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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 \[\left( \vec{b} + \vec{c} \right)\] .
Solution
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.
\[= \frac{1}{2}\left| \left( \vec{a} + \vec{b} \right) \times \left( \vec{b} + \vec{c} \right) \right|\]
\[ = \frac{1}{2}\left| - 4 \hat{ i } - 2 \hat{ j } - \hat{ k } \right|\]
\[ = \frac{1}{2}\sqrt{\left( - 4 \right)^2 + \left( - 2 \right)^2 + \left( - 1 \right)^2}\]
\[ = \frac{\sqrt{21}}{2} \text{ square units } \]
Thus, the required area of the parallelogram is \[\frac{\sqrt{21}}{2}\] square units.
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