Question

Find the area of the parallelogram whose diagonals are \[2 \hat{ i }  + 3 \hat{ j } + 6 \hat{ k } \text{ and }  3 \hat{ i }  - 6 \hat{ j }  + 2 \hat{ k } \]

 

Answer 1

\[ \text{ Let } : \]
\[ \vec{a} = 2 \hat{ i }  + 3 \hat{ j } + 6 \hat{ k } \]
\[ \vec{b} = 3 \hat{ i }  - 6 \hat{ j } + 2 \hat{ k } \]
\[ \therefore \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i }  & \hat{ j } & \hat{ k }  \\ 2 & 3 & 6 \\ 3 & - 6 & 2\end{vmatrix}\]
\[ = \left( 6 + 36 \right) \hat{ i  } - \left( 4 - 18 \right) \hat[{ j }  + \left( - 12 - 9 \right) \hat{ k } \]
\[ = 42 \hat{ i }  + 14 \hat{ j }  - 21 \hat{ k }  \]
\[ \Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{{42}^2 + {14}^2 + \left( - 21 \right)^2}\]
\[ = \sqrt{2401}\]
\[ = 49\]
\[\text{ Area of the parallelogram }  =\frac{1}{2}\left| \vec{a} \times \vec{b} \right|\]
\[ = \frac{49}{2} \text{ sq. units } \]