Question

Let $a$, $b$ and $c$ be three non - zero vectors such that no two of these are collinear. If the vector $a+2b$ is collinear with $c$, $b+3c$ is collinear with $a$ then $a+2b+6c$ is equals to

• A. $\lambda a(\lambda \ne 0,\text{a}\text{scalar})$
• B. $\lambda b(\lambda \ne 0,\text{a scalar})$
• C. $\lambda c(\lambda \ne 0,\text{a}\text{scalar})$
• D. $0$

Answer 1

The correct option is D

$0$

Explanation for correct answer:

Finding $a+2b+6c$:

If any two vectors $a\text{and}b$ are collinear, then $a=\lambda b(\lambda \text{is scalar)}$

Given,

$a$, $b$ and $c$ be three non-zero vectors such that no two of these are collinear

$a+2b$ is collinear with $c$

$\Rightarrow a+2b=\beta c\to \left(i\right)$

$b+3c$ is collinear with $a$

$\Rightarrow b+3c=\mu a\to \left(ii\right)$

Finding $a+2b+6c$

Add $6c$ in equation $\left(i\right)$

$a+2b+6c=\beta c+6c\to \left(iii\right)$

Multiplying by $2$and then $a$ in equation $\left(ii\right)$

$$\begin{gathered} 2\left(b+3c\right)=2\mu a \\ a+2b+6c=2\mu a+a\to \left(iv\right) \end{gathered}$$

Equating equation $\left(iii\right)\text{and}\left(iv\right)$

$$\begin{gathered} 2\mu a+a=\beta c+6c \\ 2a\left(\mu +1\right)=c\left(\beta +6\right) \end{gathered}$$

Since, $a\text{and}c$ are non-collinear.

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

substitute this value in equation $\left(iii\right)\text{or}\left(iv\right)$

Then, $a+2b+6c=0$

Hence, option (D) is the correct answer