. } \]