Vector Algebra Question 92
Question: Let $ \vec{a},\vec{b} $ and $ \vec{c} $ be three non-zero vectors such that no two of these are collinear. If the vector $ \vec{a}+2\vec{b} $ is collinear with $ \vec{c} $ and $ \vec{b}+3\vec{c} $ is collinear with $ \vec{a} $ ( $ \lambda $ being some non-zero scalar) then $ \vec{a}+2\vec{b}+6\vec{c} $ equals
Options:
A) 0
B) $ \lambda \vec{b} $
C) $ \lambda \vec{c} $
D) $ \lambda \vec{a} $
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Answer:
Correct Answer: C
Solution:
- [c] Let $ \vec{a}+2\vec{b}=t\vec{c} $ and $ \vec{b}+3\vec{c}=s\vec{a} $ , where t and s are scalars. Adding, we get $ \vec{a}+3\vec{b}+3\vec{c}=t\vec{c}+s\vec{a}\Rightarrow \vec{a}+2\vec{b}+6\vec{c} $ $ =t\vec{c}+s\vec{a}-\vec{b}+3\vec{c} $ $ =t\vec{c}+(\vec{b}+3\vec{c})-\vec{b}+3\vec{c}=(t+6)\vec{c} $ [using $ s\vec{a}=\vec{b}+3\vec{c} $ ] $ =\lambda \vec{c} $ , where $ \lambda =t+6 $
BETA
