Vector Algebra Question 228
Question: If $ \vec{a} $ satisfies $ \vec{a}\times (\hat{i}+2\hat{j}+\hat{k})=\hat{i}-\hat{k} $ , then $ \vec{a} $ is equal to
Options:
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 $
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Answer:
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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