Q135 / 225 · JEE Main 2004 · Mathematics

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

  • A
    $\lambda \vec{a}$
  • B
    $\lambda \vec{b}$
  • C
    $\lambda \overrightarrow{\mathbf{c}}$
  • Correct answer
    D
    0

Solution

$(\vec{a}+2 b)=t_1 \vec{c}$ and $\vec{b}+3 \vec{c}=t_2 \vec{a}$ $(1)-2 \times(2) \Rightarrow \vec{a}\left(1+2 t_2\right)+\vec{c}\left(-t_1-6\right)=0 \Rightarrow 1+2 t_2=0 \Rightarrow t_2=-1 / 2 \& t_1=-6$. Since $\vec{a}$ and $\vec{c}$ are non-collinear. Putting the value of $t_1$ and $t_2$ in (1) and (2), we get $\vec{a}+2 \vec{b}+6 \vec{c}=\overrightarrow{0}$

Hence, the correct answer is option: 'D'.

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