Question

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) \(\vec{\gamma}\)
(B) \(\overrightarrow{0}\)
(C) \(\vec{\alpha}+\vec{\gamma}\)
(D) \(\vec{\alpha}\)

Answer 1

Ans: (B)
$$
\text { Hint } \begin{aligned}
& \vec{\alpha}+3 \vec{\beta}=\mathrm{k}_{1} \vec{\gamma} \Rightarrow \vec{\beta}=\frac{\mathrm{k}_{1}}{3} \vec{\gamma}-\frac{\vec{\alpha}}{3} \\
& \vec{\beta}+2 \vec{\gamma}=\mathrm{k}_{2} \vec{\alpha} \Rightarrow \vec{\beta}=\mathrm{k}_{2} \vec{\alpha}-2 \vec{\gamma} \\
& \Rightarrow \frac{\mathrm{k}_{1} \vec{\gamma}}{3}-\frac{\vec{\alpha}}{3}=\mathrm{k}_{2} \vec{\alpha}-2 \vec{\gamma} \Rightarrow \vec{\alpha}\left(\mathrm{k}_{2}+\frac{1}{3}\right)=\vec{\gamma}\left(\frac{\mathrm{k}_{1}}{3}+2\right) \Rightarrow \mathrm{k}_{2}=-\frac{1}{3} \text { and } \frac{\mathrm{k}_{1}}{3}=-2 \Rightarrow \mathrm{k}_{1}=-6 \\
& \Rightarrow \vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}=\overrightarrow{0}
\end{aligned}
$$

Answer 2

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Answer 3

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