A) \[5\hat{i}+5\hat{j}-15\hat{k}\]
B) 0
C) \[12\hat{j}+4\hat{k}\]
D) \[-3\hat{i}+6\hat{j}-3\hat{k}\]
E) \[7\hat{i}-7\hat{j}-19\hat{k}\]
Correct Answer: A
Solution :
\[\overrightarrow{a}=-\hat{i}+2\hat{j}-\hat{k},\] \[\overrightarrow{b}=\hat{i}+\hat{j}-3\hat{k},\overrightarrow{c}=4\hat{i}-\hat{k}\] \[\therefore \]\[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})+(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}=(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b}\] \[=(4+1)(\hat{i}+\hat{j}-3\hat{k})\] \[=5(\hat{i}+\hat{j}-3\hat{k})\] \[=5\hat{i}+5\hat{j}-15\hat{k}\]
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