A) \[\frac{1}{2}\sqrt{7}\]
B) \[\sqrt{41}\]
C) \[\frac{1}{2}\sqrt{41}\]
D) \[\frac{1}{2}\sqrt{17}\]
Correct Answer: B
Solution :
Key Idea: If \[\vec{a},\vec{b}\]are the sides of a parallelogram, then area\[=\,|\vec{a}\times \vec{b}|\] We have,\[\vec{a}=3\hat{i}-\hat{k},\vec{b}=\hat{i}+2\hat{j}\] Now, \[\vec{a}\times \vec{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 0 & -1 \\ 1 & 2 & 0 \\ \end{matrix} \right|\] \[=\hat{i}(0+2)-\hat{j}(0+1)+\hat{k}(6-0)\] \[=2\hat{i}-\hat{j}+6\hat{k}\] \[\therefore \] Area of parallelogram \[=|\vec{a}\times \vec{b}|\] \[=|2\hat{i}-\hat{j}+6\hat{k}|\] \[=\sqrt{4+1+36}\] \[=\sqrt{41}\] Note: If \[\vec{a}\]and \[\vec{b}\]are the diagonals of a parallelogram, then area \[=\frac{1}{2}|\vec{a}\times \vec{b}|.\]
You need to login to perform this action.
You will be redirected in
3 sec
