Chapter 11 · Three Dimensional Geometry

Example 1: Direction cosines of a line at $90^\circ$ ⁠, $60^\circ$ ⁠, $30^\circ$ to the axes

ExamplesDirection cosines · Section 11.2

The question

If a line makes angle $90^\circ$ ⁠, $60^\circ$ and $30^\circ$ with the positive direction of $x$ ⁠, $y$ and $z$ -axis respectively, find its direction cosines.

Exam answer

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

\begin{aligned} l &= \cos 90^\circ = 0 \\ m &= \cos 60^\circ = \frac12 \\ n &= \cos 30^\circ = \frac{\sqrt3}{2} \end{aligned}

$\therefore$ The direction cosines are $0, \dfrac12, \dfrac{\sqrt3}{2}$ ⁠.

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

The one idea this problem is built on

The direction cosines of a directed line are, by definition, the cosines of the angles it makes with the positive $x$ ⁠, $y$ and $z$ -axes. Here the three angles are given, so we only have to take their cosines. The rule $l^2 + m^2 + n^2 = 1$ then checks the answer.

Given, to find, and the plan

Given: A line making angles $\alpha = 90^\circ$ ⁠, $\beta = 60^\circ$ ⁠, $\gamma = 30^\circ$ with the positive directions of the $x$ ⁠, $y$ and $z$ -axes.
To find: Its direction cosines $l, m, n$ ⁠.
Plan: (1) Name the direction angles. (2) Take their cosines. (3) Check that $l^2 + m^2 + n^2 = 1$ ⁠. (4) See where the line points.

Basics you need

1Direction angles and direction cosines

If a directed line makes angles $\alpha, \beta, \gamma$ with the positive $x$ ⁠, $y$ and $z$ -axes, then $l = \cos\alpha$ ⁠, $m = \cos\beta$ ⁠, $n = \cos\gamma$ are its direction cosines (Section 11.2). Basics: direction cosines

2Cosines of the standard angles

$\cos 90^\circ = 0$ ⁠, $\cos 60^\circ = \frac12$ and $\cos 30^\circ = \frac{\sqrt3}{2}$ ⁠, from the half-equilateral triangle with sides $1, \sqrt3, 2$ ⁠.

3The squares add up to 1

For every line, $l^2 + m^2 + n^2 = 1$ ⁠: the vector $l\hat i + m\hat j + n\hat k$ is a unit vector along the line, and its length is found by Pythagoras in three dimensions. Basics: distance in 3D

4Points in space

$(x, y, z)$ means $x$ units along $OX$ ⁠, $y$ along $OY$ and $z$ along $OZ$ ⁠. A point with $x = 0$ lies in the $yz$ -plane. Basics: three numbers fix a point

Solution

Name the direction angles

The angles with the positive $x$ ⁠, $y$ and $z$ -axes are
$\alpha = 90^\circ, \quad \beta = 60^\circ, \quad \gamma = 30^\circ.$
The question gives the angles with the positive directions of the axes, which are exactly the direction angles of Section 11.2. They describe the line taken in one direction. The same line taken the other way makes the supplementary angles $180^\circ - \alpha$ ⁠, $180^\circ - \beta$ ⁠, $180^\circ - \gamma$ ⁠, that is $90^\circ$ ⁠, $120^\circ$ ⁠, $150^\circ$ ⁠.

Direction angles of a directed line (Section 11.2, recalled from Chapter 10). Basics: direction cosines
Take the cosines
$\begin{aligned} l &= \cos 90^\circ = 0, \\ m &= \cos 60^\circ = \frac12, \\ n &= \cos 30^\circ = \frac{\sqrt3}{2}. \end{aligned}$
This is the definition: direction cosines are the cosines of the direction angles, in the order $x$ ⁠, $y$ ⁠, $z$ ⁠. The values come from the half-equilateral triangle: cut an equilateral triangle of side 2 in half to get a right triangle with sides $1, \sqrt3, 2$ and angles $30^\circ, 60^\circ, 90^\circ$ ⁠; then $\cos 60^\circ = \frac12$ and $\cos 30^\circ = \frac{\sqrt3}{2}$ ⁠. The cosine of a right angle is 0.

$l = \cos\alpha$ ⁠, $m = \cos\beta$ ⁠, $n = \cos\gamma$ ⁠. Lesson: direction cosines and direction ratios
Check that the squares add up to 1
$l^2 + m^2 + n^2 = 0 + \frac14 + \frac34 = 1.$
Every set of direction cosines must pass this test, because $l\hat i + m\hat j + n\hat k$ is a unit vector. It is also a real check on the question: not every three angles belong to a line. For example, $60^\circ, 60^\circ, 60^\circ$ would give $\frac14 + \frac14 + \frac14 = \frac34 \ne 1$ ⁠, so no line makes $60^\circ$ with all three axes. Our angles pass, so such a line exists.

$l^2 + m^2 + n^2 = 1$ (Section 11.2). Basics: Pythagoras twice
See where the line points

Draw the line through the origin, in the given direction. The unit vector along it is
$0\,\hat i + \frac12\,\hat j + \frac{\sqrt3}{2}\,\hat k,$
so it ends at $P\left(0, \frac12, \frac{\sqrt3}{2}\right)$ ⁠. Its $x$ -coordinate is 0, so $OP$ lies in the $yz$ -plane.

The unit arrow $OP$ in an oblique projection (not to scale: the $x$ -axis is drawn shortened and slanting). $OP$ lies in the $yz$ -plane, which is drawn true to shape, so the $60^\circ$ with $OY$ and the $30^\circ$ with $OZ$ look right; $OP$ is perpendicular to $OX$ ⁠.

A line at $90^\circ$ to the $x$ -axis is perpendicular to it, so (through the origin) it lies in the $yz$ -plane. Inside that plane the angles with $OY$ and $OZ$ must add up to $90^\circ$ ⁠, and indeed $60^\circ + 30^\circ = 90^\circ$ ⁠. The dashed lines show the cosines as lengths: $OP$ has length 1, its shadow on $OY$ is $\cos 60^\circ = \frac12$ and its shadow on $OZ$ is $\cos 30^\circ = \frac{\sqrt3}{2}$ ⁠.

The components of a unit vector are its direction cosines. Basics: a unit arrow's shadows on the axes

Result

The direction cosines of the line are $l = 0$ ⁠, $m = \frac12$ ⁠, $n = \frac{\sqrt3}{2}$ ⁠. (The same line taken in the opposite direction has $0, -\frac12, -\frac{\sqrt3}{2}$ ⁠.)

(Textbook answer: $l = 0$ ⁠, $m = \frac12$ ⁠, $n = \frac{\sqrt3}{2}$ ⁠.)

Check it with numbers

To eight places, $\cos 60^\circ = 0.50000000$ and $\cos 30^\circ = 0.86602540$ ⁠. Then $0.50000000^2 = 0.25$ and $0.86602540^2 \approx 0.7500000$ ⁠, so $l^2 + m^2 + n^2 \approx 0 + 0.25 + 0.75 = 1$ ⁠.

Going back from the cosines to the angles: $\cos^{-1}0 = 90^\circ$ ⁠, $\cos^{-1}0.5 = 60^\circ$ and $\cos^{-1}0.86602540 \approx 30.000000^\circ$ ⁠.

Common mistakes

Writing $\cos 90^\circ = 1$ ⁠. It is $\cos 0^\circ$ that equals 1; $\cos 90^\circ = 0$ ⁠.
Swapping the values of $\cos 30^\circ$ and $\cos 60^\circ$ ⁠. The larger angle has the smaller cosine: $\cos 60^\circ = \frac12 < \frac{\sqrt3}{2} = \cos 30^\circ$ ⁠.
Taking sines instead of cosines. That gives $\sin 90^\circ, \sin 60^\circ, \sin 30^\circ = 1, \frac{\sqrt3}{2}, \frac12$ ⁠, whose squares add up to $1 + \frac34 + \frac14 = 2$ ⁠, not 1; the check $l^2 + m^2 + n^2 = 1$ catches this slip. (It does not catch the swap in mistake 2, because $0, \frac{\sqrt3}{2}, \frac12$ also passes it.)
Using the angles with the coordinate planes. Direction angles are measured from the axes, not from the planes $xy$ ⁠, $yz$ ⁠, $zx$ ⁠.

What to take away

Given the angles with the axes, the direction cosines are just their cosines, in the order $x$ ⁠, $y$ ⁠, $z$ ⁠.
Always check $l^2 + m^2 + n^2 = 1$ ⁠; three angles that fail it belong to no line.
A direction cosine 0 means the line is perpendicular to that axis.

Every fact used, and where it comes from

Fact	Used in	Source
Direction angles and $l = \cos\alpha$ ⁠, $m = \cos\beta$ ⁠, $n = \cos\gamma$	Steps 1, 2	Section 11.2; Section 10.2 (Basics)
Reversing a line replaces each angle by its supplement and flips the signs	Step 1, Result	Section 11.2
$\cos 90^\circ = 0$ ⁠, $\cos 60^\circ = \frac12$ ⁠, $\cos 30^\circ = \frac{\sqrt3}{2}$	Step 2	Trigonometry (Class XI)
$l^2 + m^2 + n^2 = 1$	Step 3	Section 11.2 (Basics)
$l\hat i + m\hat j + n\hat k$ is a unit vector along the line	Step 4	Section 10.5.1 (Lesson)

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Example 1: Direction cosines of a line at 90∘90^\circ90∘⁠, 60∘60^\circ60∘⁠, 30∘30^\circ30∘ to the axes

Basics you need

Solution

Name the direction angles

Take the cosines

Check that the squares add up to 1

See where the line points

Every fact used, and where it comes from

Example 1: Direction cosines of a line at $90^\circ$ ⁠, $60^\circ$ ⁠, $30^\circ$ to the axes