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Chapter 10 · Vector Algebra

Miscellaneous · Q11: Direction cosines of a vector equally inclined to the axes are ±13\pm\frac{1}{\sqrt3}

Miscellaneous ExerciseDirection cosines, l² + m² + n² = 1 · Section 10.2
The question

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are ±(13,13,13)\pm\left(\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}\right)⁠.

Exam answer

Let the vector make the same angle α\alpha with each of OX, OY and OZ.

Then its direction cosines are l=m=n=cos⁡αl = m = n = \cos\alpha⁠.

Using l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠,

cos⁡2α+cos⁡2α+cos⁡2α=1⇒3cos⁡2α=1⇒cos⁡2α=13⇒cos⁡α=±13\begin{aligned} &\cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \\ &\Rightarrow 3\cos^2\alpha = 1 \\ &\Rightarrow \cos^2\alpha = \frac13 \\ &\Rightarrow \cos\alpha = \pm\frac{1}{\sqrt3} \end{aligned}

Hence the direction cosines are ±(13,13,13)\pm\left(\dfrac{1}{\sqrt3}, \dfrac{1}{\sqrt3}, \dfrac{1}{\sqrt3}\right)⁠.

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

The one idea this problem is built on

The direction cosines are the cosines of the angles with the three axes. Equal angles give equal cosines, l=m=nl = m = n⁠, and every set of direction cosines obeys l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠. Together these leave only 3l2=13l^2 = 1⁠.

Given, to show, and the plan
Given
A vector that makes equal angles with OX, OY and OZ.
To show
Its direction cosines are 13,13,13\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3} or −13,−13,−13-\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}⁠.
Plan
(1) Turn "equally inclined" into l=m=nl = m = n⁠. (2) Use l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠. (3) Show that both signs really occur. (4) Conclude.

Basics you need

1Direction angles and direction cosines

If r⃗\vec r makes angles α,β,γ\alpha, \beta, \gamma with OX, OY and OZ, then l=cos⁡αl = \cos\alpha⁠, m=cos⁡βm = \cos\beta⁠, n=cos⁡γn = \cos\gamma are its direction cosines (Section 10.2). Basics: cos θ is a shadow, sin θ is a height

2The sum of their squares is 1

l=xrl = \frac{x}{r}⁠, m=yrm = \frac{y}{r}⁠, n=zrn = \frac{z}{r} with r=x2+y2+z2r = \sqrt{x^2 + y^2 + z^2}⁠, so l2+m2+n2=1l^2 + m^2 + n^2 = 1 (Section 10.2, Note). Basics: use Pythagoras twice

3Square roots of a positive number

x2=13x^2 = \frac13 has the two solutions x=13x = \frac{1}{\sqrt3} and x=−13x = -\frac{1}{\sqrt3} (school algebra).

Solution

  1. Equally inclined means equal direction cosines

    Let the vector make angles α,β,γ\alpha, \beta, \gamma with OX, OY and OZ. "Equally inclined" means α=β=γ\alpha = \beta = \gamma⁠, so

    l=cos⁡α=cos⁡β=cos⁡γ=m=n.l = \cos\alpha = \cos\beta = \cos\gamma = m = n.
    The direction cosines are not just any three numbers: by definition they are the cosines of the three angles with the axes. Equal angles therefore give equal cosines, and the whole question reduces to finding the common value ll⁠.
    Direction cosines (Section 10.2). Basics: cos θ is a shadow

  2. Use l2+m2+n2=1l^2 + m^2 + n^2 = 1

    l2+l2+l2=13l2=1l2=13l=±13.\begin{aligned} l^2 + l^2 + l^2 &= 1 \\ 3l^2 &= 1 \\ l^2 &= \frac13 \\ l &= \pm\frac{1}{\sqrt3}. \end{aligned}

    So l=m=n=13l = m = n = \frac{1}{\sqrt3} or l=m=n=−13l = m = n = -\frac{1}{\sqrt3}.

    For a vector xi^+yj^+zk^x\hat i + y\hat j + z\hat k of length rr⁠, l2+m2+n2=x2+y2+z2r2=1l^2 + m^2 + n^2 = \frac{x^2 + y^2 + z^2}{r^2} = 1⁠. With mm and nn both equal to ll⁠, the identity has only one unknown. A square equal to 13\frac13 has two square roots, one positive and one negative, and since l=m=nl = m = n the three cosines carry the same sign: we cannot mix +13+\frac{1}{\sqrt3} and −13-\frac{1}{\sqrt3}⁠.
    l2+m2+n2=1l^2 + m^2 + n^2 = 1 (Section 10.2, Note). Basics: use Pythagoras twice

  3. Both signs really occur

    i^+j^+k^\hat i + \hat j + \hat k has length 3\sqrt3⁠, so its direction cosines are 13,13,13\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}⁠: each angle is cos⁡−113\cos^{-1}\frac{1}{\sqrt3}⁠, an acute angle.

    −i^−j^−k^-\hat i - \hat j - \hat k has direction cosines −13,−13,−13-\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}⁠: each angle is cos⁡−1(−13)\cos^{-1}\left(-\frac{1}{\sqrt3}\right)⁠, an obtuse angle.

    O Z i + j direction P(1, 1, 1) Q(−1, −1, −1) γ OP = i + j + k cos γ = 1/√3 γ ≈ 54.74° OQ = −(i + j + k) 180° − γ ≈ 125.26°
    The plane through OZ and the vector i^+j^+k^\hat i + \hat j + \hat k⁠, drawn true to scale (one scale). The horizontal line is the direction of i^+j^\hat i + \hat j in the XY-plane. OP→=i^+j^+k^\overrightarrow{OP} = \hat i + \hat j + \hat k makes γ=cos⁡−113\gamma = \cos^{-1}\frac{1}{\sqrt3} with OZ, and the opposite vector OQ→\overrightarrow{OQ} makes 180∘−γ180^\circ - \gamma⁠. By symmetry the same two angles appear with OX and OY.
    Both vectors make equal angles with the three axes, so both signs are needed in the answer; that is what the ±\pm records. The two directions are opposite, and the angles γ\gamma and 180∘−γ180^\circ - \gamma have opposite cosines.
    l=xrl = \frac{x}{r}⁠, m=yrm = \frac{y}{r}⁠, n=zrn = \frac{z}{r} (Section 10.2). Lesson: components

  4. Conclude

    Every vector equally inclined to OX, OY and OZ has

    (l,m,n)=±(13,13,13).(l, m, n) = \pm\left(\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}\right).

    Hence proved.

    Step 2 showed that no other values are possible, and Step 3 that both values occur. Every vector equally inclined to the axes is a positive multiple of i^+j^+k^\hat i + \hat j + \hat k or of −i^−j^−k^-\hat i - \hat j - \hat k⁠; compare Exercise 10.2 Q14, where i^+j^+k^\hat i + \hat j + \hat k is shown to be equally inclined to the axes.
    Direction cosines fix the direction (Section 10.2). Lesson: the miscellaneous exercises
Result

A vector equally inclined to OX, OY and OZ has direction cosines ±(13,13,13)\pm\left(\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}\right)⁠: the plus sign when the common angle is acute, the minus sign when it is obtuse. 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. The common angle is cos⁡−113≈54.7356∘\cos^{-1}\frac{1}{\sqrt3} \approx 54.7356^\circ for the plus sign and cos⁡−1(−13)≈125.2644∘\cos^{-1}\left(-\frac{1}{\sqrt3}\right) \approx 125.2644^\circ for the minus sign; the two add up to 180∘180^\circ⁠.

The vector 2i^+2j^+2k^2\hat i + 2\hat j + 2\hat k has length 12=23\sqrt{12} = 2\sqrt3⁠, so l=223=13l = \frac{2}{2\sqrt3} = \frac{1}{\sqrt3}⁠, the same as for i^+j^+k^\hat i + \hat j + \hat k⁠. And exactly, 13+13+13=1\frac13 + \frac13 + \frac13 = 1⁠.

Common mistakes
  1. Dropping the minus sign. l2=13l^2 = \frac13 has two roots; the vector −i^−j^−k^-\hat i - \hat j - \hat k is also equally inclined to the axes.
  2. Mixing the signs. (13,−13,13)\left(\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}, \frac{1}{\sqrt3}\right) is not allowed: the cosines are equal, so they share one sign.
  3. Writing l=13l = \frac13⁠. The equation is 3l2=13l^2 = 1⁠, not 3l=13l = 1⁠.
What to take away
  1. Equal angles with the axes means l=m=nl = m = n⁠.
  2. Any condition on direction cosines is combined with l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠.
  3. A square root brings a ±\pm⁠: here the two opposite directions along i^+j^+k^\hat i + \hat j + \hat k⁠.

Every fact used, and where it comes from

FactUsed inSource
Direction cosines l=cos⁡αl = \cos\alpha⁠, m=cos⁡βm = \cos\beta⁠, n=cos⁡γn = \cos\gammaSteps 1, 3Section 10.2 (Basics: cos θ is a shadow)
l=xrl = \frac{x}{r}⁠, m=yrm = \frac{y}{r}⁠, n=zrn = \frac{z}{r}Steps 2, 3, CheckSection 10.2
l2+m2+n2=1l^2 + m^2 + n^2 = 1Step 2Section 10.2, Note
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Step 3, CheckSection 10.5.1
cos⁡(180∘−γ)=−cos⁡γ\cos(180^\circ - \gamma) = -\cos\gammaStep 3School trigonometry
The answerResult"Show that" question; no key entry (Lesson: the miscellaneous exercises)
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