NDA Maths · 3D Geometry

Direction Cosines & Direction Ratios

The numbers that capture a line's direction in space — direction cosines are the unit version (squaring to 1), direction ratios are any proportional set.

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Why this matters

Twenty-four PYQs across 2017–2026 — the engine room of the chapter. Almost every line, plane, and angle question reduces to direction cosines and ratios. The identity l² + m² + n² = 1 and the angle-between-lines formula together unlock the bulk of the bank. Difficulty here runs a touch higher (25% HARD), so the seven concepts below earn their keep.

Test yourself — a quick 12-question 3D Geometry recall quiz, instant score.

Concept 1 of 7: Direction ratios, direction cosines, and the unit identity

A line's direction can be described by any vector along it — those components are its DIRECTION RATIOS, and there are infinitely many proportional sets. Normalise that vector to unit length and the components become the DIRECTION COSINES

\langle l, m, n\rangle

— the cosines of the angles the line makes with the three axes. Because they're a unit vector's components, they always satisfy

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

Definition

If a line has direction ratios $\langle a, b, c\rangle$ , its direction cosines are obtained by dividing by the magnitude: $l = a/\sqrt{a^2+b^2+c^2}$ , and similarly for $m, n$ . They are the cosines of the angles $\alpha, \beta, \gamma$ the line makes with the positive $x$ -, $y$ -, $z$ -axes. The defining identity is below.

Direction cosines square-sum to 1

l^2 + m^2 + n^2 = 1, \quad l = \cos\alpha,\ m = \cos\beta,\ n = \cos\gamma

$l,m,n$ direction cosines
$\alpha,\beta,\gamma$ angles with the x, y, z axes

Diagram · direction cosines (drag to rotate)

α ≈ 49° · l = 0.66β ≈ 62° · m = 0.48γ ≈ 54° · n = 0.58

l, m, n are the cosines of the angles r makes with the x-, y-, z-axes — and the components of the unit vector along r. So l² + m² + n² = 1.00 = 1, always.

Worked example

Find the direction cosines of the line with direction ratios

\langle 2, -1, 2\rangle

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2020 · Q63Easy

Example 1 · 3D Geometry · Direction Cosines and Ratios

If a line has direction ratios

\langle a+b, b+c, c+a\rangle

, then what is the sum of the squares of its direction cosines?

Direction RATIOS are not unique; direction COSINES (almost) are

\langle 2,-1,2\rangle

and

\langle 4,-2,4\rangle

are the same direction. Only after normalising do you get direction cosines — and even then a line has TWO sets (

\pm

) for its two orientations. Sum of squares of ratios is NOT 1; only the cosines satisfy that.

You must NORMALISE direction ratios into cosines — and the identity is = 1, not = 0

Direction ratios

\langle a,b,c\rangle

are NOT direction cosines until you divide each by the magnitude

\sqrt{a^2+b^2+c^2}

. Treating raw ratios like

\langle 2,-1,2\rangle

as cosines is wrong — their square-sum is

9

, not

1

. And the defining identity is

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

(a UNIT vector), never

= 0

;

= 0

would force all three to vanish.

Drill 4 more on direction ratios, direction cosines, and the unit identity

Concept 2 of 7: Direction cosines of the axes and special lines

The axes themselves are the simplest directions. The x-axis points purely along

x

, so its direction cosines are

\langle 1, 0, 0\rangle

— and a line perpendicular to an axis has a 0 in that slot. Memorise this tiny table and the recall questions become instant.

Definition

A line parallel to an axis has that axis's direction cosines; a line perpendicular to an axis has a zero in the corresponding component. A line in (or parallel to) the XY-plane is perpendicular to the z-axis, so $n = 0$ .

Line	Direction cosines	Note
x-axis	$\langle 1, 0, 0 \rangle$	makes 0° with x, 90° with y and z
y-axis	$\langle 0, 1, 0 \rangle$	DCs $\langle 0,1,0\rangle$ ; DRs e.g. $\langle 0,4,0\rangle$
z-axis	$\langle 0, 0, 1 \rangle$	perpendicular to the whole XY-plane
⊥ to z-axis	$n = 0$ , e.g. $\langle 5, 6, 0\rangle$	lies in / parallel to the XY-plane A line perpendicular to the z-axis just needs its z-component zero — the x, y parts are free.

Parallel to an axis → that axis's DCs. Perpendicular to an axis → a zero in that slot.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2019 · Q65Easy

Example 2 · 3D Geometry · Direction Cosines and Ratios

What are the direction cosines of z-axis?

Drill 2 more on direction cosines of the axes and special lines

Concept 3 of 7: Reading direction ratios off a line

When a line is given in symmetric form

\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}

, the denominators

\langle a, b, c\rangle

ARE the direction ratios. The catch: the form must have coefficient 1 on each variable in the numerator — rewrite things like

2(y+3)

1 - z

first, watching the sign.

Definition

Put the line into true symmetric form (each numerator $x - x_0$ , coefficient 1). The denominators are the direction ratios. A coordinates-of-a-point form $(x_0 + at,\ y_0 + bt,\ z_0 + ct)$ gives direction ratios $\langle a, b, c\rangle$ directly (the coefficients of the parameter). Normalise to get direction cosines.

Worked example

Find the direction cosines of the line

x - 1 = 2(y+3) = 1 - z

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q56Moderate

Example 3 · 3D Geometry · Direction Cosines and Ratios

A point on a line has coordinates

(p+1,\, p-3,\, \sqrt{2}p)

where

p

is any real number. What are the direction cosines of the line ?

Mind the coefficient and the sign before reading denominators

2(y+3)

is NOT denominator 2 — it's

\frac{y+3}{1/2}

, so the ratio component is

\tfrac12

. And

1 - z

flips the sign:

\frac{z-1}{-1}

. Read symmetric form only after each variable has coefficient

+1

Drill 3 more on reading direction ratios off a line

Concept 4 of 7: Angle between two lines

Two lines' directions are vectors; the angle between them comes straight from the dot product. Use direction ratios in the numerator and the product of magnitudes below — the same

\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a||\vec b|}

you know from vectors.

Definition

For lines with direction ratios $\langle a_1,b_1,c_1\rangle$ and $\langle a_2,b_2,c_2\rangle$ , the acute angle $\theta$ between them satisfies the formula below. With direction cosines the denominator is 1, so $\cos\theta = l_1l_2 + m_1m_2 + n_1n_2$ .

Angle between two lines (direction ratios)

\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}

Diagram · angle between two lines (drag to rotate)

d₁ = ⟨2, 2, 1⟩, d₂ = ⟨2, −1, 2⟩ · cos θ = |d₁·d₂| / (|d₁||d₂|) = 4/9 · θ ≈ 64°.

Worked example

Find the angle between the two lines with direction ratios

\langle 1, 1, 0\rangle

and

\langle 0, 1, 1\rangle

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2021 · Q66Moderate

Example 4 · 3D Geometry · Direction Cosines and Ratios

What is the angle between the two lines having direction ratios

\langle 6, 3, 6\rangle

and

\langle 3, 3, 0\rangle

For two LINES use their DIRECTIONS, not normals; divide by BOTH magnitudes

The angle between two lines comes from the dot product of their DIRECTION ratios

\cos\theta = \dfrac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}

. Two common slips: (1) forgetting to divide by the product of the magnitudes (the bare dot product is only

\cos\theta

when both are unit vectors / true direction cosines), and (2) borrowing a plane's normal — that's for plane angles. Lines: directions. Planes: normals.

Drill 3 more on angle between two lines

Concept 5 of 7: Perpendicular and parallel conditions

Two lines are parallel when their direction ratios are proportional, and perpendicular when their dot product is zero. These two one-line tests answer a surprising share of the bank — and a line perpendicular to two given lines has direction ratios given by their cross product.

Definition

Lines $\langle a_1,b_1,c_1\rangle$ and $\langle a_2,b_2,c_2\rangle$ are:

Parallel iff $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ .
Perpendicular iff $a_1a_2 + b_1b_2 + c_1c_2 = 0$ .

A line perpendicular to both has direction ratios equal to their cross product $\langle a_1,b_1,c_1\rangle \times \langle a_2,b_2,c_2\rangle$ .

Line of intersection of two planes: it lies in both planes, so it is perpendicular to BOTH normals — its direction ratios are $\vec{n_1} \times \vec{n_2}$ (cross product of the plane normals).

Perpendicularity condition

a_1 a_2 + b_1 b_2 + c_1 c_2 = 0

Worked example

Find the direction ratios of a line perpendicular to both

\langle 1, 2, 1\rangle

and

\langle 2, -1, 1\rangle

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

NDA · Apr 2017 · Q65Moderate

Example 5 · 3D Geometry · Direction Cosines and Ratios

Under which one of the following conditions are the lines

x=ay+b;\; z=cy+d

and

x=ey+f;\; z=gy+h

perpendicular?

Drill 4 more on perpendicular and parallel conditions

Concept 6 of 7: Projection of a segment on an axis or line

The projection of a segment onto an axis is simply how far it reaches along that axis — the difference of the relevant coordinates. Onto a general line, it's the segment's length times the cosine of the angle, which equals the dot product of the segment vector with the line's direction cosines.

Definition

Projection of $\overrightarrow{AB}$ on the x-axis is $x_2 - x_1$ (and similarly for y, z). Projection on a line with direction cosines $\langle l, m, n\rangle$ is $(x_2-x_1)l + (y_2-y_1)m + (z_2-z_1)n$ .

Projection of AB on a line of direction cosines ⟨l, m, n⟩

\text{proj} = (x_2-x_1)\,l + (y_2-y_1)\,m + (z_2-z_1)\,n

Worked example

Find the projection of the segment joining

A(2, -1, 4)

and

B(7, 3, 1)

on the x-axis.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2021 · Q68Easy

Example 6 · 3D Geometry · Direction Cosines and Ratios

What is the projection of the line segment joining

A(1,7,-5)

and

B(-3,4,-2)

y

-axis?

Concept 7 of 7: Direction-angle identities

Because

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

, a chain of useful identities follows for the angles a line makes with the axes — for example the sines square-sum to 2, and

\cos2\alpha + \cos2\beta + \cos2\gamma = -1

. These power the chapter's hardest one-liners.

Definition

From $l^2+m^2+n^2 = 1$ with $l=\cos\alpha$ etc.:

$\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$ (subtract the cosine identity from 3).
$\cos2\alpha + \cos2\beta + \cos2\gamma = -1$ (use $\cos2\theta = 2\cos^2\theta - 1$ ).
Product form: $\cos(\alpha+\beta)\cos(\alpha-\beta) = \cos^2\alpha - \sin^2\beta$ .

Core identities

\sum \cos^2\!\theta = 1, \quad \sum \sin^2\!\theta = 2, \quad \sum \cos 2\theta = -1

Worked example

A line makes angles

\alpha, \beta, \gamma

with the axes. Show

\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2025 · Q61Hard

Example 7 · 3D Geometry · Direction Cosines and Ratios

If a line in 3 dimensions makes angles

\alpha, \beta

and

\gamma

with the positive directions of the coordinate axes, then what is

\cos(\alpha+\beta)\cos(\alpha-\beta)

equal to?

A line cannot make equal acute angles with all three axes unless it's the diagonal

\alpha=\beta=\gamma

, then

3\cos^2\alpha = 1

, so

\cos\alpha = \tfrac{1}{\sqrt3}

(

\approx 54.7°

), NOT

45°

60°

. Options offering 45°/60° for the equal-angle case are distractors.

Drill 2 more on direction-angle identities

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Direction ratios, direction cosines, and the unit identity
Direction cosines square-sum to 1
$l^2 + m^2 + n^2 = 1, \quad l = \cos\alpha,\ m = \cos\beta,\ n = \cos\gamma$
Angle between two lines
Angle between two lines (direction ratios)
$\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}$
Perpendicular and parallel conditions
Perpendicularity condition
$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$
Projection of a segment on an axis or line
Projection of AB on a line of direction cosines ⟨l, m, n⟩
$\text{proj} = (x_2-x_1)\,l + (y_2-y_1)\,m + (z_2-z_1)\,n$
Direction-angle identities
Core identities
$\sum \cos^2\!\theta = 1, \quad \sum \sin^2\!\theta = 2, \quad \sum \cos 2\theta = -1$

Reference tables (1)

Direction cosines of the axes and special lines4 rows

Line	Direction cosines	Note
x-axis	$\langle 1, 0, 0 \rangle$	makes 0° with x, 90° with y and z
y-axis	$\langle 0, 1, 0 \rangle$	DCs $\langle 0,1,0\rangle$ ; DRs e.g. $\langle 0,4,0\rangle$
z-axis	$\langle 0, 0, 1 \rangle$	perpendicular to the whole XY-plane
⊥ to z-axis	$n = 0$ , e.g. $\langle 5, 6, 0\rangle$	lies in / parallel to the XY-plane A line perpendicular to the z-axis just needs its z-component zero — the x, y parts are free.

Parallel to an axis → that axis's DCs. Perpendicular to an axis → a zero in that slot.

Watch out for (5)

Direction RATIOS are not unique; direction COSINES (almost) are
→ Direction ratios, direction cosines, and the unit identity
You must NORMALISE direction ratios into cosines — and the identity is = 1, not = 0
→ Direction ratios, direction cosines, and the unit identity
Mind the coefficient and the sign before reading denominators
→ Reading direction ratios off a line
For two LINES use their DIRECTIONS, not normals; divide by BOTH magnitudes
→ Angle between two lines
A line cannot make equal acute angles with all three axes unless it's the diagonal
→ Direction-angle identities

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