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Chapter 10 · Vector Algebra

Exercise 10.4 · Q11 (MCQ): Angle making a⃗×b⃗\vec a\times\vec b a unit vector, ∣a⃗∣=3|\vec a|=3⁠, ∣b⃗∣=23|\vec b|=\frac{\sqrt2}{3}

Exercise 10.4Length of the vector product · Section 10.6.3
The question

Let the vectors a⃗\vec a and b⃗\vec b be such that ∣a⃗∣=3|\vec a| = 3 and ∣b⃗∣=23|\vec b| = \frac{\sqrt2}{3}⁠, then a⃗×b⃗\vec a\times\vec b is a unit vector, if the angle between a⃗\vec a and b⃗\vec b is

(A) π/6\pi/6

(B) π/4\pi/4

(C) π/3\pi/3

(D) π/2\pi/2

Exam answer ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ=3⋅23sin⁡θ=2sin⁡θ\begin{aligned} |\vec a\times\vec b| &= |\vec a||\vec b|\sin\theta \\ &= 3 \cdot \frac{\sqrt2}{3}\sin\theta = \sqrt2\sin\theta \end{aligned}

For a unit vector, 2sin⁡θ=1\sqrt2\sin\theta = 1⁠, so sin⁡θ=12\sin\theta = \frac{1}{\sqrt2} and θ=π4\theta = \frac{\pi}{4}⁠.

Answer: (B) π/4\pi/4

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

The one idea this problem is built on

The length of the vector product is

∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ.|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta.

"Unit vector" means this length is 1, which fixes sin⁡θ\sin\theta⁠.

Given, to find, and the plan
Given
∣a⃗∣=3|\vec a| = 3⁠, ∣b⃗∣=23|\vec b| = \frac{\sqrt2}{3}⁠, and a⃗×b⃗\vec a\times\vec b is a unit vector.
To find
The angle θ\theta between a⃗\vec a and b⃗\vec b⁠, among the options.
Plan
(1) Write ∣a⃗×b⃗∣|\vec a\times\vec b| in terms of θ\theta⁠. (2) Set it equal to 1 and solve for θ\theta in [0,π][0, \pi]⁠. (3) Test every option.

Basics you need

1The length of a⃗×b⃗\vec a\times\vec b

a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θ n^\vec a\times\vec b = |\vec a||\vec b|\sin\theta\,\hat n with ∣n^∣=1|\hat n| = 1 and 0≤θ≤π0 \le \theta \le \pi⁠, so ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta (Definition 3, Section 10.6.3). Lesson: the cross product

2Sines of special angles

sin⁡π6=12\sin\frac{\pi}{6} = \frac12⁠, sin⁡π4=12\sin\frac{\pi}{4} = \frac{1}{\sqrt2}⁠, sin⁡π3=32\sin\frac{\pi}{3} = \frac{\sqrt3}{2}⁠, sin⁡π2=1\sin\frac{\pi}{2} = 1⁠. Basics: cos θ is a shadow, sin θ is a height

Solution

  1. The length of a⃗×b⃗\vec a\times\vec b

    ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ=3⋅23sin⁡θ=2sin⁡θ.\begin{aligned} |\vec a\times\vec b| &= |\vec a||\vec b|\sin\theta \\ &= 3 \cdot \frac{\sqrt2}{3}\sin\theta = \sqrt2\sin\theta. \end{aligned}
    n^\hat n is a unit vector and sin⁡θ≥0\sin\theta \ge 0 on [0,π][0, \pi]⁠, so the length of ∣a⃗∣∣b⃗∣sin⁡θ n^|\vec a||\vec b|\sin\theta\,\hat n is the number ∣a⃗∣∣b⃗∣sin⁡θ|\vec a||\vec b|\sin\theta⁠. The 3s cancel.
    Definition 3, Section 10.6.3. Lesson: the cross product

  2. Set the length equal to 1

    2sin⁡θ=1  ⇒  sin⁡θ=12.\sqrt2\sin\theta = 1 \;\Rightarrow\; \sin\theta = \frac{1}{\sqrt2}.

    On 0≤θ≤π0 \le \theta \le \pi this gives θ=π4\theta = \frac{\pi}{4} or θ=3π4\theta = \frac{3\pi}{4}⁠. Only π4\frac{\pi}{4} is among the options.

    A unit vector has length 1. The sine takes the value 12\frac{1}{\sqrt2} twice on [0,π][0, \pi]⁠, at π4\frac{\pi}{4} and at π−π4=3π4\pi - \frac{\pi}{4} = \frac{3\pi}{4}⁠; both make a⃗×b⃗\vec a\times\vec b a unit vector, but the options offer only the first.
    sin⁡π4=12\sin\frac{\pi}{4} = \frac{1}{\sqrt2}⁠; sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin x⁠. Basics: cos θ is a shadow, sin θ is a height

  3. Test every option

    Optionsin⁡θ\sin\theta∣a⃗×b⃗∣=2sin⁡θ|\vec a\times\vec b| = \sqrt2\sin\thetaUnit?
    (A) π/6\pi/612\frac1222=12\frac{\sqrt2}{2} = \frac{1}{\sqrt2}No
    (B) π/4\pi/412\frac{1}{\sqrt2}1Yes
    (C) π/3\pi/332\frac{\sqrt3}{2}62\frac{\sqrt6}{2}No
    (D) π/2\pi/212\sqrt2No
    Only (B) gives length exactly 1. (A) gives a vector shorter than 1, while (C) and (D) give vectors longer than 1. Note (D) is the angle that makes ∣a⃗×b⃗∣|\vec a\times\vec b| largest; a unit vector does not need the vectors to be perpendicular.
    Values of sine at special angles. Lesson: the self-check quiz
Result

The answer is (B), θ=π4\theta = \frac{\pi}{4}⁠. (Textbook answer: (B).)

Check it with numbers

sin⁡π4=12≈0.707107\sin\frac{\pi}{4} = \frac{1}{\sqrt2} \approx 0.707107 and 2≈1.414214\sqrt2 \approx 1.414214⁠; their product is exactly 1.

The other options: 2×12≈0.707107\sqrt2 \times \frac12 \approx 0.707107⁠, 2×32=62≈1.224745\sqrt2 \times \frac{\sqrt3}{2} = \frac{\sqrt6}{2} \approx 1.224745⁠, and 2×1≈1.414214\sqrt2 \times 1 \approx 1.414214⁠; none is 1.

Common mistakes
  1. Using cos⁡θ\cos\theta⁠. ∣a⃗∣∣b⃗∣cos⁡θ|\vec a||\vec b|\cos\theta is the scalar product, a number, not the length of a⃗×b⃗\vec a\times\vec b⁠. Here 2cos⁡θ=1\sqrt2\cos\theta = 1 happens to give π4\frac{\pi}{4} too, but only because sin⁡π4=cos⁡π4\sin\frac{\pi}{4} = \cos\frac{\pi}{4}⁠; with other data the two disagree.
  2. Thinking a unit vector product needs θ=π2\theta = \frac{\pi}{2}⁠. At π2\frac{\pi}{2} the length is ∣a⃗∣∣b⃗∣=2|\vec a||\vec b| = \sqrt2⁠, not 1.
  3. Forgetting that 3π4\frac{3\pi}{4} also works. It is not an option here, but in general sin⁡θ=12\sin\theta = \frac{1}{\sqrt2} has two solutions in [0,π][0, \pi]⁠.
What to take away
  1. ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta⁠.
  2. "a⃗×b⃗\vec a\times\vec b is a unit vector" means ∣a⃗∣∣b⃗∣sin⁡θ=1|\vec a||\vec b|\sin\theta = 1⁠.
  3. On [0,π][0, \pi]⁠, each sine value below 1 comes from two angles, θ\theta and π−θ\pi - \theta⁠.

Every fact used, and where it comes from

FactUsed inSource
a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θ n^\vec a\times\vec b = |\vec a||\vec b|\sin\theta\,\hat n⁠, 0≤θ≤π0 \le \theta \le \piStep 1Section 10.6.3, Definition 3 (Lesson: the cross product)
Unit vector: length 1Step 2Section 10.3
Sines of π6,π4,π3,π2\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}⁠; sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin xSteps 2, 3Trigonometry (Basics: cos θ is a shadow)
The question and the answerResultExercise 10.4 Q11; NCERT answer key (Lesson: the self-check quiz)
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