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Chapter 10 · Vector Algebra

Example 13: Angle between a⃗\vec a⁠, b⃗\vec b with ∣a⃗∣=1|\vec a|=1⁠, ∣b⃗∣=2|\vec b|=2⁠, a⃗⋅b⃗=1\vec a\cdot\vec b=1

ExamplesAngle between two vectors · Section 10.6.1
The question

Find the angle between two vectors a⃗\vec a and b⃗\vec b with magnitudes 1 and 2 respectively and when a⃗⋅b⃗=1\vec a\cdot\vec b = 1⁠.

Exam answer

Given a⃗⋅b⃗=1\vec a\cdot\vec b = 1⁠, ∣a⃗∣=1|\vec a| = 1 and ∣b⃗∣=2|\vec b| = 2⁠. Let θ\theta be the angle between a⃗\vec a and b⃗\vec b⁠. Then

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=11×2=12\begin{aligned} \cos\theta &= \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} \\ &= \frac{1}{1 \times 2} = \frac12 \end{aligned}

Since 0≤θ≤π0 \le \theta \le \pi⁠,

θ=cos⁡−1(12)\theta = \cos^{-1}\Big(\frac12\Big)

∴ θ=π3\therefore\ \theta = \dfrac{\pi}{3}⁠.

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

The one idea this problem is built on

The scalar product is defined by a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta⁠, where θ\theta is the angle between the vectors, 0≤θ≤π0 \le \theta \le \pi⁠. Knowing the scalar product and both magnitudes, we can solve for cos⁡θ\cos\theta:

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣.\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}.

Because θ\theta is restricted to [0,π][0, \pi]⁠, each value of cos⁡θ\cos\theta gives exactly one angle.

Given, to find, and the plan
Given
∣a⃗∣=1|\vec a| = 1⁠, ∣b⃗∣=2|\vec b| = 2⁠, a⃗⋅b⃗=1\vec a\cdot\vec b = 1⁠.
To find
The angle θ\theta between a⃗\vec a and b⃗\vec b⁠.
Plan
(1) Write cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠. (2) Substitute. (3) Find the one angle in [0,π][0, \pi] with that cosine. (4) Check by putting θ\theta back into the definition.

Basics you need

1The scalar product

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta⁠, with θ\theta the angle between a⃗\vec a and b⃗\vec b⁠, 0≤θ≤π0 \le \theta \le \pi (Section 10.6.1, Definition 2, Fig 10.19). Lesson: the dot product

2The angle formula

For nonzero vectors, cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠, or θ=cos⁡−1(a⃗⋅b⃗∣a⃗∣∣b⃗∣)\theta = \cos^{-1}\left(\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}\right) (Section 10.6.1, Observation 6). Lesson: Examples 13–21

3A standard cosine

cos⁡π3=12\cos\frac{\pi}{3} = \frac12⁠, and cos⁡\cos takes each value in [−1,1][-1, 1] exactly once on [0,π][0, \pi]⁠. Basics: cos θ is a shadow

Solution

  1. Write the angle formula

    Both vectors are nonzero (their magnitudes are 1 and 2), so the angle between them is defined and

    cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣.\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}.
    Dividing the definition a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta by ∣a⃗∣∣b⃗∣|\vec a||\vec b| isolates cos⁡θ\cos\theta⁠. The division is allowed because ∣a⃗∣∣b⃗∣=2≠0|\vec a||\vec b| = 2 \ne 0⁠; for a zero vector the angle would not be defined.
    Definition 2 and Observation 6 (Section 10.6.1). Lesson: the dot product

  2. Substitute the given values

    cos⁡θ=11×2=12.\cos\theta = \frac{1}{1 \times 2} = \frac12.
    Only three numbers are needed: the scalar product 1 on top, and the product of the magnitudes 1×2=21 \times 2 = 2 underneath. The components of the vectors are never needed.
    cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} (Section 10.6.1, Observation 6). Lesson: Examples 13–21

  3. Find the angle in [0,π][0, \pi]

    θ=cos⁡−1(12)=π3.\theta = \cos^{-1}\Big(\frac12\Big) = \frac{\pi}{3}.

    So θ=π3\theta = \frac{\pi}{3}, that is 60∘60^\circ⁠.

    l 60° O a b, length 2 shadow of b: 2 cos 60° = 1 a · b = |a| × shadow = 1 × 1 = 1 |a| = 1, |b| = 2
    Drawn to scale, one scale on both axes: a⃗\vec a of length 1, b⃗\vec b of length 2 at θ=60∘\theta = 60^\circ to it. The perpendicular from the tip of b⃗\vec b meets the line of a⃗\vec a at the tip of a⃗\vec a⁠: the shadow of b⃗\vec b on a⃗\vec a has length ∣b⃗∣cos⁡θ=1|\vec b|\cos\theta = 1⁠, and a⃗⋅b⃗=∣a⃗∣×1=1\vec a\cdot\vec b = |\vec a| \times 1 = 1⁠.
    The angle between two vectors is taken in [0,π][0, \pi]⁠, which is also the principal range of cos⁡−1\cos^{-1}⁠. On that interval cosine decreases from 1 to −1-1⁠, so it takes the value 12\frac12 exactly once, at π3\frac{\pi}{3}⁠. Angles such as −π3-\frac{\pi}{3} or 5π3\frac{5\pi}{3} have the same cosine but are not in [0,π][0, \pi]⁠.
    0≤θ≤π0 \le \theta \le \pi (Section 10.6.1, Definition 2); cos⁡π3=12\cos\frac{\pi}{3} = \frac12⁠. Basics: cos θ is a shadow

  4. Check in the definition

    ∣a⃗∣∣b⃗∣cos⁡π3=1×2×12=1=a⃗⋅b⃗.|\vec a||\vec b|\cos\frac{\pi}{3} = 1 \times 2 \times \frac12 = 1 = \vec a\cdot\vec b.
    Putting the angle back into the definition must return the given scalar product. The figure shows the same fact: ∣b⃗∣cos⁡θ=2×12=1|\vec b|\cos\theta = 2 \times \frac12 = 1 is the shadow of b⃗\vec b on the line of a⃗\vec a⁠, and the scalar product is ∣a⃗∣|\vec a| times that shadow.
    a⃗⋅b⃗=∣a⃗∣ (∣b⃗∣cos⁡θ)\vec a\cdot\vec b = |\vec a|\,(|\vec b|\cos\theta)⁠: length of a⃗\vec a times the projection of b⃗\vec b on a⃗\vec a (Sections 10.6.1 and 10.6.2). Lesson: the dot product
Result

The angle between a⃗\vec a and b⃗\vec b is θ=π3\theta = \frac{\pi}{3} (60∘60^\circ⁠).

(Textbook answer: θ=cos⁡−1(a⃗⋅b⃗∣a⃗∣∣b⃗∣)=cos⁡−1(12)=π3\theta = \cos^{-1}\left(\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}\right) = \cos^{-1}\left(\frac12\right) = \frac{\pi}{3}⁠.)

Check it with numbers

π3≈1.047198\frac{\pi}{3} \approx 1.047198 radians, and cos⁡(1.047198)≈0.500000\cos(1.047198) \approx 0.500000⁠.

A concrete pair with these data: a⃗=i^\vec a = \hat i and b⃗=i^+3 j^\vec b = \hat i + \sqrt3\,\hat j⁠. Then ∣a⃗∣=1|\vec a| = 1⁠, ∣b⃗∣=1+3=2|\vec b| = \sqrt{1 + 3} = 2 and a⃗⋅b⃗=1\vec a\cdot\vec b = 1⁠, and the angle of b⃗\vec b with the x-axis is tan⁡−13=60∘\tan^{-1}\sqrt3 = 60^\circ⁠.

Common mistakes
  1. Forgetting to divide by both magnitudes. cos⁡θ=a⃗⋅b⃗∣a⃗∣=1\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a|} = 1 would give θ=0\theta = 0⁠; the denominator is ∣a⃗∣∣b⃗∣=2|\vec a||\vec b| = 2⁠.
  2. Giving an angle outside [0,π][0, \pi]⁠. 5π3\frac{5\pi}{3} also has cosine 12\frac12⁠, but the angle between two vectors is never more than π\pi⁠.
  3. Using sine instead of cosine. sin⁡θ=12\sin\theta = \frac12 would give π6\frac{\pi}{6}⁠; the scalar product carries cos⁡θ\cos\theta⁠.
What to take away
  1. cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} needs only the scalar product and the two lengths.
  2. The angle between two vectors lies in [0,π][0, \pi]⁠, so θ=cos⁡−1(…)\theta = \cos^{-1}(\ldots) is the answer, with no other choices.
  3. a⃗⋅b⃗\vec a\cdot\vec b is the length of a⃗\vec a times the shadow of b⃗\vec b on a⃗\vec a⁠.

Every fact used, and where it comes from

FactUsed inSource
a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta⁠, 0≤θ≤π0 \le \theta \le \piSteps 1, 3, 4Section 10.6.1, Definition 2, Fig 10.19 (Lesson: the dot product)
cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}Steps 1, 2Section 10.6.1, Observation 6
cos⁡π3=12\cos\frac{\pi}{3} = \frac12⁠; principal range of cos⁡−1\cos^{-1} is [0,π][0, \pi]Step 3Trigonometry (Basics: cos θ is a shadow)
Projection of b⃗\vec b on a⃗\vec a is ∣b⃗∣cos⁡θ|\vec b|\cos\thetaStep 4Section 10.6.2
The book's working and answerResultExample 13, Section 10.6.2 (Lesson: Examples 13–21)
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