Vector-Algebra Question 417
Question: If $ \overrightarrow{p}=\lambda (\overrightarrow{u}\times \overrightarrow{v})+\mu (\overrightarrow{v}\times \overrightarrow{w})+v(\overrightarrow{w}\times \overrightarrow{u}) $ and $ [\overset{\to }{\mathop{u}}\overset{\to }{\mathop{v}}\overset{\to }{\mathop{w}},]=\frac{1}{5} $ , then $ \lambda +\mu +v $ is equal to
Options:
A) 5
B) 10
C) 15
D) None of these
Correct Answer: DShow Answer
Answer:
Solution:
$ \Rightarrow \overset{\to }{\mathop{p}},.\overset{\to }{\mathop{w}},=\lambda (\overset{\to }{\mathop{u}},\times \overset{\to }{\mathop{v}},).\overset{\to }{\mathop{w}},+\mu (\overset{\to }{\mathop{v}},\times \overset{\to }{\mathop{w}},).\overset{\to }{\mathop{w}},+v(\overset{\to }{\mathop{w}},\times \overset{\to }{\mathop{u}},).\overset{\to }{\mathop{w}}, $ $ =\lambda [\overset{\to }{\mathop{u}},\overset{\to }{\mathop{v}},\overset{\to }{\mathop{w}},]+0+0=\frac{\lambda }{5}\Rightarrow \lambda =5(\overset{\to }{\mathop{p}},.\overset{\to }{\mathop{w}},) $ Similarly, $ \mu =5(\overset{\to }{\mathop{p}},.\overset{\to }{\mathop{u}},) $ and $ v=5(\overset{\to }{\mathop{p}},.\overset{\to }{\mathop{v}},) $
$ \therefore \lambda +\mu +v=5(\overset{\to }{\mathop{p}},.\overset{\to }{\mathop{w}},)+5(\overset{\to }{\mathop{p}},.\overset{\to }{\mathop{u}},)+5(\overset{\to }{\mathop{p}},.\overset{\to }{\mathop{v}},) $ $ =5\overset{\to }{\mathop{p}},.(\overset{\to }{\mathop{u}},+\overset{\to }{\mathop{v}},+\overset{\to }{\mathop{w}},) $ Hence, $ \lambda +\mu +v $ depends on the vectors.
BETA
