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Chapter 11 · Three Dimensional Geometry

Exercise 11.1 · Q2: Direction cosines of a line equally inclined to the axes

Exercise 11.1Direction cosines and l² + m² + n² = 1 · Section 11.2
The question

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Exam answer

Let the line make an angle α\alpha with each of the axes. Then its direction cosines are l=m=n=cos⁡αl = m = n = \cos\alpha⁠.

l2+m2+n2=1⇒3cos⁡2α=1⇒cos⁡2α=13⇒cos⁡α=±13\begin{aligned} l^2 + m^2 + n^2 &= 1 \\ \Rightarrow 3\cos^2\alpha &= 1 \\ \Rightarrow \cos^2\alpha &= \frac13 \\ \Rightarrow \cos\alpha &= \pm\frac{1}{\sqrt3} \end{aligned}

∴\therefore The direction cosines are 13,13,13\dfrac{1}{\sqrt3}, \dfrac{1}{\sqrt3}, \dfrac{1}{\sqrt3} or −13,−13,−13-\dfrac{1}{\sqrt3}, -\dfrac{1}{\sqrt3}, -\dfrac{1}{\sqrt3}⁠.

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

The one idea this problem is built on

Equal angles with the axes mean equal direction cosines: l=m=nl = m = n⁠. One equation is then enough to find them, the identity l2+m2+n2=1l^2 + m^2 + n^2 = 1 that every set of direction cosines satisfies.

Given, to find, and the plan
Given
A line making the same angle α\alpha with the xx⁠, yy and zz-axes.
To find
Its direction cosines l,m,nl, m, n⁠.
Plan
(1) Equal angles give l=m=n=cos⁡αl = m = n = \cos\alpha⁠. (2) Put this into l2+m2+n2=1l^2 + m^2 + n^2 = 1 and solve. (3) Decide the signs. (4) See the line: the diagonal of a cube.

Basics you need

1Direction cosines

A directed line making angles α,β,γ\alpha, \beta, \gamma with the positive xx⁠, yy and zz-axes has direction cosines cos⁡α,cos⁡β,cos⁡γ\cos\alpha, \cos\beta, \cos\gamma (Section 11.2). Basics: direction cosines

2The squares add up to 1

l2+m2+n2=1l^2 + m^2 + n^2 = 1 for every line, because li^+mj^+nk^l\hat i + m\hat j + n\hat k is a unit vector. Basics: Pythagoras twice

3Square roots come in pairs

t2=13t^2 = \frac13 has two solutions, t=13t = \frac{1}{\sqrt3} and t=−13t = -\frac{1}{\sqrt3}⁠.

4Reversing a line

The same line taken the other way has the supplementary angles and the direction cosines with all signs changed (Section 11.2). Lesson: direction cosines and direction ratios

Solution

  1. Equal angles give equal direction cosines

    Let the line make the angle α\alpha with each axis: α=β=γ\alpha = \beta = \gamma⁠. Then

    l=m=n=cos⁡α.l = m = n = \cos\alpha.
    Direction cosines are the cosines of the direction angles. If the three angles are equal, so are their cosines. Here "the angle with an axis" means the direction angle with the positive direction of that axis, as in Section 11.2, so the three equal angles really do give three equal numbers, with the same sign.
    l=cos⁡αl = \cos\alpha⁠, m=cos⁡βm = \cos\beta⁠, n=cos⁡γn = \cos\gamma⁠. Basics: direction cosines

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

    cos⁡2α+cos⁡2α+cos⁡2α=13cos⁡2α=1cos⁡2α=13cos⁡α=±13.\begin{aligned} \cos^2\alpha + \cos^2\alpha + \cos^2\alpha &= 1 \\ 3\cos^2\alpha &= 1 \\ \cos^2\alpha &= \frac13 \\ \cos\alpha &= \pm\frac{1}{\sqrt3}. \end{aligned}
    Replacing each of l,m,nl, m, n by cos⁡α\cos\alpha turns the identity into one equation in one unknown. Dividing by 3 and taking square roots gives two values, because a positive number has two square roots, one positive and one negative.
    l2+m2+n2=1l^2 + m^2 + n^2 = 1 (Section 11.2). Lesson: direction cosines

  3. Decide the signs

    Since l=m=nl = m = n⁠, all three take the same sign:

    (13,13,13)or (−13,−13,−13).\begin{aligned} &\left(\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}\right) \\ \text{or } &\left(-\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}\right). \end{aligned}
    The plus sign gives α=cos⁡−113\alpha = \cos^{-1}\frac{1}{\sqrt3}⁠, an acute angle with every axis. The minus sign gives the supplementary angle 180∘−α180^\circ - \alpha with every axis: this is the same line taken in the opposite direction, and reversing a line changes all three signs together. A mixed choice such as 13,−13,13\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}, \frac{1}{\sqrt3} is not allowed: its angles are α\alpha⁠, 180∘−α180^\circ - \alpha⁠, α\alpha⁠, which are not equal. The answer key's "±13,±13,±13\pm\frac{1}{\sqrt3}, \pm\frac{1}{\sqrt3}, \pm\frac{1}{\sqrt3}" is to be read in this way, with one sign for all three.
    Reversing the direction of a line reverses the signs of its direction cosines (Section 11.2). Exercise 11.1 player

  4. See the line: the diagonal of a cube

    Through the origin, the unit vector 13(i^+j^+k^)\frac{1}{\sqrt3}(\hat i + \hat j + \hat k) points along the diagonal OPOP of the unit cube, from OO to P(1,1,1)P(1, 1, 1)⁠.

    X Y Z P(1, 1, 1) 1 1 1 O
    Oblique projection, not to scale. The diagonal OPOP of the unit cube has direction ratios 1,1,11, 1, 1⁠. By symmetry it makes the same angle α=cos⁡−113\alpha = \cos^{-1}\frac{1}{\sqrt3} with each axis.
    The direction cosines 13,13,13\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3} are proportional to 1,1,11, 1, 1⁠, so 1,1,11, 1, 1 are direction ratios of the line. The cube diagonal from OO to (1,1,1)(1, 1, 1) has exactly these direction ratios and length 1+1+1=3\sqrt{1 + 1 + 1} = \sqrt3⁠; dividing by this length gives 13\frac{1}{\sqrt3} for each direction cosine, as found.
    Direction cosines of the line through two points: x2−x1PQ\frac{x_2 - x_1}{PQ}⁠, and so on (Section 11.2.1). Basics: three numbers fix a point
Result

The 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} (the same sign in all three places).

(Textbook answer: ±13,±13,±13\pm\frac{1}{\sqrt3}, \pm\frac{1}{\sqrt3}, \pm\frac{1}{\sqrt3}⁠.)

Check it with numbers

13≈0.57735027\frac{1}{\sqrt3} \approx 0.57735027⁠, and 3×0.577350272≈1.00000003 \times 0.57735027^2 \approx 1.0000000⁠.

The angle is α=cos⁡−10.57735027≈54.7356∘\alpha = \cos^{-1} 0.57735027 \approx 54.7356^\circ with each axis; the opposite direction makes 180∘−54.7356∘=125.2644∘180^\circ - 54.7356^\circ = 125.2644^\circ with each axis. The equal angle is not 45∘45^\circ⁠: 3cos⁡245∘=32≠13\cos^2 45^\circ = \frac32 \ne 1⁠.

Common mistakes
  1. Answering 45∘45^\circ by analogy with the plane, where a line equally inclined to the two axes makes 45∘45^\circ with each. In space there are three axes, and 3cos⁡245∘=32≠13\cos^2 45^\circ = \frac32 \ne 1⁠.
  2. Dropping the negative root. cos⁡2α=13\cos^2\alpha = \frac13 gives cos⁡α=±13\cos\alpha = \pm\frac{1}{\sqrt3}⁠; the minus sign is the same line in the opposite direction.
  3. Mixing signs, such as 13,−13,13\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}, \frac{1}{\sqrt3}⁠. The angles would then be unequal; all three signs must be the same.
What to take away
  1. Equal angles with the axes mean l=m=nl = m = n⁠; then l2+m2+n2=1l^2 + m^2 + n^2 = 1 gives 3l2=13l^2 = 1⁠.
  2. The two answers ±13\pm\frac{1}{\sqrt3} (one sign for all three) are the two directions of the same line.
  3. Such a line is parallel to a diagonal of a cube, at about 54.74∘54.74^\circ to each axis.

Every fact used, and where it comes from

FactUsed inSource
Direction cosines are the cosines of the direction anglesStep 1Section 11.2 (Basics)
l2+m2+n2=1l^2 + m^2 + n^2 = 1Step 2Section 11.2 (Lesson)
Reversing a line replaces the angles by their supplements and reverses the signs of the direction cosinesStep 3Section 11.2
Direction cosines of the line through two pointsStep 4Section 11.2.1 (Exercise 11.1 player)
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