Three Dimensional Geometry · Lesson 1 of 5

Directions and Coordinates in Space

“Use coordinate changes and vectors to describe the direction of a line in three dimensions.”

Learning Objectives

• Explain direction angles, direction cosines, and direction ratios of a directed line. • Normalize a nonzero direction vector and handle reversal of direction consistently. • Derive direction cosines from the coordinates of two distinct points. • Use proportional direction ratios to recognise collinear points.

From points to directions

A point tells us where something is, but a line also needs a direction. In a room, the same direction can be followed from the floor, a wall, or the ceiling. Vector algebra gives us a compact way to describe that direction independently of where the line is placed.

We use three mutually perpendicular coordinate axes, with unit vectors $\hat{i}$ , $\hat{j}$ , and $\hat{k}$ pointing along their positive directions. A displacement $a \hat{i} + b \hat{j} + c \hat{k}$ changes the three coordinates by $a, b, c$ . Its components can therefore describe the direction of travel along a line. The chapter builds from this idea to line equations, angles, and distances.

A geometric viewpoint

Mathematical imagination works alongside reasoning. Picture a direction or a perpendicular connector before calculating it, then justify the relationship using coordinates and vectors. The diagrams support the calculations; they do not replace them.

Three angles for one directed line

First place the line through the origin and choose which way along it is positive. Imagine an arrow pointing from the origin in that direction. We compare this arrow with the positive direction of each coordinate axis, so even an angle greater than a right angle can carry useful information.

Definition

Direction angles and direction cosines

If the chosen direction makes angles $α, β, γ$ with the positive $x$ -, $y$ -, and $z$ -axes, these are its direction angles. Their cosines $l = cos α$ , $m = cos β$ , and $n = cos γ$ are its direction cosines. Each angle lies between $0$ and $π$ .

A direction compared with three axes— The dashed projections separate coordinate components; the true angles are measured in space.

A positive direction cosine means that the arrow has a positive component along that axis. A negative one means that it points partly against the positive axis. A zero means there is no change in that coordinate: the direction is perpendicular to that axis. Thus direction cosines tell us more than three angle sizes; they also encode the signs of the motion.

Let a point on the chosen ray have coordinates $(x, y, z)$ and be a distance $R > 0$ from the origin. Projection onto the axes gives $x = R l$ , $y = R m$ , and $z = R n$ . Since $x^{2} + y^{2} + z^{2} = R^{2}$ , dividing by $R^{2}$ produces a restriction on the three direction cosines.

The direction-cosine identityLaTeX

l^{2} + m^{2} + n^{2} = 1

Here

l, m, n

are the components of a unit vector along the chosen direction. Their squares add to one, so three direction angles cannot be chosen independently.

Example — Read direction from angles

Problem
A line is directed at $9 0^{\circ}, 6 0^{\circ}, 3 0^{\circ}$ to the positive coordinate axes. Find its direction cosines.

1.Take the cosine of each stated angle: $l = cos 9 0^{\circ} = 0$ , $m = cos 6 0^{\circ} = 1/2$ , and $n = cos 3 0^{\circ} = 3 /2$ .
2.Check the identity: $0^{2} + (1/2)^{2} + (3 /2)^{2} = 0 + 1/4 + 3/4 = 1$ .
3.The triple is $(0, 1/2, 3 /2)$ . The zero first component means that travel along this line leaves the $x$ -coordinate unchanged.

Example — The coordinate axes

Problem
Find the direction cosines of the positive coordinate axes.

1.The positive $x$ -axis makes angles $0^{\circ}, 9 0^{\circ}, 9 0^{\circ}$ with the three positive axes. Its direction cosines are $(1, 0, 0)$ .
2.The same reasoning gives $(0, 1, 0)$ for the positive $y$ -axis and $(0, 0, 1)$ for the positive $z$ -axis.
3.For a negative coordinate-axis direction, reverse the corresponding nonzero component. For example, the negative $x$ -axis has direction cosines $(- 1, 0, 0)$ .

If the original line does not pass through the origin, draw a parallel line through the origin and orient it the same way. Parallel arrows have the same direction angles. Moving a line without rotating it therefore preserves its direction cosines.

Reverse all components together

Reversing the arrow replaces each direction angle by its supplement. Since $cos (π - α) = - cos α$ , the triple $(l, m, n)$ becomes $(- l, - m, - n)$ . These describe opposite directions on the same undirected line. Changing just one nonzero sign usually gives a different line direction.

Direction ratios and unit length

Angles are not always the easiest starting point. Often we already know a vector parallel to the line. Any nonzero scalar multiple of that vector points along the same undirected line, so its components need not have unit length.

Definition

Direction ratios

Any nonzero triple $(a, b, c)$ proportional to the direction-cosine triple is a set of direction ratios of the line. Direction ratios are also called direction numbers. The associated vector $v = a \hat{i} + b \hat{j} + c \hat{k}$ is parallel to the line.

Suppose $(l, m, n) = k (a, b, c)$ . Substitute into the unit-length identity to get $k^{2} (a^{2} + b^{2} + c^{2}) = 1$ . Because $(a, b, c)$ is not the zero triple, its length is positive. Hence $k = \pm 1/ a^{2} + b^{2} + c^{2}$ . The sign selects one of the two orientations.

Normalize direction ratiosLaTeX

(l, m, n) = \pm \frac{( a , b , c )}{a ^{2} + b ^{2} + c ^{2}}

Use the same sign for the whole triple. Dividing by the positive length gives the orientation of

(a, b, c)

; the negative gives its reverse.

For example, $(1, 2, 3)$ , $(2, 4, 6)$ , and $(- 1, - 2, - 3)$ are direction ratios for the same undirected direction. There are infinitely many such triples. They are proportional, but only the normalized triples are direction cosines. Proportionality is best written as one vector being a nonzero scalar multiple of another; this avoids dividing by a component that might be zero.

Example — Normalize a signed triple

Problem
A line has direction ratios $(2, - 1, - 2)$ . Find its direction cosines in that orientation.

1.The length of the direction vector is $2^{2} + (- 1)^{2} + (- 2)^{2} = 9 = 3$ .
2.Divide each component by $3$ . The required direction cosines are $(2/3, - 1/3, - 2/3)$ .
3.Their squares sum to $(4 + 1 + 4) /9 = 1$ . Reversing the orientation gives $(- 2/3, 1/3, 2/3)$ , with every sign reversed.

A zero triple gives no direction

The vector $(0, 0, 0)$ has no direction and cannot be normalized. Individual direction ratios may be zero, but all three cannot be zero. Also, do not normalize by $a + b + c$ : a vector length uses the square root of the sum of squares.

Find a direction from two points

Two distinct points provide a natural arrow: travel from the first point to the second. Each coordinate difference records the displacement along one axis. Once we know the length of this displacement, we can normalize it exactly as before.

Let $P = (x_{1}, y_{1}, z_{1})$ and $Q = (x_{2}, y_{2}, z_{2})$ . The vector $P Q$ has components $(x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1})$ . Projecting onto an axis gives, for example, the signed vertical component $z_{2} - z_{1}$ . In a right triangle, the corresponding cosine is that signed component divided by $P Q$ . The same argument works for the other two axes.

A right triangle inside a spatial displacement— The drawn triangle illustrates a positive vertical difference. Signed components handle every orientation.

Direction cosines from two pointsLaTeX

R = P Q (l, m, n) = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2} = \frac{( x _{2} - x _{1} , y _{2} - y _{1} , z _{2} - z _{1} )}{R}

The points must be distinct, so

R > 0

. This triple follows the direction from

P

Q

. Reversing their order changes all three signs.

Example — A line joining two points

Problem
Find direction cosines from $P = (- 2, 4, - 5)$ towards $Q = (1, 2, 3)$ .

1.Subtract the coordinates of $P$ from those of $Q$ : $P Q = (1 + 2, 2 - 4, 3 + 5) = (3, - 2, 8)$ .
2.Its length is $P Q = 3^{2} + (- 2)^{2} + 8^{2} = 77$ .
3.The direction cosines are $(3/ 77, - 2/ 77, 8/ 77)$ . The line moves positively in $x$ and $z$ and negatively in $y$ in this orientation.

Proportional displacements and collinearity

When three points lie on a straight line, travelling between any two of them follows the same undirected direction. This suggests comparing displacement vectors. The shared point is essential: two separate parallel lines need not be the same line.

Example — Prove three points are collinear

Problem
Show that $A = (2, 3, - 4)$ , $B = (1, - 2, 3)$ , and $C = (3, 8, - 11)$ lie on one line.

1.Compute $A B = (- 1, - 5, 7)$ and $B C = (2, 10, - 14)$ .
2.Every component of $B C$ is $- 2$ times the corresponding component of $A B$ . Thus $B C = - 2 A B$ .
3.The two displacements are parallel and both lines contain $B$ . They therefore belong to the same line, proving collinearity. The negative factor says that the two chosen arrows point in opposite directions.

Example — Check a proposed set of angles

Problem
Can a line make an angle of $6 0^{\circ}$ with each positive coordinate axis?

1.Those angles would give direction cosines $(1/2, 1/2, 1/2)$ .
2.Their squared sum is $3/4$ , not $1$ . Therefore no such line exists.
3.For equal direction angles, write $l = m = n = t$ . Then $3 t^{2} = 1$ , so $t = \pm 1/ 3$ . The two signs describe opposite orientations with equal angles to all three positive axes.

Check your understanding

Distinguish a direction from a location, and a proportional triple from a unit triple. In the questions below, every stated direction is a directed one unless an undirected line is mentioned.

Quick check

Which triple can be a set of direction cosines?

Quick check

Direction ratios are $(0, - 6, 8)$ . Which normalized triple has the same orientation?

Quick check

What happens to direction cosines when a directed line is reversed?

Quick check

For $P = (1, 2, 3)$ and $Q = (1, - 1, 7)$ , which is a direction vector from $P$ to $Q$ ?

Quick check

A line has $l = 0$ . What follows?

Quick check

For distinct $A, B, C$ , which observation proves collinearity?

Practice Problems

Find direction cosines for direction angles $9 0^{\circ}, 13 5^{\circ}, 4 5^{\circ}$ .
Find the possible direction cosines when a directed line makes equal angles with the three positive axes.
Normalize the direction ratios $(- 18, 12, - 4)$ , keeping their orientation.
Prove that $(2, 3, 4)$ , $(- 1, - 2, 1)$ , and $(5, 8, 7)$ are collinear.
For $A = (3, 5, - 4)$ , $B = (- 1, 1, 2)$ , and $C = (- 5, - 5, - 2)$ , find direction cosines along $A B$ , $B C$ , and $C A$ .
Find direction cosines from $P = (2, - 1, 4)$ towards $Q = (2, 2, 0)$ .
Explain why $(2, 4, - 4)$ and $(- 3, - 6, 6)$ describe the same undirected direction.
Can a line have direction cosines $(2/3, 2/3, 1/3)$ ? Can it have $(2/3, 2/3, 2/3)$ ? Justify both answers.

Take the cosines: $(0, - 1/ 2, 1/ 2)$ . The squared sum is one.

Key Takeaways

• Direction cosines are the components of a unit direction vector, so $l^{2} + m^{2} + n^{2} = 1$ . • Direction ratios are any nonzero proportional triple; normalization produces direction cosines. • Reversing orientation changes all three signs together. • Two distinct points give a direction by coordinate subtraction and a length by the distance formula. • Proportional displacements through a common point establish collinearity.

Start of chapter

Next · Lesson 2

Representing a Line in Space

From points to directions

Three angles for one directed line

Direction ratios and unit length

Find a direction from two points

Proportional displacements and collinearity

Check your understanding

Quiz

Practice Problems

Key Takeaways