# Let $\dpi{100} \vec{a},\vec{b}\; \; and\; \; \vec{c}$ be three non­-zero vectors such that no two of these are collinear. If the vector $\dpi{100} \vec{a}+2\vec{b}\;$ is collinear with $\dpi{100} \vec{c}\; \; and\; \; \vec{b}\;+3\vec{c}\;$ is collinear with $\dpi{100} \vec{a}\;$ ($\dpi{100} \lambda$ being some non-­zero scalar) then $\dpi{100} \vec{a}+2\vec{b}+6\vec{c}$ equals Option 1) $\lambda \vec{c}\;$ Option 2) $\; \; \lambda \vec{b}\;$ Option 3) $\; \; \lambda \vec{a}$ Option 4) $0$

As we learnt in

Collinear Vectors -

Two vectors are said to be collinear if their directed line segments are parallel disregards to their direction.

- wherein

If $\vec{a}$ and $\vec{b}$ are collinear , then $\vec{a}= K\vec{b}$ where K $\epsilon$ R

$\vec{a} +\vec{2b}$ is collinear with $\vec{c}$

$(\vec{a}+\vec{2b}) =p\vec{c}$

Similarily $(\vec{b}+\vec{3b}) =q\vec{a}$

Eliminating $\vec{b}$

we get $\vec{a} +\vec{2b} -\vec{2b} -\vec{6c} = p\vec{c} -2q\vec{a}$

$\vec{a}\left ( 1+2q \right )+\vec{c}(-p-6)=0$

Thus $p= -6, q=\frac{-1}{2}$

Now

$\vec{a}+\vec{2b} = p\vec{c}$

$\Rightarrow \vec{a}+\vec{2b}+\vec{6c} = 0$

Option 1)

$\lambda \vec{c}\;$

Incorrect

Option 2)

$\; \; \lambda \vec{b}\;$

Incorrect

Option 3)

$\; \; \lambda \vec{a}$

Incorrect

Option 4)

$0$

Correct

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