A) \[\hat{i}+\hat{j}+\hat{k}\]
B) \[-\hat{i}+\hat{j}+\hat{k}\]
C) \[\hat{i}-\hat{j}+\hat{k}\]
D) \[\hat{i}+\hat{j}-\hat{k}\]
Correct Answer: B
Solution :
Given,\[\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{b}=\hat{j}+\hat{k}\]and\[\overrightarrow{c}=\hat{i}-\hat{k}\] \[\overrightarrow{a}+\overrightarrow{b}=2\hat{i}+0\hat{j}+2\hat{k}=2\hat{i}+2\hat{k}\] \[\overrightarrow{b}+\overrightarrow{c}=\hat{i}+\hat{j}+0\hat{k}=\hat{i}+\hat{j}\] Since,\[\overrightarrow{a}+\overrightarrow{b}\]and\[\overrightarrow{c}+\overrightarrow{d}\]are the diagonals of parallelogram. \[\therefore \] Area\[=\frac{1}{2}|(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{b}+\overrightarrow{c})|\] \[=\frac{1}{2}\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & 0 & 2 \\ 1 & 1 & 0 \\ \end{matrix} \right|\] \[=\frac{1}{2}[\hat{i}(0-2)-\hat{j}(0-2)+\hat{k}(2-0)]\] \[=\frac{1}{2}[-2\hat{i}+2\hat{j}+2\hat{k}]\] \[=-\hat{i}+\hat{j}+\hat{k}\]
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