Question

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

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

Answer 1

Answer: (A)

0

Answer 2

As $\mathbf{a} + 2\mathbf{b}$ and $\mathbf{c}$ are collinear $\mathbf{a} + 2\mathbf{b} = \lambda\mathbf{c}$ ......(i)

Again $\mathbf{b} + 3\mathbf{c}$ is collinear with $\mathbf{a}$

$\therefore$ $\mathbf{b} + 3\mathbf{c}$ = $\mu \mathbf { a }$ .....(ii)

$( 2 \mu + 1 ) \mathbf { a }$ ......(iv)

From (iii) and (iv), $( \lambda + 6 ) \mathbf { c } = ( 2 \mu + 1 ) \mathbf { a }$

But $\mathbf { a }$ and $\mathbf { C }$ are non-zero , non-collinear vectors,

$\therefore$ $\lambda + 6 = 0 = 2 \mu + 1$. Hence, $\mathbf { a } + 2 \mathbf { b } + 6 \mathbf { c } = \mathbf { 0 }$