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Chapter 10 · Vector Algebra

Exercise 10.2 · Q11: 2i^−3j^+4k^2\hat i-3\hat j+4\hat k and −4i^+6j^−8k^-4\hat i+6\hat j-8\hat k are collinear

Exercise 10.2Collinear vectors · Sections 10.3, 10.5 and 10.5.1
The question

Show that the vectors 2i^−3j^+4k^2\hat i - 3\hat j + 4\hat k and −4i^+6j^−8k^-4\hat i + 6\hat j - 8\hat k are collinear.

Exam answer

Let a⃗=2i^−3j^+4k^\vec a = 2\hat i - 3\hat j + 4\hat k and b⃗=−4i^+6j^−8k^\vec b = -4\hat i + 6\hat j - 8\hat k⁠.

b⃗=−4i^+6j^−8k^=−2(2i^−3j^+4k^)=−2a⃗\begin{aligned} \vec b &= -4\hat i + 6\hat j - 8\hat k \\ &= -2(2\hat i - 3\hat j + 4\hat k) \\ &= -2\vec a \end{aligned}

So b⃗=λa⃗\vec b = \lambda\vec a with λ=−2\lambda = -2⁠, a nonzero scalar.

Hence the given vectors are collinear.

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

The one idea this problem is built on

Two vectors are collinear exactly when one is a nonzero number times the other: b⃗=λa⃗\vec b = \lambda\vec a⁠. In components this means the ratios b1a1,b2a2,b3a3\frac{b_1}{a_1}, \frac{b_2}{a_2}, \frac{b_3}{a_3} are all the same number λ\lambda⁠. So we compute the three ratios and check that they agree.

Given, to find, and the plan
Given
a⃗=2i^−3j^+4k^\vec a = 2\hat i - 3\hat j + 4\hat k and b⃗=−4i^+6j^−8k^\vec b = -4\hat i + 6\hat j - 8\hat k⁠.
To show
a⃗\vec a and b⃗\vec b are collinear.
Plan
(1) Recall the test b⃗=λa⃗\vec b = \lambda\vec a⁠. (2) Compute the three ratios of components. (3) Conclude, and read off what λ=−2\lambda = -2 says about direction and length.

Basics you need

1Collinear vectors

Two or more vectors are collinear if they are parallel to the same line, whatever their magnitudes and directions (Section 10.3). Lesson: six kinds of vectors worth naming

2The test in components

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, that is, b1=λa1b_1 = \lambda a_1⁠, b2=λa2b_2 = \lambda a_2⁠, b3=λa3b_3 = \lambda a_3 (Remarks (i), Section 10.5.1). Lesson: components

3What the sign of λ\lambda says

λa⃗\lambda\vec a has the same direction as a⃗\vec a if λ>0\lambda > 0 and the opposite direction if λ<0\lambda < 0⁠, and ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a| (Section 10.5). Lesson: components

Solution

  1. The test for collinear vectors

    We must find one scalar λ≠0\lambda \ne 0 with b⃗=λa⃗\vec b = \lambda\vec a⁠, which in components means

    −4=2λ,6=−3λ,−8=4λ.\begin{aligned} -4 &= 2\lambda, \\ 6 &= -3\lambda, \\ -8 &= 4\lambda. \end{aligned}
    "Parallel to the same line" is a picture; b⃗=λa⃗\vec b = \lambda\vec a turns it into something we can compute. Multiplying a⃗\vec a by λ\lambda stretches or flips it along its own line, and every vector along that line arises this way. One and the same λ\lambda must work for all three components.
    Collinear if and only if b⃗=λa⃗\vec b = \lambda\vec a⁠, λ≠0\lambda \ne 0 (Remarks (i), Section 10.5.1). Lesson: components

  2. Compute the three ratios

    −42=−2,6−3=−2,−84=−2.\begin{aligned} \frac{-4}{2} &= -2, \\ \frac{6}{-3} &= -2, \\ \frac{-8}{4} &= -2. \end{aligned}

    All three ratios equal −2-2⁠, so λ=−2\lambda = -2 works for every component.

    Each equation of Step 1 gives λ\lambda as a ratio bkak\frac{b_k}{a_k} (no aka_k is 0, so we may divide). If even one ratio were different, no single λ\lambda would exist and the vectors would not be collinear.
    b1a1=b2a2=b3a3=λ\frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} = \lambda (Remarks (i), Section 10.5.1). Lesson: components

  3. Conclude, and read the meaning of λ=−2\lambda = -2

    −4i^+6j^−8k^=−2(2i^−3j^+4k^),that is,b⃗=−2a⃗.\begin{aligned} -4\hat i + 6\hat j - 8\hat k &= -2(2\hat i - 3\hat j + 4\hat k), \\ \text{that is,}\quad \vec b &= -2\vec a. \end{aligned}

    Since λ=−2≠0\lambda = -2 \ne 0⁠, a⃗\vec a and b⃗\vec b are collinear.

    O a, length √29 b = −2a, length 2√29 b = λa with λ = −2: one line, opposite ways
    a⃗\vec a and b⃗=−2a⃗\vec b = -2\vec a drawn from one point O with one scale. They lie on one line; b⃗\vec b points the opposite way and is twice as long (∣a⃗∣=29|\vec a| = \sqrt{29}⁠, ∣b⃗∣=229|\vec b| = 2\sqrt{29}⁠).
    The negative λ\lambda means b⃗\vec b points opposite to a⃗\vec a⁠, and ∣λ∣=2|\lambda| = 2 means it is twice as long. Collinear vectors may differ in both direction and magnitude; they only have to lie along the same line.
    λa⃗\lambda\vec a points opposite to a⃗\vec a for λ<0\lambda < 0⁠, and ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a| (Section 10.5). Lesson: six kinds of vectors worth naming
Result

−4i^+6j^−8k^=−2(2i^−3j^+4k^)-4\hat i + 6\hat j - 8\hat k = -2(2\hat i - 3\hat j + 4\hat k)⁠, so the vectors are collinear (with λ=−2\lambda = -2⁠: opposite directions, the second twice as long). Hence proved.

(Textbook answer: the answer key has no entry for this "show that" question.)

Check it with numbers

∣a⃗∣=4+9+16=29≈5.385165|\vec a| = \sqrt{4 + 9 + 16} = \sqrt{29} \approx 5.385165 and ∣b⃗∣=16+36+64=116=229≈10.770330|\vec b| = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29} \approx 10.770330⁠, exactly twice as long, as ∣λ∣=2|\lambda| = 2 says.

Adding the vectors gives a⃗+b⃗=−2i^+3j^−4k^=−a⃗\vec a + \vec b = -2\hat i + 3\hat j - 4\hat k = -\vec a⁠, again a multiple of a⃗\vec a⁠, as it must be for vectors on one line.

Common mistakes
  1. Thinking collinear vectors must point the same way. λ\lambda may be negative; here it is −2-2⁠.
  2. Checking only one or two ratios. All three must agree. For example, 2i^−3j^+4k^2\hat i - 3\hat j + 4\hat k and −4i^+6j^+8k^-4\hat i + 6\hat j + 8\hat k have ratios −2,−2,2-2, -2, 2 and are not collinear.
  3. Mixing up the order of a ratio. Take bkak\frac{b_k}{a_k} every time (or akbk\frac{a_k}{b_k} every time); mixing them gives −2-2 and −12-\frac12 and a false "not collinear".
What to take away
  1. Collinear test: b⃗=λa⃗\vec b = \lambda\vec a for one nonzero λ\lambda⁠, that is, equal ratios of components.
  2. The sign of λ\lambda gives the relative direction; ∣λ∣|\lambda| gives the ratio of the lengths.
  3. Look for a common factor first: −4,6,−8-4, 6, -8 is −2-2 times 2,−3,42, -3, 4⁠.

Every fact used, and where it comes from

FactUsed inSource
Collinear vectors: parallel to the same lineStep 1Section 10.3 (Lesson: six kinds of vectors)
Collinear if and only if b⃗=λa⃗\vec b = \lambda\vec a⁠, λ≠0\lambda \ne 0⁠; equal ratios of componentsSteps 1, 2, 3Remarks (i), Section 10.5.1
Direction of λa⃗\lambda\vec a by the sign of λ\lambda⁠; ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|Step 3Section 10.5
∣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: use Pythagoras twice)
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