Exercise 10.2 · Q19 (MCQ): Which statements about collinear $\vec a$⁠, $\vec b$ are incorrect

Basics you need

Solution

1. What "collinear" allows

   $\vec a$ and $\vec b$ are collinear exactly when

   $$\vec b = \lambda\vec a \quad \text{for some scalar } \lambda \ne 0.$$
   $\lambda$ is free: $\lambda = 2$ gives a longer vector the same way, $\lambda = -1$ the same length the opposite way, $\lambda = 1$ the same vector. All of these pairs are collinear, so a statement that fails for even one of them is incorrect as a statement about collinear vectors.
   Collinear if and only if $\vec b = \lambda\vec a$ (Remarks (i), Section 10.5.1); collinear vectors may differ in magnitude and direction (Section 10.3). Lesson: six kinds of vectors worth naming

2. Option (A): correct

   (A) says $\vec b = \lambda\vec a$ for some scalar $\lambda$⁠. That is the test of Step 1 itself, so (A) is correct.

   The book states the test as "if and only if there exists a nonzero scalar $\lambda$ such that $\vec b = \lambda\vec a$", for a nonzero $\vec a$⁠. (A) does not mention "nonzero", but it only claims that some scalar exists, which is true.
   Remarks (i), Section 10.5.1. Lesson: components

3. Option (C): incorrect

   From $\vec b = \lambda\vec a$⁠, comparing components,

   $$\begin{aligned} b_1 &= \lambda a_1, \\ b_2 &= \lambda a_2, \\ b_3 &= \lambda a_3. \end{aligned}$$

   So the components of $\vec b$ are always $\lambda$ times those of $\vec a$⁠: they are proportional. (C) says they are not, so (C) is incorrect, for every collinear pair.

   Two vectors are equal exactly when their components are equal, so $\vec b = \lambda\vec a$ gives the three equations. Where no $a_k$ is zero this is the book's $\frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} = \lambda$⁠.
   $\lambda\vec a$ by components (Section 10.5.1, (iv)); equal vectors have equal components (Section 10.5.1, (iii)). Lesson: components

4. Options (B) and (D): incorrect

   (B) says $\vec a = \pm\vec b$⁠. Take $\vec b = 2\vec a$ with $\vec a \ne \vec 0$⁠: the vectors are collinear, but $\vec b$ is twice as long as $\vec a$⁠, so $\vec a \ne \vec b$ and $\vec a \ne -\vec b$⁠. (B) is incorrect.

   (D) says the vectors have the same direction but different magnitudes. Take $\vec b = -\vec a$⁠: collinear, but opposite in direction (and equal in magnitude). (D) is incorrect.

   (B) is true only when $|\lambda| = 1$⁠, and (D) only when $\lambda > 0$ and $\lambda \ne 1$⁠. Collinear vectors need neither, so one counterexample for each is enough.
   $\lambda\vec a$ points the same way for $\lambda > 0$⁠, the opposite way for $\lambda < 0$⁠, and $|\lambda\vec a| = |\lambda||\vec a|$ (Section 10.5). Compare Exercise 10.1 Q5 and Exercise 10.2 Q11.

5. Collect the answers

   Option	Always true?
   (A) $\vec b = \lambda\vec a$	Yes, it is the definition: correct
   (B) $\vec a = \pm\vec b$	No, fails for $\vec b = 2\vec a$⁠: incorrect
   (C) components not proportional	No, they are always proportional: incorrect
   (D) same direction, different magnitudes	No, fails for $\vec b = -\vec a$⁠: incorrect

   So (B), (C) and (D) are incorrect: three answers, which is why the question says "which of the following are incorrect".

   This MCQ is unusual: it has more than one answer, and the key lists all three, (B), (C), (D). That is mathematically right. (C) is false for every collinear pair; (B) and (D) can hold for particular pairs ($\vec b = -\vec a$ satisfies (B); $\vec b = 3\vec a$ satisfies (D)), but they are not true of collinear vectors in general, so as statements about collinear vectors they are incorrect.
   A general statement is shown false by one counterexample. Lesson: six kinds of vectors worth naming

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