Question

# The position vectors of three points are $$2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } ,\overrightarrow { a } -2\overrightarrow { b } +\lambda \overrightarrow { c }$$ and $$\mu \overrightarrow { a } -5\overrightarrow { b }$$, where $$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c }$$ are non-coplanar vectors. The points are coliinear when

A
λ=2,μ=94
B
λ=94,μ=2
C
λ=94,μ=2
D
None of these

Solution

## The correct option is D None of theseLet the points $$A,B$$ and $$C$$ respectively, then $$\overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } \\ =\overrightarrow { a } -2\overrightarrow { b } +\lambda \overrightarrow { c } -\left( 2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } \right) \\ =-\overrightarrow { a } -\overrightarrow { b } +\left( \lambda -3 \right) \overrightarrow { c }$$$$\overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } \\ =\mu \overrightarrow { a } -5\overrightarrow { b } -\left( 2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } \right) \\ =\left( \mu -2 \right) \overrightarrow { a } -4\overrightarrow { b } -3\overrightarrow { c }$$The points are collinear if $$\overrightarrow { AB } =t\overrightarrow { AC }$$$$\Rightarrow -1=t\left( \mu -2 \right) ,-1=4t,\lambda -3=-3t$$$$\displaystyle \Rightarrow -1=t\left( \mu -2 \right) ,-1=4t,\lambda -3=-3t$$ and $$\displaystyle -1=\frac { -1 }{ 4 } \left( \mu -2 \right) ,\lambda -3=\frac { 3 }{ 4 }$$$$\displaystyle \Rightarrow \mu =6,\lambda =\frac {15 }{ 4 }$$Mathematics

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