Question

Let $\stackrel{\rightharpoonup }{a}$,$\stackrel{\rightharpoonup }{b}$ and $\stackrel{\rightharpoonup }{c}$be three non - zero vectors that are pairwise non-collinear. If $\stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}$ is collinear with $\stackrel{\rightharpoonup }{c}$ and $\stackrel{\rightharpoonup }{b}+2\stackrel{\rightharpoonup }{c}$ is collinear with $\stackrel{\rightharpoonup }{a}$, then $\stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}+6\stackrel{\rightharpoonup }{c}$ is equal to ?

• A. $0$
• B. $a$
• C. $a+b$
• D. $c$

Answer 1

The correct option is A

$0$

Step 1: Apply condition for given vectors to be collinear

Let $\stackrel{\rightharpoonup }{a}$,$\stackrel{\rightharpoonup }{b}$ and $\stackrel{\rightharpoonup }{c}$be three non-zero vectors.

If any two vectors are collinear vectors then they are parallel to each other.

If $\stackrel{\rightharpoonup }{a}$and $\stackrel{\rightharpoonup }{b}$are collinear vectors, then $\stackrel{\rightharpoonup }{a}=\lambda \stackrel{\rightharpoonup }{b}$

Given that, $\stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}$ is collinear with $\stackrel{\rightharpoonup }{c}$

$\Rightarrow \stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}=\lambda \stackrel{\rightharpoonup }{c}\to \left(i\right)$

$\stackrel{\rightharpoonup }{b}+2\stackrel{\rightharpoonup }{c}$ is collinear with $\stackrel{\rightharpoonup }{a}$,

$\Rightarrow \stackrel{\rightharpoonup }{b}+2\stackrel{\rightharpoonup }{c}=\mu \stackrel{\rightharpoonup }{a}\to \left(ii\right)$

Step 2: Determine $\stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}+6\stackrel{\rightharpoonup }{c}$:

Add and subtract $6\stackrel{\rightharpoonup }{c}\text{in equation}\left(i\right)$

$$\begin{gathered} \stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}+6\stackrel{\rightharpoonup }{c}-6\stackrel{\rightharpoonup }{c}=\lambda \stackrel{\rightharpoonup }{c} \\ \Rightarrow \stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}+6\stackrel{\rightharpoonup }{c}=\lambda \stackrel{\rightharpoonup }{c}+6\stackrel{\rightharpoonup }{c} \\ \stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}+6\stackrel{\rightharpoonup }{c}=\left(\lambda +6\right)\stackrel{\rightharpoonup }{c}\to \left(iii\right) \end{gathered}$$

Multiply by $3$ on both sides in equation $\left(ii\right)$

$$\begin{gathered} 3\left(\stackrel{\rightharpoonup }{b}+2\stackrel{\rightharpoonup }{c}\right)=3\mu \stackrel{\rightharpoonup }{a} \\ 3\stackrel{\rightharpoonup }{b}+6\stackrel{\rightharpoonup }{c}+\stackrel{\rightharpoonup }{a}=3\mu \stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{a}\left[\because Addandsubtract\stackrel{\rightharpoonup }{a}onbothsides\right] \\ \stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}+6\stackrel{\rightharpoonup }{c}=\left(3\mu +1\right)\stackrel{\rightharpoonup }{a}\to \left(iv\right) \end{gathered}$$

equating equation $\left(iii\right)$ and $\left(iv\right)$

$\left(3\mu +1\right)\stackrel{\rightharpoonup }{a}=\left(\lambda +6\right)\stackrel{\rightharpoonup }{c}$is not possible.

since,$\stackrel{\rightharpoonup }{a}$,$\stackrel{\rightharpoonup }{b}$ and $\stackrel{\rightharpoonup }{c}$be three non-zero vectors and pairwise collinear.

Therefore,$\left(3\mu +1\right)=\left(\lambda +6\right)=0$

Substitute this in equation $\left(iii\right)$ or $\left(iv\right)$

we get, $\stackrel{\rightharpoonup }{a}+3\stackrel{\rightharpoonup }{b}+6\stackrel{\rightharpoonup }{c}=0$

Hence, option (A) is the correct answer