A) \[\Sigma \hat{i}\times (\hat{j}+\hat{k})=\vec{0}\]
B) \[\Sigma \hat{i}\times (\hat{j}\times \hat{k})=\vec{0}\]
C) \[\Sigma \,\hat{i}.(\hat{j}\times \hat{k})=\vec{0}\]
D) \[\Sigma \,\hat{i}.(\hat{j}+\hat{k})=\vec{0}\]
Correct Answer: C
Solution :
Given, \[\hat{i},\hat{j},\hat{k}\] are unit vectors along the positive direction of x, y and z-axes, then [a] \[\Sigma \hat{i}\times (\hat{j}+\hat{k})\] \[=\hat{i}\times (\hat{j}+\hat{k})+\hat{j}\times (\hat{k}+\hat{i})+(\hat{i}+\hat{j})\] \[=\hat{k}-\hat{j}+\hat{i}-\hat{k}+\hat{j}-\hat{i}\] \[=0\] [b] \[\Sigma \hat{i}\times (\hat{j}\times \hat{k})=\hat{i}\times (\hat{j}\times \hat{k})+\hat{j}\times (\hat{k}\times \hat{i})\] \[+\hat{k}\times (\hat{i}+\hat{j})\] \[=(\hat{i}\times \hat{i})+(\hat{j}\times \hat{j})+(\hat{k}\times \hat{k})\] \[=0+0+0\] \[=0\] [c] \[\Sigma \hat{i}.(\hat{j}\times \hat{k})=\Sigma (\hat{i}.\hat{i})=\Sigma (1)\] \[=1+1+1=3\] [d] \[\Sigma \hat{i}.(\hat{j}+\hat{k})=\Sigma (\hat{i}.\hat{j}+\hat{i}.\hat{k})\] \[=\Sigma (0+0)=0\]
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