A) \[5\hat{i}-7\hat{j}-3\hat{k}\]
B) \[5\hat{i}+7\hat{j}-3\hat{k}\]
C) \[5\hat{i}-7\hat{j}+3\hat{k}\]
D) zero
Correct Answer: A
Solution :
Key Idea: \[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}\] Given \[\overrightarrow{a}=2\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-2\hat{j}-\hat{k}\], and \[\overrightarrow{c}=\hat{i}+\hat{j}+\hat{k}\] \[\therefore \] \[\overrightarrow{a}.\overrightarrow{c}=2+1+1=4\] and \[\overrightarrow{a}.\overrightarrow{b}=2-2-1=-1\] \[\therefore \]\[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}\] \[=4(\hat{i}-2\hat{j}-\hat{k})+1(\hat{i}+\hat{j}+\hat{k})\] \[=5\hat{i}-7\hat{j}-3\hat{k}\] Alternative Method \[\overrightarrow{a}=2\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-2\hat{j}-\hat{k},\] \[\overrightarrow{c}=\hat{i}+\hat{j}+\hat{k}\] \[\therefore \]\[\overrightarrow{b}\times \overrightarrow{c}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -2 & -1 \\ 1 & 1 & 1 \\ \end{matrix} \right|\] \[=-\hat{i}-2\hat{j}+3\hat{k}\] \[\therefore \] \[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & 1 & 1 \\ -1 & -2 & 3 \\ \end{matrix} \right|\] \[=5\hat{i}-7\hat{j}-3\hat{k}\]
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