Discussion
Let the three vectors \(\vec{a}, \vec{b},\) and \(\vec{c}\) be pairwise non-collinear vectors. If \(\vec{a} + 2\vec{b}\) is collinear with \(\vec{c}\) and if \(\vec{b} + 2\vec{c}\) is collinear with \(\vec{a}\) , then \(\vec{a} + 2\vec{b} + 5\vec{c}\) is equal to
Answer: D
\(\vec{a} + 2\vec{b} + 5\vec{c}\) equals \(\vec{c}\).
Write \(\vec{a} + 2\vec{b} = \lambda\vec{c}\) and \(\vec{b} + 2\vec{c} = \mu\vec{a}\). Substituting \(\vec{a} = \lambda\vec{c} - 2\vec{b}\) into the second:
\[ \begin{aligned} \vec{b} + 2\vec{c} &= \mu\lambda\vec{c} - 2\mu\vec{b} \\ (1 + 2\mu)\vec{b} &= (\mu\lambda - 2)\vec{c} \end{aligned} \]
Since \(\vec{b}\) and \(\vec{c}\) are not collinear, both coefficients vanish: \(\mu = -\frac{1}{2}\) and \(\lambda = -4\). So \(\vec{a} + 2\vec{b} = -4\vec{c}\), and \(\vec{a} + 2\vec{b} + 5\vec{c} = \vec{c}\). The zero vector would be \(\vec{a} + 2\vec{b} + 4\vec{c}\).
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