question_answer

Find the area of the parallelogram determined by the vectors \[\hat{i}+2\hat{j}+3\hat{k}\] and \[3\hat{i}-2\hat{j}+\hat{k}.\]

Answer:

Let \[\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\] and \[\vec{b}=3\hat{i}-2\hat{j}+\hat{k}\] Then, \[\vec{a}\times \vec{b}=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    1 & 2 & 3  \\    3 & -\,2 & 1  \\ \end{matrix} \right|\]             \[=\hat{i}(2+6)-\hat{j}(1-9)+\hat{k}(-\,2-6)\]             \[=8\hat{i}-8\hat{j}\,-8\hat{k}\] \[\therefore \] Area of parallelogram whose adjacent sides are \[\vec{a}\] and \[\vec{b},\]             \[A=|\vec{a}\times \vec{b}|\,\,=\sqrt{{{8}^{2}}+{{8}^{2}}+{{(-\,8)}^{2}}}\]             \[=8\sqrt{3}\,\,\text{sq}\,\,\text{units}\]