A) \[88\hat{i}+176\hat{j}-22\hat{k}\]
B) \[22\hat{i}+44\hat{j}-11\hat{k}\]
C) \[\hat{i}+4\hat{j}-\hat{k}\]
D) None of these
Correct Answer: A
Solution :
Given, \[\overrightarrow{A}=4\hat{i}+8\hat{j}-\hat{k},\] \[\overrightarrow{B}=3\hat{i}+2\hat{j}-\hat{k}\] and \[\overrightarrow{C}=5\hat{i}+4\hat{j}+\hat{k}\] Now, \[\overrightarrow{B}.\overrightarrow{C}=(3\hat{i}+2\hat{j}-\hat{k}).(5\hat{i}+4\hat{j}+\hat{k})\] \[=15+8-1\] \[=22\] \[\therefore \] \[\overrightarrow{A}.(\overrightarrow{B}.\overrightarrow{C})=(4\hat{i}+8\hat{j}-\hat{k}).22\] \[=88\hat{i}+176\hat{j}-22\hat{k}\]
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