A) \[\lambda \hat{i}+(2\lambda -1)\hat{j}\,+\lambda \hat{k},\lambda \in R\]
B) \[\lambda \hat{i}+(1-2\lambda )\hat{j}\,+\lambda \hat{k},\lambda \in R\]
C) \[\lambda \hat{i}+(2\lambda +1)\hat{j}\,+\lambda \hat{k},\lambda \in R\]
D) \[\lambda \hat{i}-(1+2\lambda )\hat{j}\,+\lambda \hat{k},\lambda \in R\]
Correct Answer: C
Solution :
[c] \[\vec{a}\times (\hat{i}+2\hat{j}+\hat{k})=\hat{i}-\hat{k}=(\hat{j}\times 2\hat{j}+\hat{k}))\] Or \[(\vec{a}-\hat{j})\times (\hat{i}+2\hat{j}+\hat{k})=\vec{0}\] Or \[\vec{a}-\hat{j}=\lambda (\hat{i}+2\hat{j}+\hat{k})\] Or \[\vec{a}=\lambda \hat{i}+(2\lambda +1)\hat{j}+\lambda \hat{k},\lambda \in R\]
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