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 noncoplanar vectors. The points are collinear when :
A.
$$\lambda = - 2,\,\mu = \frac{9}{4}$$
B.
$$\lambda = - \frac{9}{4},\,\mu = 2$$
C.
$$\lambda = \frac{9}{4},\,\mu = - 2$$
D.
none of these
Answer :
$$\lambda = \frac{9}{4},\,\mu = - 2$$
Solution :
If the points be $$A,\,B$$ and $$C$$ respectively then
$$\eqalign{
& \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = \overrightarrow a - 2\overrightarrow b + \lambda \overrightarrow c - \left( {2\overrightarrow a - \overrightarrow b + 3\overrightarrow c } \right) = - \overrightarrow a - \overrightarrow b + \left( {\lambda - 3} \right)\overrightarrow c \cr
& \overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} = \mu \overrightarrow a - 5\overrightarrow b - \left( {2\overrightarrow a - \overrightarrow b + 3\overrightarrow c } \right) = \left( {\mu - 2} \right)\overrightarrow a - 4\overrightarrow b - 3\overrightarrow c \cr} $$
The points are collinear if $$\overrightarrow {AB} = t\overrightarrow {AC} $$
$$\eqalign{
& \Rightarrow - 1 = t\left( {\mu - 2} \right),\,\, - 1 = - 4t,\,\,\lambda - 3 = - 3t \cr
& \therefore \,t = \frac{1}{4},{\text{ and }} - 1 = \frac{1}{4}\left( {\mu - 2} \right),\,\lambda - 3 = - \frac{3}{4} \cr} $$