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Chapter 10 · Vector Algebra

Exercise 10.2 · Q14: i^+j^+k^\hat i+\hat j+\hat k is equally inclined to OX, OY and OZ

Exercise 10.2Direction cosines and direction angles · Sections 10.2 and 10.5.1
The question

Show that the vector i^+j^+k^\hat i + \hat j + \hat k is equally inclined to the axes OX, OY and OZ.

Exam answer

Let a⃗=i^+j^+k^\vec a = \hat i + \hat j + \hat k⁠.

∣a⃗∣=12+12+12=3\begin{aligned} |\vec a| &= \sqrt{1^2 + 1^2 + 1^2} \\ &= \sqrt3 \end{aligned}

So the direction cosines of a⃗\vec a are

l=m=n=13l = m = n = \frac{1}{\sqrt3}

Let α,β,γ\alpha, \beta, \gamma be the angles that a⃗\vec a makes with OX, OY and OZ. Then

cos⁡α=cos⁡β=cos⁡γ=13\cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt3}

As α,β,γ∈[0,π]\alpha, \beta, \gamma \in [0, \pi]⁠, where cosine takes each value only once, α=β=γ=cos⁡−113\alpha = \beta = \gamma = \cos^{-1}\frac{1}{\sqrt3}⁠.

Hence i^+j^+k^\hat i + \hat j + \hat k is equally inclined to OX, OY and OZ.

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

The one idea this problem is built on

The direction cosines of a vector are the cosines of the angles it makes with OX, OY and OZ. If the three cosines are equal, the three angles are equal, because an angle between 0 and π\pi is fixed by its cosine. So we only need to show that l=m=nl = m = n⁠.

Given, to find, and the plan
Given
a⃗=i^+j^+k^\vec a = \hat i + \hat j + \hat k⁠.
To show
The angles α,β,γ\alpha, \beta, \gamma that a⃗\vec a makes with OX, OY and OZ are equal.
Plan
(1) Recall that l=cos⁡αl = \cos\alpha⁠, m=cos⁡βm = \cos\beta⁠, n=cos⁡γn = \cos\gamma⁠. (2) Find ∣a⃗∣|\vec a|⁠. (3) Find l,m,nl, m, n⁠. (4) Equal cosines give equal angles; find the common angle.

Basics you need

1Direction angles and cosines

The angles α,β,γ\alpha, \beta, \gamma that r⃗\vec r makes with the positive x, y and z-axes are its direction angles; l=cos⁡α=xrl = \cos\alpha = \frac{x}{r}⁠, m=cos⁡β=yrm = \cos\beta = \frac{y}{r}⁠, n=cos⁡γ=zrn = \cos\gamma = \frac{z}{r} (Section 10.2, Fig 10.3). Basics: cos θ is a shadow, sin θ is a height

2Magnitude from components

∣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). Basics: use Pythagoras twice

3An angle in [0,π][0, \pi] is fixed by its cosine

On [0,π][0, \pi]⁠, cos⁡θ\cos\theta decreases from 1 to −1-1 and takes each value exactly once; so cos⁡α=cos⁡β\cos\alpha = \cos\beta with α,β∈[0,π]\alpha, \beta \in [0, \pi] forces α=β\alpha = \beta (school trigonometry). Basics: cos θ is a shadow

Solution

  1. What "equally inclined" means

    Let α,β,γ\alpha, \beta, \gamma be the angles that a⃗\vec a makes with OX, OY and OZ. We must show α=β=γ\alpha = \beta = \gamma⁠. Their cosines are the direction cosines:

    l=cos⁡α,m=cos⁡β,n=cos⁡γ.\begin{aligned} l &= \cos\alpha, \\ m &= \cos\beta, \\ n &= \cos\gamma. \end{aligned}
    "Inclined to an axis" is the angle between the vector and the positive direction of that axis. These are exactly the direction angles, and their cosines are what we can compute from the components.
    Direction angles and direction cosines (Section 10.2, Fig 10.3). Basics: cos θ is a shadow

  2. Find the length

    ∣a⃗∣=12+12+12=3.|\vec a| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt3.
    The direction cosines are the components divided by the length, so we need ∣a⃗∣|\vec a|⁠. Here a⃗\vec a is the diagonal of the unit cube from O to P(1, 1, 1).
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. Find the direction cosines

    l=13,m=13,n=13.\begin{aligned} l &= \frac{1}{\sqrt3}, \\ m &= \frac{1}{\sqrt3}, \\ n &= \frac{1}{\sqrt3}. \end{aligned}

    So cos⁡α=cos⁡β=cos⁡γ=13\cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt3}⁠.

    All three components of a⃗\vec a are 1, and each is divided by the same length 3\sqrt3⁠, so the three cosines are equal. As a check, l2+m2+n2=13+13+13=1l^2 + m^2 + n^2 = \frac13 + \frac13 + \frac13 = 1⁠.
    l=xrl = \frac{x}{r}⁠, m=yrm = \frac{y}{r}⁠, n=zrn = \frac{z}{r} (Section 10.2); l2+m2+n2=1l^2 + m^2 + n^2 = 1 (Note, Section 10.2). Lesson: components

  4. Equal cosines give equal angles

    Each of α,β,γ\alpha, \beta, \gamma lies in [0,π][0, \pi]⁠, and there the cosine takes each value only once. So

    α=β=γ=cos⁡−113.\alpha = \beta = \gamma = \cos^{-1}\frac{1}{\sqrt3}.

    Hence i^+j^+k^\hat i + \hat j + \hat k is equally inclined to OX, OY and OZ.

    X Y Z O P A B C O A P 1 √2 √3 α cos α = OA/OP = 1/√3 OBP and OCP: the same so α = β = γ ≈ 54.74°
    Left: the unit cube with the diagonal OP→=i^+j^+k^\overrightarrow{OP} = \hat i + \hat j + \hat k and the feet A, B, C of the perpendiculars from P to the axes (orthographic view, one scale). Right: the right triangle OAP drawn true to scale, with OA=1\text{OA} = 1⁠, AP=2\text{AP} = \sqrt2⁠, OP=3\text{OP} = \sqrt3⁠, so cos⁡α=13\cos\alpha = \frac{1}{\sqrt3}⁠. The triangles OBP and OCP have the same three sides, so β\beta and γ\gamma equal α\alpha⁠.
    Without the range [0,π][0, \pi]⁠, equal cosines would not force equal angles (for example cos⁡60∘=cos⁡300∘\cos 60^\circ = \cos 300^\circ⁠). The angle between a vector and an axis is always taken in [0,π][0, \pi]⁠, so the step is safe. The figure shows the same fact geometrically: each foot of a perpendicular is 1 from O and 2\sqrt2 from P.
    The angle between two vectors is taken with 0≤θ≤π0 \le \theta \le \pi (Definition 2, Section 10.6.1, later in this chapter); cos⁡\cos is one-one on [0,π][0, \pi] (school trigonometry). Basics: cos θ is a shadow
Result

The direction cosines of i^+j^+k^\hat i + \hat j + \hat k are 13,13,13\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}⁠, so it makes the same angle cos⁡−113\cos^{-1}\frac{1}{\sqrt3} with each of OX, OY and OZ: it is equally inclined to the axes. Hence proved.

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

Check it with numbers

13≈0.577350\frac{1}{\sqrt3} \approx 0.577350⁠, and cos⁡−1(0.577350)≈54.736∘\cos^{-1}(0.577350) \approx 54.736^\circ⁠: the common angle is a little less than 55∘55^\circ⁠.

In the triangle OAP of the figure, OA2+AP2=1+2=3=OP2\text{OA}^2 + \text{AP}^2 = 1 + 2 = 3 = \text{OP}^2⁠, so the angle at A is a right angle and cos⁡α=OAOP=13\cos\alpha = \frac{\text{OA}}{\text{OP}} = \frac{1}{\sqrt3} again.

Common mistakes
  1. Stopping at "the components are equal". The question is about angles; go from the components to the cosines and then to the angles.
  2. Thinking the angle is 45∘45^\circ⁠. That is the angle in a square; in the cube the diagonal makes cos⁡−113≈54.74∘\cos^{-1}\frac{1}{\sqrt3} \approx 54.74^\circ with each edge.
  3. Leaving out why equal cosines give equal angles. Say that the angles lie in [0,π][0, \pi]⁠, where the cosine takes each value once.
What to take away
  1. Equally inclined to the axes means equal direction cosines: l=m=nl = m = n⁠.
  2. For l=m=nl = m = n⁠, l2+m2+n2=1l^2 + m^2 + n^2 = 1 gives l=±13l = \pm\frac{1}{\sqrt3}⁠; the vector i^+j^+k^\hat i + \hat j + \hat k takes the plus sign.
  3. Any positive multiple λ(i^+j^+k^)\lambda(\hat i + \hat j + \hat k) is equally inclined to the axes too.

Every fact used, and where it comes from

FactUsed inSource
Direction angles; l=cos⁡α=xrl = \cos\alpha = \frac{x}{r}⁠, m=cos⁡βm = \cos\beta⁠, n=cos⁡γn = \cos\gammaSteps 1, 3Section 10.2, Fig 10.3 (Basics: cos θ is a shadow)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Step 2Section 10.5.1
l2+m2+n2=1l^2 + m^2 + n^2 = 1Step 3Note, Section 10.2
The angle between two vectors lies in [0,π][0, \pi]Step 4Definition 2, Section 10.6.1
cos⁡\cos takes each value once on [0,π][0, \pi]Step 4School trigonometry
Converse of Pythagoras' theoremCheckSchool geometry
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