Vector Algebra Question 135
Question: If $ \mathbf{d}=\lambda ,(\mathbf{a}\times \mathbf{b})+\mu ,(\mathbf{b}\times \mathbf{c})+\nu ,(\mathbf{c}\times \mathbf{a}) $ and $ [\mathbf{a},\mathbf{b},\mathbf{c}]=\frac{1}{8}, $ then $ \lambda +\mu +\nu $ is equal to
Options:
A) $ 8\mathbf{d},.,(\mathbf{a}+\mathbf{b}+\mathbf{c}) $
B) $ 8\mathbf{d},\times ,(\mathbf{a}+\mathbf{b}+\mathbf{c}) $
C) $ \frac{\mathbf{d},}{8}.,(\mathbf{a}+\mathbf{b}+\mathbf{c}) $
D) $ \frac{\mathbf{d},}{8}\times ,(\mathbf{a}+\mathbf{b}+\mathbf{c}) $
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Answer:
Correct Answer: A
Solution:
- $ \mathbf{d},.,\mathbf{c}=\lambda (\mathbf{a}\times \mathbf{b}),.,\mathbf{c}+\mu (\mathbf{b}\times \mathbf{c}),.,\mathbf{c}+\nu (\mathbf{c}\times \mathbf{a}),.\mathbf{c} $ $ =\lambda ,[\mathbf{a},\mathbf{b},\mathbf{c}]+0+0=\lambda ,[\mathbf{a},\mathbf{b},\mathbf{c}]=\frac{\lambda }{8} $ Hence $ \lambda =8(\mathbf{d},.,\mathbf{c}), $ $ \mu =8(\mathbf{d},.,\mathbf{a}) $ and $ \nu =8(\mathbf{d},.,\mathbf{b}) $ Therefore, $ \lambda +\mu +\nu =8\mathbf{d},.,\mathbf{c}+8\mathbf{d}.,\mathbf{a}+8\mathbf{d},.,\mathbf{b} $ $ =8\mathbf{d},.,(\mathbf{a}+\mathbf{b}+\mathbf{c}). $
BETA
