Question

If \[\vec{a} = 2 \hat{ i } - 3 \hat{ j  } + \hat{ k } , \vec{b} = -\hat{  i }  + \hat{ k } , \vec{c} = 2 \hat{ j }  - \hat{ k } \]  are three vectors, find the area of the parallelogram having diagonals \[\left( \vec{a} + \vec{b} \right)\]  and \[\left( \vec{b} + \vec{c} \right)\] .

 

 

Answer 1

It is given that \[\vec{a} = 2 \hat{ i }  - 3 \hat{ j } + \hat{ k }  , \vec{b} = - \hat{ i }  + \hat{ k }  , \vec{c} = 2 \hat{ j }  - \hat{ k } \]

∴ \[\vec{a} + \vec{b} = \left( 2 \hat { i }  - 3 \hat{ j }  + \hat{ k }  \right) + \left( - \hat{ i }+ \hat{ k }  \right) = \hat{ i }  - 3 \hat{ j } + 2 \hat{ k } \]

\[\vec{b} + \vec{c} = \left( - \hat{ i } + \hat{ k } \right) + \left( 2 \hat{ j }  - \hat{ k }  \right) = - \hat{ i }  + 2 \hat{ j } \]

We know that the area of parallelogram is \[\frac{1}{2}\left| \vec{d_1} \times \vec{d_2} \right|\] , where \[\vec{d_1}\]  and \[\vec{d_2}\]  are the diagonal vectors.

Now,

\[\left( \vec{a} + \vec{b} \right) \times \left( \vec{b} + \vec{c} \right) = \begin{vmatrix}\hat{ i } & \hat{ j}  & \hat{ k } \\ 1 & - 3 & 2 \\ - 1 & 2 & 0\end{vmatrix} = - 4 \hat{ i } - 2 \hat{ j }  - \hat{ k } \]

∴ Area of the parallelogram having diagonals \[\left( \vec{a} + \vec{b} \right)\] and  \[\left( \vec{b} + \vec{c} \right)\]

\[= \frac{1}{2}\left| \left( \vec{a} + \vec{b} \right) \times \left( \vec{b} + \vec{c} \right) \right|\]

\[ = \frac{1}{2}\left| - 4 \hat{ i  } - 2 \hat{ j }  - \hat{ k } \right|\]

\[ = \frac{1}{2}\sqrt{\left( - 4 \right)^2 + \left( - 2 \right)^2 + \left( - 1 \right)^2}\]

\[ = \frac{\sqrt{21}}{2} \text{ square units } \] 

Thus, the required area of the parallelogram is \[\frac{\sqrt{21}}{2}\]  square units.