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Chapter 10 · Vector Algebra

Exercise 10.1 · Q5: True or false: collinear vectors, equal magnitudes and equal vectors

Exercise 10.1Types of vectors · Section 10.3
The question

Answer the following as true or false.

(i) a⃗\vec a and −a⃗-\vec a are collinear.

(ii) Two collinear vectors are always equal in magnitude.

(iii) Two vectors having same magnitude are collinear.

(iv) Two collinear vectors having the same magnitude are equal.

Exam answer

(i) −a⃗=(−1)a⃗-\vec a = (-1)\vec a⁠, so −a⃗-\vec a is parallel to the line of a⃗\vec a⁠: True.

(ii) 2i^=2(i^)2\hat i = 2(\hat i) is collinear with i^\hat i⁠, but ∣2i^∣=2≠1=∣i^∣|2\hat i| = 2 \ne 1 = |\hat i|⁠: False.

(iii) ∣i^∣=∣j^∣=1|\hat i| = |\hat j| = 1⁠, but j^≠λi^\hat j \ne \lambda\hat i for every scalar λ\lambda (they are perpendicular), so they are not collinear: False.

(iv) i^\hat i and −i^-\hat i are collinear and ∣i^∣=∣−i^∣=1|\hat i| = |-\hat i| = 1⁠, but their directions are opposite, so −i^≠i^-\hat i \ne \hat i⁠: False.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Keep the three ideas apart. Collinear is only about the line: the vectors are parallel to the same line, "irrespective of their magnitudes and directions". Magnitude is only about length. Equal needs both the same magnitude and the same direction. A statement that is true must hold for every pair of vectors; a statement is false as soon as one pair breaks it, so each "false" is settled by one explicit counterexample.

Given, to find, and the plan
Given
Four statements about collinear vectors, magnitudes and equality.
To find
Whether each statement is true or false.
Plan
(i) Prove it for every a⃗\vec a using −a⃗=(−1)a⃗-\vec a = (-1)\vec a⁠. (ii) to (iv) Test with the simple vectors i^\hat i⁠, 2i^2\hat i⁠, −i^-\hat i and j^\hat j⁠, and give one counterexample for each false statement.

Basics you need

1Collinear and equal vectors

Collinear: parallel to the same line, irrespective of their magnitudes and directions. Equal: same magnitude and direction, regardless of the positions of their initial points (Section 10.3). Lesson: six kinds of vectors worth naming

2Multiplying by a scalar

λa⃗\lambda\vec a is collinear to a⃗\vec a⁠; it points the same way when λ>0\lambda > 0 and the opposite way when λ<0\lambda < 0⁠, and ∣λa⃗∣=∣λ∣ ∣a⃗∣|\lambda\vec a| = |\lambda|\,|\vec a|⁠. Also, b⃗\vec b and a⃗\vec a are collinear if and only if b⃗=λa⃗\vec b = \lambda\vec a for a nonzero scalar λ\lambda (Section 10.5 and Remark (i) of Section 10.5.1). Lesson: components

3The unit vectors i^\hat i and j^\hat j

i^\hat i and j^\hat j have magnitude 1 and lie along the x-axis and the y-axis, which are perpendicular (Section 10.5.1). Basics: three numbers fix a point

4True for all, false by one example

To show a general statement is false, one specific case where it fails is enough. To show it is true, the argument must work for every vector.

Part (i): a⃗\vec a and −a⃗-\vec a are collinear

  1. Write −a⃗-\vec a as a scalar multiple of a⃗\vec a

    For every vector a⃗\vec a⁠, the negative of a⃗\vec a is the scalar multiple with λ=−1\lambda = -1⁠:

    −a⃗=(−1) a⃗.-\vec a = (-1)\,\vec a.
    The book defines −a⃗-\vec a as the vector λa⃗\lambda\vec a with λ=−1\lambda = -1⁠: same magnitude as a⃗\vec a⁠, opposite direction. Writing it as a scalar multiple lets us use the collinearity test.
    When λ=−1\lambda = -1⁠, λa⃗=−a⃗\lambda\vec a = -\vec a⁠, the negative (additive inverse) of a⃗\vec a (Section 10.5). Lesson: components

  2. Apply the collinearity test

    Since −a⃗=λa⃗-\vec a = \lambda\vec a with the nonzero scalar λ=−1\lambda = -1⁠, the vectors a⃗\vec a and −a⃗-\vec a are collinear. In a picture, −a⃗-\vec a lies along the same line as a⃗\vec a⁠, pointing the other way (the rose arrow −i^-\hat i beside i^\hat i in the figure below). So statement (i) is true.

    Collinearity ignores direction along the line, so pointing the opposite way does not matter. The argument used nothing special about a⃗\vec a⁠, so it holds for every vector. (If a⃗=0⃗\vec a = \vec 0⁠, both vectors are the zero vector, which may be regarded as having any direction, so the statement still holds.)
    a⃗\vec a and b⃗\vec b are collinear if and only if b⃗=λa⃗\vec b = \lambda\vec a for a nonzero scalar λ\lambda (Remark (i), Section 10.5.1). Lesson: six kinds of vectors

Part (ii): collinear vectors are always equal in magnitude

  1. Try two vectors along the same line with different lengths

    Take i^\hat i and 2i^2\hat i⁠. Since 2i^=2(i^)2\hat i = 2(\hat i)⁠, they are collinear (both along the x-axis). Their magnitudes are

    ∣i^∣=1,∣2i^∣=2∣i^∣=2.|\hat i| = 1, \qquad |2\hat i| = 2|\hat i| = 2.
    2î î −î ĵ O x y grid: 1 unit per square
    The four test vectors drawn from O on a unit grid (the three arrows along the x-axis are drawn a little apart so that they do not cover each other). i^\hat i⁠, 2i^2\hat i and −i^-\hat i all lie along the x-axis, so they are collinear; j^\hat j has the same length as i^\hat i but lies along the y-axis.
    Multiplying by a scalar keeps a vector on the same line but changes its length by the factor ∣λ∣|\lambda|⁠. Choosing λ=2\lambda = 2 gives a collinear vector twice as long, the simplest way to break the statement.
    ∣λa⃗∣=∣λ∣ ∣a⃗∣|\lambda\vec a| = |\lambda|\,|\vec a| (Section 10.5). Lesson: components

  2. Conclude

    i^\hat i and 2i^2\hat i are collinear but 1≠21 \ne 2⁠, so collinear vectors need not have equal magnitudes. Statement (ii) is false.

    "Always" claims the property for every pair; one pair that fails is a counterexample, and it disproves the claim. The definition itself says collinear vectors may have any magnitudes.
    Collinear vectors are parallel to the same line "irrespective of their magnitudes" (Section 10.3). Lesson: six kinds of vectors

Part (iii): vectors with the same magnitude are collinear

  1. Try two vectors of the same length in different lines

    Take i^\hat i and j^\hat j⁠. Both have magnitude 1. If they were collinear we would have j^=λi^\hat j = \lambda\hat i for some scalar λ\lambda⁠, that is

    0i^+1j^+0k^=λi^+0j^+0k^.0\hat i + 1\hat j + 0\hat k = \lambda\hat i + 0\hat j + 0\hat k.

    Comparing the j^\hat j components gives 1=01 = 0⁠, which is impossible. So no such λ\lambda exists.

    Two vectors are equal only if their corresponding components are equal. The j^\hat j component of λi^\lambda\hat i is 0 for every λ\lambda⁠, so λi^\lambda\hat i can never be j^\hat j⁠. Geometrically, i^\hat i lies along the x-axis and j^\hat j along the y-axis, and these lines are perpendicular, not parallel.
    Equality of vectors in component form (Section 10.5.1, (iii)) and the collinearity test b⃗=λa⃗\vec b = \lambda\vec a (Remark (i)). Basics: three numbers fix a point

  2. Conclude

    i^\hat i and j^\hat j have the same magnitude but are not collinear. Statement (iii) is false.

    Magnitude says nothing about direction: arrows of length 1 can point anywhere (their tips fill a whole circle in the plane, a whole sphere in space). Only a few of them are parallel to a given line.
    Magnitude is the length of the directed line segment (Section 10.2); collinearity is about the line (Section 10.3). Lesson: six kinds of vectors

Part (iv): collinear vectors with the same magnitude are equal

  1. Try a vector and its negative

    Take i^\hat i and −i^-\hat i⁠. By Part (i) they are collinear, and

    ∣−i^∣=∣−1∣ ∣i^∣=1=∣i^∣.|-\hat i| = |-1|\,|\hat i| = 1 = |\hat i|.

    But −i^-\hat i points in the negative x direction and i^\hat i in the positive x direction; in components, −1≠1-1 \ne 1⁠. So −i^≠i^-\hat i \ne \hat i⁠.

    On one line there are exactly two directions. Two collinear vectors of the same length are equal when they point the same way and are negatives of each other when they point opposite ways. The statement forgets the second case.
    Equal vectors need the same magnitude and the same direction (Section 10.3); −a⃗-\vec a has the opposite direction (Negative of a vector, Section 10.3). Lesson: six kinds of vectors

  2. Conclude

    i^\hat i and −i^-\hat i are collinear with equal magnitudes but are not equal. Statement (iv) is false.

    The counterexample uses opposite directions, which collinearity allows ("irrespective of their directions") but equality does not.
    In Exercise 10.1 Q4, a⃗\vec a and c⃗=−a⃗\vec c = -\vec a on opposite sides of the square are another example. Exercise 10.1 Q4
Result

(i) True. (ii) False (for example i^\hat i and 2i^2\hat i⁠). (iii) False (for example i^\hat i and j^\hat j⁠). (iv) False (for example i^\hat i and −i^-\hat i⁠).

(Textbook answer: (i) True (ii) False (iii) False (iv) False.)

Check it with numbers

Magnitudes by the formula x2+y2+z2\sqrt{x^2 + y^2 + z^2}⁠: ∣i^∣=1=1|\hat i| = \sqrt{1} = 1⁠, ∣2i^∣=4=2|2\hat i| = \sqrt{4} = 2⁠, ∣−i^∣=(−1)2=1|-\hat i| = \sqrt{(-1)^2} = 1⁠, ∣j^∣=1=1|\hat j| = \sqrt{1} = 1⁠.

Collinearity by ratios of components: for 2i^2\hat i and i^\hat i the only nonzero components give 21=2\frac{2}{1} = 2 (so λ=2\lambda = 2⁠); for −i^-\hat i and i^\hat i⁠, −11=−1\frac{-1}{1} = -1 (so λ=−1\lambda = -1⁠). For j^\hat j and i^\hat i the x components are 0 and 1 and the y components are 1 and 0, so no single λ\lambda works.

Common mistakes
  1. Thinking collinear means "same direction". Collinear vectors may point in opposite directions, as a⃗\vec a and −a⃗-\vec a do.
  2. Thinking equal length means equal or collinear. Length alone fixes neither the line nor the direction.
  3. Answering "false" without a reason. Give one explicit pair of vectors for which the statement fails.
What to take away
  1. Equal vectors are collinear and have equal magnitudes, but neither property alone, nor both together, makes vectors equal.
  2. Collinear vectors are scalar multiples of each other: b⃗=λa⃗\vec b = \lambda\vec a⁠.
  3. i^\hat i⁠, 2i^2\hat i⁠, −i^-\hat i and j^\hat j are a handy kit of counterexamples.

Every fact used, and where it comes from

FactUsed inSource
Collinear vectors; equal vectors; negative of a vectorParts (i) to (iv)Section 10.3 (Lesson: six kinds of vectors)
λa⃗\lambda\vec a is collinear to a⃗\vec a⁠; ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|⁠; λ=−1\lambda = -1 gives −a⃗-\vec aParts (i), (ii), (iv)Section 10.5
a⃗\vec a⁠, b⃗\vec b collinear if and only if b⃗=λa⃗\vec b = \lambda\vec a⁠, λ≠0\lambda \ne 0Parts (i), (iii)Remark (i), Section 10.5.1
Equal vectors have equal components; i^\hat i⁠, j^\hat j are unit vectors along OX, OYParts (iii), (iv)Section 10.5.1
Magnitude ∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}CheckSection 10.5.1 (Basics: Pythagoras twice)
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