Question

Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ be three non -zero vectors such that no two of these are collinear. If the vector $\overrightarrow{a}+2\overrightarrow{b}$ is collinear with $\overrightarrow{c}$ and $\overrightarrow{b}+3\overrightarrow{c}$ is collinear with $\overrightarrow{a}$ $(\lambda$ being some non -zero scalar) then $\overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}$ equals

• A. $\lambda \overrightarrow{a}$
  
• B. $\lambda \overrightarrow{b}$
  
• C. $\lambda \overrightarrow{c}$
  
• D. $0$

Answer 1

The correct option is D

$0$

This question is about the collinearity of vectors. As we learnt one very useful way of expressing that two vectors are collinear is by stating that one of those vectors can be equated to a scalar multiple of the second vector.
In the question we are given that a+2b is collinear to c.. So expressing the first vector as a scalar multiple of the second we can write as,

$(\overrightarrow{a}+2\overrightarrow{b})=t_1\overrightarrow{c}...\left(1\right)$
Similarly for the second set of collinear vectors we can write,
and $\overrightarrow{b}+3\overrightarrow{c}=t_2\overrightarrow{a}...\left(2\right)$
Note that t1 and t2 are the scalar used here. Now on solving the equations (1) and (2),
$\left(1\right)-2\times \left(2\right)\Rightarrow \overrightarrow{a}(1+2t_2)+\overrightarrow{c}(-t_1-6)=0$
Look carefully, this equation is of the form,
a.(scalar 1) + c(scalar 2) =0.
In addition to this we are given in the question that a,b and c be three non-zero vectors such that no two of these are collinear. Therefore it is safe to assume that scalars 1 and 2 are zero.
$\Rightarrow 1+2t_2=0\Rightarrow t_2=-\frac{1}{2}\&t_1=-6$.
Since $\overrightarrow{a}$ and $\overrightarrow{c}$ are non -collinear.
Putting the value of $t_1$ and $t_2$ in (1) and (2), we get $\overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}=\overrightarrow{0}$.