Chapter 11 · Three Dimensional Geometry

Example 4: Direction cosines of the $x$ -, $y$ - and $z$ -axes

ExamplesDirection cosines · Section 11.2

The question

Find the direction cosines of $x$ ⁠, $y$ and $z$ -axis.

Exam answer

The $x$ -axis makes angles $0^\circ, 90^\circ, 90^\circ$ with the $x$ ⁠, $y$ and $z$ -axes. So its direction cosines are

\begin{aligned} &\cos 0^\circ, \cos 90^\circ, \cos 90^\circ \\ &\text{i.e. } 1, 0, 0 \end{aligned}

Similarly, the $y$ -axis makes angles $90^\circ, 0^\circ, 90^\circ$ and the $z$ -axis makes $90^\circ, 90^\circ, 0^\circ$ with the axes.

$\therefore$ The direction cosines of the $x$ ⁠, $y$ and $z$ -axes are $1, 0, 0$ ⁠; $0, 1, 0$ and $0, 0, 1$ ⁠.

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

The one idea this problem is built on

An axis is itself a line, so it has direction cosines like any other line: the cosines of the angles it makes with the three positive axes. An axis makes $0^\circ$ with itself and $90^\circ$ with the other two, and $\cos 0^\circ = 1$ ⁠, $\cos 90^\circ = 0$ ⁠.

Given, to find, and the plan

Given: The three coordinate axes, taken in their positive directions.
To find: The direction cosines of each axis.
Plan: (1) Find the angles the $x$ -axis makes with the three axes and take cosines. (2) Do the same for the $y$ and $z$ -axes. (3) Check with $l^2 + m^2 + n^2 = 1$ and with the unit vectors $\hat i, \hat j, \hat k$ ⁠.

Basics you need

1Direction cosines

For a directed line making angles $\alpha, \beta, \gamma$ with the positive $x$ ⁠, $y$ and $z$ -axes, the direction cosines are $\cos\alpha, \cos\beta, \cos\gamma$ (Section 11.2). Basics: direction cosines

2The axes are mutually perpendicular

In the rectangular coordinate system each axis is at $90^\circ$ to the other two. Basics: coordinates in 3D

3Two cosines

$\cos 0^\circ = 1$ and $\cos 90^\circ = 0$ ⁠.

Solution

The $x$ -axis

The positive $x$ -axis makes $0^\circ$ with the $x$ -axis, and $90^\circ$ with each of the $y$ and $z$ -axes. So its direction cosines are
$\cos 0^\circ,\ \cos 90^\circ,\ \cos 90^\circ, \quad\text{i.e. } 1, 0, 0.$
A directed line makes the angle $0^\circ$ with a line pointing the same way, so the $x$ -axis makes $0^\circ$ with the positive $x$ -axis. The other two axes are perpendicular to it, which is how the coordinate system is built. We take the positive direction of the axis; that is what "the $x$ -axis" means here.

Direction cosines are the cosines of the direction angles. Basics: direction cosines
The $y$ and $z$ -axes
$\begin{aligned} y\text{-axis:}\quad &\cos 90^\circ, \cos 0^\circ, \cos 90^\circ \\ &= 0, 1, 0, \\ z\text{-axis:}\quad &\cos 90^\circ, \cos 90^\circ, \cos 0^\circ \\ &= 0, 0, 1. \end{aligned}$
The same reasoning: each axis makes $0^\circ$ with itself and $90^\circ$ with the other two. Only the position of the 1 moves, because the order is always (angle with $x$ ⁠, angle with $y$ ⁠, angle with $z$ ⁠).

The axes are mutually perpendicular. Basics: coordinates in 3D
Check and see them

Each set passes $l^2 + m^2 + n^2 = 1$ ⁠, for example $1^2 + 0^2 + 0^2 = 1$ ⁠. And $l\hat i + m\hat j + n\hat k$ gives $\hat i$ ⁠, $\hat j$ and $\hat k$ ⁠, the unit vectors along the axes.

The unit vectors along the axes in an oblique projection (not to scale: the $x$ -axis is drawn shortened and slanting). Their components $(1, 0, 0)$ ⁠, $(0, 1, 0)$ ⁠, $(0, 0, 1)$ are the direction cosines of the axes.

The unit vector along a line is $l\hat i + m\hat j + n\hat k$ ⁠, so a line whose unit vector is $\hat i$ must have direction cosines $1, 0, 0$ ⁠. This also gives the direction cosines of the negative axes: the negative $x$ -axis is the $x$ -axis reversed, so it has $-1, 0, 0$ ⁠.

$l\hat i + m\hat j + n\hat k$ is the unit vector along the line. Lesson: direction cosines

Result

The direction cosines are $1, 0, 0$ for the $x$ -axis, $0, 1, 0$ for the $y$ -axis and $0, 0, 1$ for the $z$ -axis.

(Textbook answer: $1, 0, 0$ ⁠; $0, 1, 0$ ⁠; $0, 0, 1$ ⁠.)

Check it with numbers

Take a point on the $x$ -axis, $Q(5, 0, 0)$ ⁠, and the origin $O(0, 0, 0)$ ⁠. The direction ratios of $OQ$ are $5, 0, 0$ ⁠, and $OQ = \sqrt{25 + 0 + 0} = 5$ ⁠, so the direction cosines are $\frac55, \frac05, \frac05 = 1, 0, 0$ ⁠, as found.

Any two axes are perpendicular: for the $x$ and $y$ -axes, $1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0$ ⁠.

Common mistakes

Writing $0, 1, 1$ for the $x$ -axis. That uses $\cos 90^\circ = 1$ ⁠; in fact $\cos 90^\circ = 0$ and $\cos 0^\circ = 1$ ⁠.
Putting the 1 in the wrong place. The order is always $x$ ⁠, $y$ ⁠, $z$ ⁠: the $y$ -axis is $0, 1, 0$ ⁠.
Forgetting the direction. "The $x$ -axis" means the positive direction; the negative $x$ -axis has $-1, 0, 0$ ⁠.

What to take away

The axes have direction cosines $1, 0, 0$ ⁠; $0, 1, 0$ ⁠; $0, 0, 1$ ⁠: the components of $\hat i, \hat j, \hat k$ ⁠.
A direction cosine is 1 only when the line points exactly along that axis, and 0 when the line is perpendicular to it.

Every fact used, and where it comes from

Fact	Used in	Source
Direction cosines $\cos\alpha, \cos\beta, \cos\gamma$	Steps 1, 2	Section 11.2 (Basics)
The coordinate axes are mutually perpendicular	Steps 1, 2	Rectangular coordinate system, Section 10.2 (Basics)
$\cos 0^\circ = 1$ ⁠, $\cos 90^\circ = 0$	Steps 1, 2	Trigonometry (Class XI)
$l^2 + m^2 + n^2 = 1$ ⁠; $l\hat i + m\hat j + n\hat k$ is a unit vector	Step 3	Sections 11.2 and 10.5.1 (Lesson)

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Example 4: Direction cosines of the xxx-, yyy- and zzz-axes

Basics you need

Solution

The xxx-axis

The yyy and zzz-axes

Check and see them

Every fact used, and where it comes from

Example 4: Direction cosines of the $x$ -, $y$ - and $z$ -axes

The $x$ -axis

The $y$ and $z$ -axes