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Chapter 11 · Three Dimensional Geometry

Exercise 11.1 · Q1: Direction cosines of a line at 90∘90^\circ⁠, 135∘135^\circ⁠, 45∘45^\circ to the axes

Exercise 11.1Direction cosines from direction angles · Section 11.2
The question

If a line makes angles 90∘90^\circ⁠, 135∘135^\circ⁠, 45∘45^\circ with the xx⁠, yy and zz-axes respectively, find its direction cosines.

Exam answer

Let the direction cosines of the line be l,m,nl, m, n⁠. Then

l=cos⁡90∘=0m=cos⁡135∘=−cos⁡45∘=−12n=cos⁡45∘=12\begin{aligned} l &= \cos 90^\circ = 0 \\ m &= \cos 135^\circ = -\cos 45^\circ \\ &= -\frac{1}{\sqrt2} \\ n &= \cos 45^\circ = \frac{1}{\sqrt2} \end{aligned}

∴\therefore The direction cosines are 0,−12,120, -\dfrac{1}{\sqrt2}, \dfrac{1}{\sqrt2}⁠.

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

The one idea this problem is built on

The direction cosines of a line are simply the cosines of the angles it makes with the positive xx⁠, yy and zz-axes, in that order. The only care needed is the sign: an angle bigger than 90∘90^\circ has a negative cosine.

Given, to find, and the plan
Given
A line making angles α=90∘\alpha = 90^\circ⁠, β=135∘\beta = 135^\circ⁠, γ=45∘\gamma = 45^\circ with the xx⁠, yy and zz-axes.
To find
Its direction cosines l,m,nl, m, n⁠.
Plan
(1) Name the direction angles. (2) Take their cosines, with cos⁡135∘\cos 135^\circ from cos⁡(180∘−θ)=−cos⁡θ\cos(180^\circ - \theta) = -\cos\theta⁠. (3) Check that l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠. (4) See where the line points.

Basics you need

1Direction angles and direction cosines

If a directed line makes angles α,β,γ\alpha, \beta, \gamma with the positive xx⁠, yy and zz-axes, then l=cos⁡αl = \cos\alpha⁠, m=cos⁡βm = \cos\beta⁠, n=cos⁡γn = \cos\gamma are its direction cosines (Section 11.2). Basics: direction cosines

2Cosine of an obtuse angle

cos⁡(180∘−θ)=−cos⁡θ\cos(180^\circ - \theta) = -\cos\theta⁠. So cos⁡135∘=cos⁡(180∘−45∘)=−cos⁡45∘\cos 135^\circ = \cos(180^\circ - 45^\circ) = -\cos 45^\circ⁠. From the half-square triangle with sides 1,1,21, 1, \sqrt2⁠, cos⁡45∘=12\cos 45^\circ = \frac{1}{\sqrt2}⁠.

3The squares add up to 1

For every line, l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠, because li^+mj^+nk^l\hat i + m\hat j + n\hat k is a unit vector along the line. Basics: distance in 3D

4Points in space

A point with x=0x = 0 lies in the yzyz-plane; a negative yy means the side of the negative yy-axis. Basics: three numbers fix a point

Solution

  1. Name the direction angles

    The angles with the xx⁠, yy and zz-axes are

    α=90∘,β=135∘,γ=45∘.\alpha = 90^\circ, \quad \beta = 135^\circ, \quad \gamma = 45^\circ.
    The angles a directed line makes with the positive directions of the axes are its direction angles (Section 11.2). The question lists them in the order xx⁠, yy⁠, zz⁠, so the first cosine belongs to the xx-axis, the second to the yy-axis and the third to the zz-axis.
    Direction angles of a directed line. Basics: direction cosines

  2. Take the cosines

    l=cos⁡90∘=0,m=cos⁡135∘=−cos⁡45∘=−12,n=cos⁡45∘=12.\begin{aligned} l &= \cos 90^\circ = 0, \\ m &= \cos 135^\circ = -\cos 45^\circ = -\frac{1}{\sqrt2}, \\ n &= \cos 45^\circ = \frac{1}{\sqrt2}. \end{aligned}
    Direction cosines are, by definition, the cosines of the direction angles. The cosine of a right angle is 0. For 135∘135^\circ⁠, write it as 180∘−45∘180^\circ - 45^\circ⁠: the cosine of a supplementary angle is the negative of the cosine, so cos⁡135∘=−cos⁡45∘\cos 135^\circ = -\cos 45^\circ⁠. On the unit circle, 135∘135^\circ is in the second quadrant, where the xx-coordinate (the cosine) is negative. Rationalised, 12=22\frac{1}{\sqrt2} = \frac{\sqrt2}{2}⁠.
    l=cos⁡αl = \cos\alpha⁠, m=cos⁡βm = \cos\beta⁠, n=cos⁡γn = \cos\gamma⁠. Lesson: direction cosines and direction ratios

  3. Check that the squares add up to 1

    l2+m2+n2=0+12+12=1.l^2 + m^2 + n^2 = 0 + \frac12 + \frac12 = 1.
    Every set of direction cosines must pass this test. It also confirms the question makes sense: not every three angles belong to a line (for example 90∘,90∘,90∘90^\circ, 90^\circ, 90^\circ would give 0≠10 \ne 1⁠). These angles pass, so such a line exists. Note that squaring hides the sign of mm⁠, so the test cannot catch a sign slip; the sign comes from Step 2.
    l2+m2+n2=1l^2 + m^2 + n^2 = 1 (Section 11.2). Basics: Pythagoras twice

  4. See where the line points

    Through the origin, the unit vector along the line is

    0 i^−12 j^+12 k^,0\,\hat i - \frac{1}{\sqrt2}\,\hat j + \frac{1}{\sqrt2}\,\hat k,

    so it ends at P(0,−12,12)P\left(0, -\frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right)⁠. It lies in the yzyz-plane, halfway between the negative yy-axis and the positive zz-axis.

    X Y Z 135° 45° P(0, −1/√2, 1/√2) −1/√2 1/√2 O 90° with OX
    Oblique projection, not to scale (the yzyz-plane is drawn true to shape). The unit arrow OPOP along the line makes 135∘135^\circ with OYOY and 45∘45^\circ with OZOZ⁠, and is perpendicular to OXOX⁠.
    The xx-component is 0, so the line is perpendicular to the xx-axis and lies in the yzyz-plane (through the origin). There, 45∘45^\circ from OZOZ and 135∘135^\circ from OYOY agree, because the angle from OYOY to OZOZ is 90∘90^\circ and 90∘+45∘=135∘90^\circ + 45^\circ = 135^\circ⁠: the line leans away from OYOY⁠, past OZOZ⁠. Taking the same line the other way gives the supplements 90∘,45∘,135∘90^\circ, 45^\circ, 135^\circ and the direction cosines 0,12,−120, \frac{1}{\sqrt2}, -\frac{1}{\sqrt2}⁠, but the question fixes the direction by giving the angles.
    The point P(l,m,n)P(l, m, n) is at distance 1 from OO along the line. Basics: three numbers fix a point
Result

The direction cosines are l=0l = 0⁠, m=−12m = -\frac{1}{\sqrt2}⁠, n=12n = \frac{1}{\sqrt2}⁠.

(Textbook answer: 0,−12,120, \frac{-1}{\sqrt2}, \frac{1}{\sqrt2}⁠.)

Check it with numbers

12≈0.70710678\frac{1}{\sqrt2} \approx 0.70710678⁠, so m≈−0.70710678m \approx -0.70710678 and n≈0.70710678n \approx 0.70710678⁠. Then m2≈0.5m^2 \approx 0.5 and n2≈0.5n^2 \approx 0.5⁠, and 0+0.5+0.5=10 + 0.5 + 0.5 = 1⁠.

A calculator in degree mode gives cos⁡135∘≈−0.70710678\cos 135^\circ \approx -0.70710678⁠, the same as mm⁠.

Common mistakes
  1. Losing the minus sign of cos⁡135∘\cos 135^\circ⁠. 135∘135^\circ is obtuse, so its cosine is negative: −12-\frac{1}{\sqrt2}⁠, not 12\frac{1}{\sqrt2}⁠. The check l2+m2+n2=1l^2 + m^2 + n^2 = 1 does not catch this slip.
  2. Mixing up the order. The first direction cosine goes with the xx-axis, the second with yy⁠, the third with zz⁠.
  3. Writing the angles instead of their cosines, or giving cos⁡90∘\cos 90^\circ as 1. The cosine of a right angle is 0.
What to take away
  1. Direction cosines are the cosines of the angles with the positive axes, in the order xx⁠, yy⁠, zz⁠.
  2. An obtuse angle gives a negative direction cosine; cos⁡(180∘−θ)=−cos⁡θ\cos(180^\circ - \theta) = -\cos\theta⁠.
  3. Always check l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠; a direction cosine of 0 means the line is perpendicular to that axis.

Every fact used, and where it comes from

FactUsed inSource
Direction angles and direction cosines l=cos⁡αl = \cos\alpha⁠, m=cos⁡βm = \cos\beta⁠, n=cos⁡γn = \cos\gammaSteps 1, 2Section 11.2 (Lesson)
cos⁡(180∘−θ)=−cos⁡θ\cos(180^\circ - \theta) = -\cos\theta⁠, cos⁡45∘=12\cos 45^\circ = \frac{1}{\sqrt2}⁠, cos⁡90∘=0\cos 90^\circ = 0Step 2Trigonometric functions, Class XI
l2+m2+n2=1l^2 + m^2 + n^2 = 1Step 3Section 11.2 (Basics)
Reversing a line replaces the direction angles by their supplementsStep 4Section 11.2 (Exercise 11.1 player)
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