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Chapter 10 · Vector Algebra

Example 26: All the unit vectors in the XY-plane: cos⁡θ i^+sin⁡θ j^\cos\theta\,\hat i+\sin\theta\,\hat j

ExamplesUnit vectors in a plane · Miscellaneous Examples
The question

Write all the unit vectors in XY-plane.

Exam answer

Let r⃗=xi^+yj^\vec r = x\hat i + y\hat j be a unit vector in the XY-plane, and let θ\theta be the angle from OX to r⃗\vec r⁠.

∣r⃗∣=1⇒x2+y2=1|\vec r| = 1 \Rightarrow x^2 + y^2 = 1

So, drawn from O, r⃗=OP→\vec r = \overrightarrow{OP} where P(x, y) lies on the circle x2+y2=1x^2 + y^2 = 1⁠. Hence x=cos⁡θx = \cos\theta and y=sin⁡θy = \sin\theta⁠.

r⃗=cos⁡θ i^+sin⁡θ j^\vec r = \cos\theta\,\hat i + \sin\theta\,\hat j ∣r⃗∣=cos⁡2θ+sin⁡2θ=1\begin{aligned} |\vec r| &= \sqrt{\cos^2\theta + \sin^2\theta} \\ &= 1 \end{aligned}

As θ\theta goes from 0 to 2π2\pi⁠, P goes once round the circle, so every direction in the XY-plane is covered.

∴\therefore the unit vectors in the XY-plane are cos⁡θ i^+sin⁡θ j^\cos\theta\,\hat i + \sin\theta\,\hat j⁠, 0≤θ<2π0 \le \theta < 2\pi⁠.

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

The one idea this problem is built on

Draw every unit vector of the XY-plane from the origin. Its tip is then at distance 1 from O, so it lies on the circle x2+y2=1x^2 + y^2 = 1⁠, and a point on that circle at angle θ\theta from OX is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta)⁠. So the unit vectors are exactly cos⁡θ i^+sin⁡θ j^\cos\theta\,\hat i + \sin\theta\,\hat j⁠, one for each angle θ\theta⁠.

Given, to find, and the plan
Given
Vectors lying in the XY-plane (no k^\hat k part) of length 1.
To find
A formula that lists all of them.
Plan
(1) Write a unit vector as xi^+yj^x\hat i + y\hat j with x2+y2=1x^2 + y^2 = 1⁠. (2) Read x=cos⁡θx = \cos\theta⁠, y=sin⁡θy = \sin\theta from the unit circle. (3) Check that every such vector has length 1. (4) Check that every unit vector of the plane appears exactly once for 0≤θ<2π0 \le \theta < 2\pi⁠.

Basics you need

1Unit vector

A vector of magnitude 1 (Section 10.3); ∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Lesson: components

2Cosine and sine on the unit circle

The point of the circle x2+y2=1x^2 + y^2 = 1 at angle θ\theta from OX, measured anticlockwise, is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta)⁠; so cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1⁠. Basics: cos θ is a shadow, sin θ is a height

3Principal value of cos⁡−1\cos^{-1}

For −1≤x≤1-1 \le x \le 1⁠, cos⁡−1x\cos^{-1}x is the angle in [0,π][0, \pi] whose cosine is xx (Chapter 2). Basics: cos θ is a shadow

Solution

  1. A unit vector in the XY-plane

    A vector in the XY-plane has no k^\hat k part, so write r⃗=xi^+yj^\vec r = x\hat i + y\hat j⁠. It is a unit vector exactly when

    ∣r⃗∣=x2+y2=1,|\vec r| = \sqrt{x^2 + y^2} = 1,

    that is,

    x2+y2=1.x^2 + y^2 = 1.
    The magnitude formula with z=0z = 0 gives x2+y2\sqrt{x^2 + y^2}⁠. Setting it equal to 1 and squaring (both sides are non-negative) gives the condition on the components.
    Unit vector: magnitude 1 (Section 10.3); magnitude from components (Section 10.5.1). Lesson: components

  2. Read the components from the unit circle

    Draw r⃗\vec r from the origin: r⃗=OP→\vec r = \overrightarrow{OP} with P(x,y)(x, y)⁠. Since x2+y2=1x^2 + y^2 = 1⁠, P is on the circle of radius 1 about O. If θ\theta is the angle from OX to OP, then

    x=cos⁡θ,y=sin⁡θ,r⃗=cos⁡θ i^+sin⁡θ j^.\begin{aligned} x &= \cos\theta, \quad y = \sin\theta, \\ \vec r &= \cos\theta\,\hat i + \sin\theta\,\hat j. \end{aligned}
    This is how cosine and sine are defined on the unit circle: the foot P′ of the perpendicular from P to OX gives OP′→=cos⁡θ i^\overrightarrow{OP'} = \cos\theta\,\hat i and P′P→=sin⁡θ j^\overrightarrow{P'P} = \sin\theta\,\hat j⁠, and OP→=OP′→+P′P→\overrightarrow{OP} = \overrightarrow{OP'} + \overrightarrow{P'P} by the triangle law (the book's Fig 10.28).
    Coordinates of a point on the unit circle. Basics: cos θ is a shadow, sin θ is a height

  3. Every such vector has length 1

    ∣cos⁡θ i^+sin⁡θ j^∣=cos⁡2θ+sin⁡2θ=1=1.\begin{aligned} |\cos\theta\,\hat i + \sin\theta\,\hat j| &= \sqrt{\cos^2\theta + \sin^2\theta} \\ &= \sqrt1 = 1. \end{aligned}
    This is the converse of Step 2: whatever angle we pick, the formula gives a unit vector in the XY-plane, so the list contains nothing extra.
    cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1⁠. Basics: cos θ is a shadow

  4. Every unit vector appears, exactly once

    Take any unit vector xi^+yj^x\hat i + y\hat j⁠, so x2+y2=1x^2 + y^2 = 1 and −1≤x≤1-1 \le x \le 1⁠. Put α=cos⁡−1x\alpha = \cos^{-1}x⁠, an angle in [0,π][0, \pi] with cos⁡α=x\cos\alpha = x and sin⁡α=1−x2=∣y∣\sin\alpha = \sqrt{1 - x^2} = |y|⁠. Then

    θ=α  if y≥0,θ=2π−α  if y<0\begin{aligned} \theta &= \alpha \ \text{ if } y \ge 0, \\ \theta &= 2\pi - \alpha \ \text{ if } y < 0 \end{aligned}

    gives cos⁡θ=x\cos\theta = x and sin⁡θ=y\sin\theta = y⁠, with 0≤θ<2π0 \le \theta < 2\pi⁠. As θ\theta runs from 0 to 2π2\pi⁠, P goes once round the circle anticlockwise, so different angles in [0,2π)[0, 2\pi) give different unit vectors.

    O P(cos θ, sin θ) P′ θ X X′ Y Y′ |r| = 1 r = cos θ î + sin θ ĵ 0 ≤ θ < 2π x = cos θ, y = sin θ x² + y² = 1
    The unit circle x2+y2=1x^2 + y^2 = 1 drawn with one scale. The unit vector OP→\overrightarrow{OP} at θ=60∘\theta = 60^\circ has components cos⁡θ\cos\theta and sin⁡θ\sin\theta⁠; the violet arrows at 150∘150^\circ⁠, 225∘225^\circ and 300∘300^\circ are three more unit vectors, all ending on the same circle.
    The book lets θ\theta vary from 0 to 2π2\pi⁠; the end value 2π2\pi gives the same vector as θ=0\theta = 0⁠, namely i^\hat i⁠, so 0≤θ<2π0 \le \theta < 2\pi lists each unit vector exactly once. When y<0y < 0⁠, the angle 2π−α2\pi - \alpha has the same cosine as α\alpha and the opposite sine.
    Principal range of cos⁡−1\cos^{-1} is [0,π][0, \pi]⁠; cos⁡(2π−α)=cos⁡α\cos(2\pi - \alpha) = \cos\alpha⁠, sin⁡(2π−α)=−sin⁡α\sin(2\pi - \alpha) = -\sin\alpha⁠. Lesson: Examples 26–30
Result

The unit vectors in the XY-plane are exactly cos⁡θ i^+sin⁡θ j^\cos\theta\,\hat i + \sin\theta\,\hat j⁠, one for each θ\theta with 0≤θ<2π0 \le \theta < 2\pi⁠.

(Textbook answer: r⃗ (=OP→)=cos⁡θ i^+sin⁡θ j^\vec r\,(= \overrightarrow{OP}) = \cos\theta\,\hat i + \sin\theta\,\hat j ... (1), with ∣r⃗∣=cos⁡2θ+sin⁡2θ=1|\vec r| = \sqrt{\cos^2\theta + \sin^2\theta} = 1⁠; "as θ\theta varies from 0 to 2π2\pi⁠, the point P (Fig 10.28) traces the circle x2+y2=1x^2 + y^2 = 1 counterclockwise", so (1) gives every unit vector in the XY-plane.)

Check it with numbers

θ=2π3\theta = \frac{2\pi}{3} gives −12i^+32j^-\frac12\hat i + \frac{\sqrt3}{2}\hat j⁠, with 32≈0.866025\frac{\sqrt3}{2} \approx 0.866025 and 14+34=1\frac14 + \frac34 = 1⁠.

The unit vector 35i^−45j^\frac35\hat i - \frac45\hat j has y<0y < 0⁠, so θ=2π−cos⁡−135\theta = 2\pi - \cos^{-1}\frac35⁠. Here cos⁡−135≈53.13∘\cos^{-1}\frac35 \approx 53.13^\circ⁠, so θ≈306.87∘\theta \approx 306.87^\circ⁠, and indeed cos⁡θ=35\cos\theta = \frac35⁠, sin⁡θ=−45\sin\theta = -\frac45⁠.

Common mistakes
  1. Stopping at x2+y2=1x^2 + y^2 = 1⁠. That is the condition, not the list; the question asks you to write the vectors, as cos⁡θ i^+sin⁡θ j^\cos\theta\,\hat i + \sin\theta\,\hat j⁠.
  2. Letting θ\theta run only over [0,π][0, \pi]⁠. That gives only the vectors with y≥0y \ge 0⁠; the lower half of the circle needs π<θ<2π\pi < \theta < 2\pi⁠.
  3. Adding a k^\hat k part. A vector in the XY-plane has zero k^\hat k component.
What to take away
  1. Unit vectors in a plane drawn from one point end on a circle of radius 1.
  2. In the XY-plane they are cos⁡θ i^+sin⁡θ j^\cos\theta\,\hat i + \sin\theta\,\hat j⁠, 0≤θ<2π0 \le \theta < 2\pi⁠.
  3. The angle θ\theta is measured anticlockwise from OX, and each direction has exactly one such θ\theta⁠.

Every fact used, and where it comes from

FactUsed inSource
Unit vector: magnitude 1Step 1Section 10.3 (Lesson: components)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 1, 3Section 10.5.1
A point of the unit circle is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta)⁠; triangle lawStep 2Trigonometric functions (Class XI); Section 10.4
cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1Step 3Trigonometry (Basics: cos θ is a shadow)
Principal range of cos⁡−1\cos^{-1} is [0,π][0, \pi]⁠; cos⁡(2π−α)=cos⁡α\cos(2\pi - \alpha) = \cos\alpha⁠, sin⁡(2π−α)=−sin⁡α\sin(2\pi - \alpha) = -\sin\alphaStep 4Inverse trigonometric functions (Chapter 2); trigonometry
The book's working, answer and figureSteps 2, 4, ResultExample 26 and Fig 10.28, Miscellaneous Examples (Lesson: Examples 26–30)
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