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Chapter 10 · Vector Algebra

Miscellaneous · Q17 (MCQ): Angle making a⃗+b⃗\vec a+\vec b a unit vector, for unit vectors a⃗\vec a⁠, b⃗\vec b

Miscellaneous ExerciseLength of a sum, scalar product · Section 10.6.1
The question

Let a⃗\vec a and b⃗\vec b be two unit vectors and θ\theta is the angle between them. Then a⃗+b⃗\vec a + \vec b is a unit vector if

(A) θ=π4\theta = \frac{\pi}{4}

(B) θ=π3\theta = \frac{\pi}{3}

(C) θ=π2\theta = \frac{\pi}{2}

(D) θ=2π3\theta = \frac{2\pi}{3}

Exam answer ∣a⃗+b⃗∣2=∣a⃗∣2+∣b⃗∣2+2 a⃗⋅b⃗=1+1+2cos⁡θ\begin{aligned} |\vec a + \vec b|^2 &= |\vec a|^2 + |\vec b|^2 + 2\,\vec a\cdot\vec b \\ &= 1 + 1 + 2\cos\theta \end{aligned}

∣a⃗+b⃗∣=1  ⟺  2+2cos⁡θ=1|\vec a + \vec b| = 1 \iff 2 + 2\cos\theta = 1   ⟺  cos⁡θ=−12  ⟺  θ=2π3\iff \cos\theta = -\frac12 \iff \theta = \frac{2\pi}{3}⁠.

Answer: (D) θ=2π3\theta = \dfrac{2\pi}{3}

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

The one idea this problem is built on

The length of a sum is found from its square: ∣a⃗+b⃗∣2=∣a⃗∣2+∣b⃗∣2+2 a⃗⋅b⃗|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2 + 2\,\vec a\cdot\vec b⁠. For unit vectors this is 2+2cos⁡θ2 + 2\cos\theta⁠, and setting it equal to 1 fixes cos⁡θ\cos\theta⁠, and so θ\theta⁠.

Given, to find, and the plan
Given
∣a⃗∣=∣b⃗∣=1|\vec a| = |\vec b| = 1⁠, and θ\theta is the angle between a⃗\vec a and b⃗\vec b⁠, 0≤θ≤π0 \le \theta \le \pi⁠.
To find
The value of θ\theta⁠, among (A) to (D), for which ∣a⃗+b⃗∣=1|\vec a + \vec b| = 1⁠.
Plan
(1) Express ∣a⃗+b⃗∣2|\vec a + \vec b|^2 in terms of θ\theta⁠. (2) Set it equal to 1 and solve on [0,π][0, \pi]⁠. (3) Examine every option. (4) See the picture.

Basics you need

1Length of a sum

∣a⃗+b⃗∣2=(a⃗+b⃗)⋅(a⃗+b⃗)|\vec a + \vec b|^2 = (\vec a + \vec b)\cdot(\vec a + \vec b) =∣a⃗∣2+2 a⃗⋅b⃗+∣b⃗∣2= |\vec a|^2 + 2\,\vec a\cdot\vec b + |\vec b|^2 (Section 10.6.1, Property 1, Observations 3 and 7). Basics: dot products obey the same rule

2Scalar product of unit vectors

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta = \cos\theta when ∣a⃗∣=∣b⃗∣=1|\vec a| = |\vec b| = 1 (Section 10.6.1, Definition 2). Lesson: the dot product

3The angle with a given cosine

On [0,π][0, \pi] the cosine takes each value once; cos⁡2π3=−12\cos\frac{2\pi}{3} = -\frac12 (school trigonometry). Basics: cos θ is a shadow

Solution

  1. The square of the length of a⃗+b⃗\vec a + \vec b

    ∣a⃗+b⃗∣2=(a⃗+b⃗)⋅(a⃗+b⃗)=∣a⃗∣2+2 a⃗⋅b⃗+∣b⃗∣2=1+2cos⁡θ+1=2+2cos⁡θ.\begin{aligned} |\vec a + \vec b|^2 &= (\vec a + \vec b)\cdot(\vec a + \vec b) \\ &= |\vec a|^2 + 2\,\vec a\cdot\vec b + |\vec b|^2 \\ &= 1 + 2\cos\theta + 1 = 2 + 2\cos\theta. \end{aligned}
    A length is found most easily through its square, which is a scalar product of the vector with itself. Expanding uses distributivity and commutativity; the unit lengths turn ∣a⃗∣2|\vec a|^2 and ∣b⃗∣2|\vec b|^2 into 1 and a⃗⋅b⃗\vec a\cdot\vec b into cos⁡θ\cos\theta⁠.
    Section 10.6.1, Property 1, Observations 3 and 7, Definition 2. Basics: dot products obey the same rule

  2. Set it equal to 1 and solve

    2+2cos⁡θ=1cos⁡θ=−12θ=2π3.\begin{aligned} 2 + 2\cos\theta &= 1 \\ \cos\theta &= -\frac12 \\ \theta &= \frac{2\pi}{3}. \end{aligned}
    ∣a⃗+b⃗∣=1|\vec a + \vec b| = 1 is the same as ∣a⃗+b⃗∣2=1|\vec a + \vec b|^2 = 1⁠, since lengths are not negative. The angle lies in [0,π][0, \pi]⁠, where 2π3\frac{2\pi}{3} is the only angle with cosine −12-\frac12⁠. So the condition is both necessary and sufficient: θ=2π3\theta = \frac{2\pi}{3}.
    cos⁡2π3=−12\cos\frac{2\pi}{3} = -\frac12 (school trigonometry). Basics: cos θ is a shadow

  3. Examine every option

    (A) θ=π4: ∣a⃗+b⃗∣2=2+2,(B) θ=π3: ∣a⃗+b⃗∣2=2+1=3,(C) θ=π2: ∣a⃗+b⃗∣2=2+0=2,(D) θ=2π3: ∣a⃗+b⃗∣2=2−1=1.\begin{aligned} \text{(A) } \theta = \tfrac{\pi}{4}&:\ |\vec a + \vec b|^2 = 2 + \sqrt2, \\ \text{(B) } \theta = \tfrac{\pi}{3}&:\ |\vec a + \vec b|^2 = 2 + 1 = 3, \\ \text{(C) } \theta = \tfrac{\pi}{2}&:\ |\vec a + \vec b|^2 = 2 + 0 = 2, \\ \text{(D) } \theta = \tfrac{2\pi}{3}&:\ |\vec a + \vec b|^2 = 2 - 1 = 1. \end{aligned}

    Only (D) gives length 1; (A), (B) and (C) give 2+2\sqrt{2 + \sqrt2}⁠, 3\sqrt3 and 2\sqrt2⁠, all more than 1.

    Each option puts a value of cos⁡θ\cos\theta into 2+2cos⁡θ2 + 2\cos\theta⁠: cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt2} gives 2+22=2+22 + \frac{2}{\sqrt2} = 2 + \sqrt2⁠, cos⁡π3=12\cos\frac{\pi}{3} = \frac12⁠, cos⁡π2=0\cos\frac{\pi}{2} = 0 and cos⁡2π3=−12\cos\frac{2\pi}{3} = -\frac12⁠. For any acute or right angle the sum is at least 2\sqrt2 long; it can shrink to 1 only when the angle is obtuse.
    Section 10.6.1. Lesson: the multiple-choice questions

  4. The picture: two equilateral triangles

    Put b⃗\vec b at the tip of a⃗\vec a⁠. The triangle with sides a⃗\vec a⁠, b⃗\vec b and a⃗+b⃗\vec a + \vec b has all three sides 1, so it is equilateral. Its angle between the sides a⃗\vec a and b⃗\vec b (head to tail) is 60∘60^\circ⁠, and the angle between the vectors a⃗\vec a and b⃗\vec b placed tail to tail is 180∘−60∘=120∘=2π3180^\circ - 60^\circ = 120^\circ = \frac{2\pi}{3}⁠.

    O a b a + b 120° |a| = |b| = 1 |a + b| = 1 angle 120° = 2π/3
    The rhombus on the unit vectors a⃗\vec a and b⃗\vec b at 120∘120^\circ⁠, drawn with one scale. Its short diagonal a⃗+b⃗\vec a + \vec b also has length 1 and splits the rhombus into two equilateral triangles.
    The parallelogram on two equal vectors is a rhombus, and its diagonal from the common tail is a⃗+b⃗\vec a + \vec b⁠. When the angle is 120∘120^\circ⁠, the diagonal is exactly as long as the sides.
    Parallelogram law and triangle law (Section 10.4). Lesson: adding arrows, head to tail
Result

∣a⃗+b⃗∣2=2+2cos⁡θ=1|\vec a + \vec b|^2 = 2 + 2\cos\theta = 1 gives cos⁡θ=−12\cos\theta = -\frac12⁠, so a⃗+b⃗\vec a + \vec b is a unit vector when (D) θ=2π3\theta = \frac{2\pi}{3}⁠.

(Textbook answer: (D).)

Check it with numbers

a⃗=i^\vec a = \hat i⁠, b⃗=−12i^+32j^\vec b = -\frac12\hat i + \frac{\sqrt3}{2}\hat j⁠: both have length 1, a⃗⋅b⃗=−12\vec a\cdot\vec b = -\frac12⁠, and a⃗+b⃗=12i^+32j^\vec a + \vec b = \frac12\hat i + \frac{\sqrt3}{2}\hat j has length 14+34=1\sqrt{\frac14 + \frac34} = 1⁠. In decimals, 32≈0.866025\frac{\sqrt3}{2} \approx 0.866025⁠.

The wrong options: 2+2≈1.847759\sqrt{2 + \sqrt2} \approx 1.847759⁠, 3≈1.732051\sqrt3 \approx 1.732051⁠, 2≈1.414214\sqrt2 \approx 1.414214⁠.

Common mistakes
  1. Choosing (B), π3\frac{\pi}{3}⁠. 60∘60^\circ is the angle inside the equilateral triangle, between a⃗\vec a and b⃗\vec b placed head to tail; the angle between the vectors is measured tail to tail, 120∘120^\circ⁠.
  2. Adding lengths. ∣a⃗+b⃗∣|\vec a + \vec b| is not ∣a⃗∣+∣b⃗∣=2|\vec a| + |\vec b| = 2 unless θ=0\theta = 0⁠.
  3. Dropping the 2 in 2cos⁡θ2\cos\theta⁠. That gives cos⁡θ=−1\cos\theta = -1 and θ=π\theta = \pi⁠, where a⃗+b⃗=0⃗\vec a + \vec b = \vec 0⁠.
What to take away
  1. For unit vectors, ∣a⃗+b⃗∣2=2+2cos⁡θ|\vec a + \vec b|^2 = 2 + 2\cos\theta and ∣a⃗−b⃗∣2=2−2cos⁡θ|\vec a - \vec b|^2 = 2 - 2\cos\theta⁠.
  2. Find a length through its square.
  3. The angle between vectors is measured tail to tail.

Every fact used, and where it comes from

FactUsed inSource
Distributivity; x⃗⋅y⃗=y⃗⋅x⃗\vec x\cdot\vec y = \vec y\cdot\vec x⁠; x⃗⋅x⃗=∣x⃗∣2\vec x\cdot\vec x = |\vec x|^2Step 1Section 10.6.1, Property 1, Observations 3 and 7 (Basics: dot products obey the same rule)
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, 2Section 10.6.1, Definition 2
Values of cos⁡\cos at π4,π3,π2,2π3\frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}Steps 2, 3School trigonometry (Basics: cos θ is a shadow)
Triangle and parallelogram lawsStep 4Section 10.4 (Lesson: adding arrows, head to tail)
The answerResultNCERT answer key, Miscellaneous Exercise Q17 (Lesson: the multiple-choice questions)
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