Question

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

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

Answer 1

Answer: (D)

If $ \vec{a} + 2\vec{b}$ is collinear with $\vec{c}$, then
$\vec{a} + 2\vec{b} = t \vec{c}\quad\ldots\left(i\right)$
Also, if $\vec{b} + 3 \vec{c}$ is collinear with $\vec{a}$, then
$\vec{b} + 3 \vec{c} = \lambda \vec{a}\quad \ldots \left(ii\right)$
$\Rightarrow \vec{b} =\lambda \vec{a} - 3 \vec{c} $
On putting, this value in eq. $\left(i\right)$, we get
$\vec{a} + 2\left(\lambda \vec{a} - 3 \vec{c}\right) = t \vec{c}$
$\vec{a} + 2\lambda \vec{a} - 6 \vec{c} = t \vec{c}$
$\Rightarrow \left(\vec{a} - 6 \vec{c}\right) = t \vec{c} - 2\lambda \vec{a} $
On comparing, we get
$1 = -2\lambda$
$\lambda = -1/2$ and $-6 = t$
$\Rightarrow t = -6$
From eq. $\left(i\right)$
$\vec{a} + 2\vec{b} = - 6 \vec{c}$
$\Rightarrow \vec{a} + 2\vec{b} + 6 \vec{c} = \vec{0}$