Q47 / 155 · WBJEE 2021 · Mathematics

Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be three non-zero vectors which are pairwise non-collinear. If $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$ and $\vec{\beta}+2 \vec{\gamma}$ is collinear with $\vec{\alpha}$, then $\vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}$ is

  • A
    $\vec{\gamma}$
  • Correct answer
    B
    $\overrightarrow{0}$
  • C
    $\vec{\alpha}+\vec{\gamma}$
  • D
    $\vec{\alpha}$

Solution

$\vec{\alpha}+3 \vec{\beta}=\mathrm{k}_{1} \vec{\gamma} \Rightarrow \vec{\beta}=\frac{\mathrm{k}_{1}}{3} \vec{\gamma}-\frac{\vec{\alpha}}{3}$ $\vec{\beta}+2 \vec{\gamma}=\mathrm{k}_{2} \vec{\alpha} \Rightarrow \vec{\beta}=\mathrm{k}_{2} \vec{\alpha}-2 \vec{\gamma}$ $\Rightarrow \frac{\mathrm{k}_{1} \vec{\gamma}}{3}-\frac{\vec{\alpha}}{3}=\mathrm{k}_{2} \vec{\alpha}-2 \vec{\gamma} \Rightarrow \vec{\alpha}\left(\mathrm{k}_{2}+\frac{1}{3}\right)=\vec{\gamma}\left(\frac{\mathrm{k}_{1}}{3}+2\right) \Rightarrow \mathrm{k}_{2}=-\frac{1}{3}$ and $\frac{\mathrm{k}_{1}}{3}=-2 \Rightarrow \mathrm{k}_{1}=-6$ $\Rightarrow \vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}=\overrightarrow{0}$

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

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