Vector Algebra Question 16
Question: If $ 3\lambda \vec{c}+2\mu (\vec{a}\times \vec{b})=0 $ , then
Options:
A) $ 3\lambda +2\mu =0 $
B) $ 3\lambda =2\mu $
C) $ \lambda =\mu $
D) $ \lambda +\mu =0 $
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Answer:
Correct Answer: B
Solution:
- [b] $ 3\lambda \vec{c}+2\mu (\vec{a}\times \vec{b})=0 $
$ \Rightarrow 3\lambda \vec{c}=-2\mu (\vec{a}\times \vec{b}) $ So, one possibility is $ \vec{c} $ and $ (\vec{a}\times \vec{b}) $ are non-collinear.
$ \Rightarrow 3\lambda =2\mu $
BETA
