JEE MAIN - Mathematics (2004 - No. 27)
Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be three non-zero vectors such that no two of these are collinear. If the vector $$\overrightarrow a + 2\overrightarrow b $$ is collinear with $$\overrightarrow c $$ and $$\overrightarrow b + 3\overrightarrow c $$ is collinear with $$\overrightarrow a $$ ($$\lambda $$ being some non-zero scalar) then $$\overrightarrow a + 2\overrightarrow b + 6\overrightarrow c $$ equals to :
$\overrightarrow{0}$
$$\lambda \overrightarrow b $$
$$\lambda \overrightarrow c $$
$$\lambda \overrightarrow a $$
Explanation
If $\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}$ is collinear with $\overrightarrow{\mathbf{c}}$, then
$$ \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}=t \overrightarrow{\mathbf{c}} $$
Also, if $\overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}$ is collinear with $\overrightarrow{\mathbf{a}}$, then
$$ \begin{aligned} & \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}=\lambda \overrightarrow{\mathbf{a}} \\\\ & \Rightarrow \overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}} \end{aligned} $$
On putting this value in Eq. (i), we get
$$ \overrightarrow{\mathbf{a}}+2(\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}} $$
$$ \Rightarrow \overrightarrow{\mathbf{a}}+2 \lambda \overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}}=t \overrightarrow{\mathbf{c}} $$
$$ \Rightarrow (\overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}}-2 \lambda \overrightarrow{\mathbf{a}} $$
On comparing, we get
and $-6=t $
$\Rightarrow t=-6$
From Eq. (i),
$$\vec{a}+2 \vec{b}=-6 \vec{c} $$
$$\Rightarrow \vec{a}+2 \vec{b}+6 \vec{c}=\overrightarrow{0}$$
$$ \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}=t \overrightarrow{\mathbf{c}} $$
Also, if $\overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}$ is collinear with $\overrightarrow{\mathbf{a}}$, then
$$ \begin{aligned} & \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}=\lambda \overrightarrow{\mathbf{a}} \\\\ & \Rightarrow \overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}} \end{aligned} $$
On putting this value in Eq. (i), we get
$$ \overrightarrow{\mathbf{a}}+2(\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}} $$
$$ \Rightarrow \overrightarrow{\mathbf{a}}+2 \lambda \overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}}=t \overrightarrow{\mathbf{c}} $$
$$ \Rightarrow (\overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}}-2 \lambda \overrightarrow{\mathbf{a}} $$
On comparing, we get
and $-6=t $
$\Rightarrow t=-6$
From Eq. (i),
$$\vec{a}+2 \vec{b}=-6 \vec{c} $$
$$\Rightarrow \vec{a}+2 \vec{b}+6 \vec{c}=\overrightarrow{0}$$
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