A) 5
B) 10
C) 15
D) None of these
Correct Answer: D
Solution :
[d] \[\overset{\to }{\mathop{p}}\,=\lambda (\overset{\to }{\mathop{u}}\,\times \overset{\to }{\mathop{v}}\,)+\mu (\overset{\to }{\mathop{v}}\,\times \overset{\to }{\mathop{w}}\,)+v(\overset{\to }{\mathop{w}}\,\times \overset{\to }{\mathop{u}}\,)\] \[\Rightarrow \overset{\to }{\mathop{p}}\,.\overset{\to }{\mathop{w}}\,=\lambda (\overset{\to }{\mathop{u}}\,\times \overset{\to }{\mathop{v}}\,).\overset{\to }{\mathop{w}}\,+\mu (\overset{\to }{\mathop{v}}\,\times \overset{\to }{\mathop{w}}\,).\overset{\to }{\mathop{w}}\,+v(\overset{\to }{\mathop{w}}\,\times \overset{\to }{\mathop{u}}\,).\overset{\to }{\mathop{w}}\,\] \[=\lambda [\overset{\to }{\mathop{u}}\,\overset{\to }{\mathop{v}}\,\overset{\to }{\mathop{w}}\,]+0+0=\frac{\lambda }{5}\Rightarrow \lambda =5(\overset{\to }{\mathop{p}}\,.\overset{\to }{\mathop{w}}\,)\] 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.
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