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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.

Concept 1 of 7

Direction ratios, direction cosines, and the unit identity

Intuition

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 ⟨l,m,n⟩\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 l2+m2+n2=1l^2 + m^2 + n^2 = 1.

Definition

If a line has direction ratios ⟨a,b,c⟩\langle a, b, c\rangle, its direction cosines are obtained by dividing by the magnitude: l=a/a2+b2+c2l = a/\sqrt{a^2+b^2+c^2}, and similarly for m,nm, n. They are the cosines of the angles α,β,γ\alpha, \beta, \gamma the line makes with the positive xx-, yy-, zz-axes. The defining identity is below.

Direction cosines square-sum to 1

l2+m2+n2=1,l=cos⁡α, m=cos⁡β, n=cos⁡γl^2 + m^2 + n^2 = 1, \quad l = \cos\alpha,\ m = \cos\beta,\ n = \cos\gamma
  • l,m,nl,m,ndirection cosines
  • α,β,γ\alpha,\beta,\gammaangles with the x, y, z axes

Diagram · direction cosines (drag to rotate)

xyzr
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 ⟨2,−1,2⟩\langle 2, -1, 2\rangle.
  1. Magnitude: 22+(−1)2+22=4+1+4=9=3\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3.
  2. Divide each ratio by 3: l=23, m=−13, n=23l = \tfrac{2}{3},\ m = -\tfrac{1}{3},\ n = \tfrac{2}{3}.
  3. Check: 49+19+49=99=1\tfrac{4}{9} + \tfrac{1}{9} + \tfrac{4}{9} = \tfrac{9}{9} = 1. ✓
Answer:⟨23,−13,23⟩\left\langle \tfrac{2}{3}, -\tfrac{1}{3}, \tfrac{2}{3} \right\rangle.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 13D GeometryEASY
If a line has direction ratios ⟨a+b,b+c,c+a⟩\langle a+b, b+c, c+a\rangle, then what is the sum of the squares of its direction cosines?

[Q63 · Apr · 2020]

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

⟨2,−1,2⟩\langle 2,-1,2\rangle and ⟨4,−2,4⟩\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 ⟨a,b,c⟩\langle a,b,c\rangle are NOT direction cosines until you divide each by the magnitude a2+b2+c2\sqrt{a^2+b^2+c^2}. Treating raw ratios like ⟨2,−1,2⟩\langle 2,-1,2\rangle as cosines is wrong — their square-sum is 99, not 11. And the defining identity is l2+m2+n2=1l^2 + m^2 + n^2 = 1 (a UNIT vector), never =0= 0; =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

Intuition

The axes themselves are the simplest directions. The x-axis points purely along xx, so its direction cosines are ⟨1,0,0⟩\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=0n = 0.

LineDirection cosinesNote
x-axis⟨1,0,0⟩\langle 1, 0, 0 \ranglemakes 0° with x, 90° with y and z
y-axis⟨0,1,0⟩\langle 0, 1, 0 \rangleDCs ⟨0,1,0⟩\langle 0,1,0\rangle; DRs e.g. ⟨0,4,0⟩\langle 0,4,0\rangle
z-axis⟨0,0,1⟩\langle 0, 0, 1 \rangleperpendicular to the whole XY-plane
⊥ to z-axisn=0n = 0, e.g. ⟨5,6,0⟩\langle 5, 6, 0\ranglelies 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

From the bank · past-year question

Example 23D GeometryEASY
What are the direction cosines of z-axis?

[Q65 · Apr · 2019]

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

Concept 3 of 7

Reading direction ratios off a line

Intuition

When a line is given in symmetric form x−x0a=y−y0b=z−z0c\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}, the denominators ⟨a,b,c⟩\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)2(y+3) or 1−z1 - z first, watching the sign.

Definition

Put the line into true symmetric form (each numerator x−x0x - x_0, coefficient 1). The denominators are the direction ratios. A coordinates-of-a-point form (x0+at, y0+bt, z0+ct)(x_0 + at,\ y_0 + bt,\ z_0 + ct) gives direction ratios ⟨a,b,c⟩\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−zx - 1 = 2(y+3) = 1 - z.
  1. Rewrite each piece as var−constcoeff\frac{\text{var} - \text{const}}{\text{coeff}}: set each equal to tt.
  2. x−1=tx - 1 = t; 2(y+3)=t⇒y+31/2=t2(y+3) = t \Rightarrow \frac{y+3}{1/2} = t; 1−z=t⇒z−1−1=t1 - z = t \Rightarrow \frac{z-1}{-1} = t.
  3. Direction ratios: ⟨1, 12, −1⟩\left\langle 1,\ \tfrac12,\ -1 \right\rangle, or clearing fractions ⟨2,1,−2⟩\langle 2, 1, -2\rangle.
  4. Magnitude 4+1+4=3\sqrt{4+1+4} = 3 → direction cosines ⟨23,13,−23⟩\left\langle \tfrac23, \tfrac13, -\tfrac23\right\rangle.
Answer:⟨23,13,−23⟩\left\langle \tfrac23, \tfrac13, -\tfrac23 \right\rangle.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 33D GeometryMODERATE
A point on a line has coordinates (p+1, p−3, 2p)(p+1,\, p-3,\, \sqrt{2}p) where pp is any real number. What are the direction cosines of the line ?

[Q56 · Sep · 2019]

Mind the coefficient and the sign before reading denominators

2(y+3)2(y+3) is NOT denominator 2 — it's y+31/2\frac{y+3}{1/2}, so the ratio component is 12\tfrac12. And 1−z1 - z flips the sign: z−1−1\frac{z-1}{-1}. Read symmetric form only after each variable has coefficient +1+1.
Drill 3 more on reading direction ratios off a line

Concept 4 of 7

Angle between two lines

Intuition

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⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a||\vec b|} you know from vectors.

Definition

For lines with direction ratios ⟨a1,b1,c1⟩\langle a_1,b_1,c_1\rangle and ⟨a2,b2,c2⟩\langle a_2,b_2,c_2\rangle, the acute angle θ\theta between them satisfies the formula below. With direction cosines the denominator is 1, so cos⁡θ=l1l2+m1m2+n1n2\cos\theta = l_1l_2 + m_1m_2 + n_1n_2.

Angle between two lines (direction ratios)

cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12 a22+b22+c22\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)

xyzd₁d₂

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 ⟨1,1,0⟩\langle 1, 1, 0\rangle and ⟨0,1,1⟩\langle 0, 1, 1\rangle.
  1. Dot product: 1(0)+1(1)+0(1)=11(0) + 1(1) + 0(1) = 1.
  2. Magnitudes: 1+1+0=2\sqrt{1+1+0} = \sqrt2 and 0+1+1=2\sqrt{0+1+1} = \sqrt2.
  3. cos⁡θ=12⋅2=12\cos\theta = \dfrac{1}{\sqrt2 \cdot \sqrt2} = \dfrac{1}{2}.
  4. So θ=60°\theta = 60°.
Answer:θ=60°\theta = 60° (i.e. π3\tfrac{\pi}{3}).
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 43D GeometryMODERATE
What is the angle between the two lines having direction ratios ⟨6,3,6⟩\langle 6, 3, 6\rangle and ⟨3,3,0⟩\langle 3, 3, 0\rangle?

[Q66 · Apr · 2021]

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⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12 a22+b22+c22\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⁡θ\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

Intuition

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 ⟨a1,b1,c1⟩\langle a_1,b_1,c_1\rangle and ⟨a2,b2,c2⟩\langle a_2,b_2,c_2\rangle are:

  • Parallel iff a1a2=b1b2=c1c2\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}.
  • Perpendicular iff a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0.

A line perpendicular to both has direction ratios equal to their cross product ⟨a1,b1,c1⟩×⟨a2,b2,c2⟩\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 n1⃗×n2⃗\vec{n_1} \times \vec{n_2} (cross product of the plane normals).

Perpendicularity condition

a1a2+b1b2+c1c2=0a_1 a_2 + b_1 b_2 + c_1 c_2 = 0

Worked example

Find the direction ratios of a line perpendicular to both ⟨1,2,1⟩\langle 1, 2, 1\rangle and ⟨2,−1,1⟩\langle 2, -1, 1\rangle.
  1. Take the cross product ⟨1,2,1⟩×⟨2,−1,1⟩\langle 1,2,1\rangle \times \langle 2,-1,1\rangle.
  2. i:(2)(1)−(1)(−1)=3;j:−[(1)(1)−(1)(2)]=−(−1)=1;k:(1)(−1)−(2)(2)=−5i: (2)(1) - (1)(-1) = 3;\quad j: -[(1)(1) - (1)(2)] = -(-1) = 1;\quad k: (1)(-1) - (2)(2) = -5.
  3. Direction ratios: ⟨3,1,−5⟩\langle 3, 1, -5\rangle.
  4. Check perpendicular to the first: 3+2−5=03 + 2 - 5 = 0 ✓.
Answer:⟨3,1,−5⟩\langle 3, 1, -5\rangle.
Practice this conceptself-check · 5 quick reps

From the bank · past-year question

Example 53D GeometryMODERATE
Under which one of the following conditions are the lines x=ay+b;  z=cy+dx=ay+b;\; z=cy+d and x=ey+f;  z=gy+hx=ey+f;\; z=gy+h perpendicular?

[Q65 · Apr · 2017]

Drill 4 more on perpendicular and parallel conditions

Concept 6 of 7

Projection of a segment on an axis or line

Intuition

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 AB→\overrightarrow{AB} on the x-axis is x2−x1x_2 - x_1 (and similarly for y, z). Projection on a line with direction cosines ⟨l,m,n⟩\langle l, m, n\rangle is (x2−x1)l+(y2−y1)m+(z2−z1)n(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⟩

proj=(x2−x1) l+(y2−y1) m+(z2−z1) 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)A(2, -1, 4) and B(7,3,1)B(7, 3, 1) on the x-axis.
  1. Projection on the x-axis = difference of x-coordinates.
  2. x2−x1=7−2=5x_2 - x_1 = 7 - 2 = 5.
Answer:55 (length 55 along the x-axis).
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 63D GeometryEASY
What is the projection of the line segment joining A(1,7,−5)A(1,7,-5) and B(−3,4,−2)B(-3,4,-2) on yy-axis?

[Q68 · Apr · 2021]

Concept 7 of 7

Direction-angle identities

Intuition

Because cos⁡2α+cos⁡2β+cos⁡2γ=1\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 cos⁡2α+cos⁡2β+cos⁡2γ=−1\cos2\alpha + \cos2\beta + \cos2\gamma = -1. These power the chapter's hardest one-liners.

Definition

From l2+m2+n2=1l^2+m^2+n^2 = 1 with l=cos⁡αl=\cos\alpha etc.:

  • sin⁡2α+sin⁡2β+sin⁡2γ=2\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2 (subtract the cosine identity from 3).
  • cos⁡2α+cos⁡2β+cos⁡2γ=−1\cos2\alpha + \cos2\beta + \cos2\gamma = -1 (use cos⁡2θ=2cos⁡2θ−1\cos2\theta = 2\cos^2\theta - 1).
  • Product form: cos⁡(α+β)cos⁡(α−β)=cos⁡2α−sin⁡2β\cos(\alpha+\beta)\cos(\alpha-\beta) = \cos^2\alpha - \sin^2\beta.

Core identities

∑cos⁡2 ⁣θ=1,∑sin⁡2 ⁣θ=2,∑cos⁡2θ=−1\sum \cos^2\!\theta = 1, \quad \sum \sin^2\!\theta = 2, \quad \sum \cos 2\theta = -1

Worked example

A line makes angles α,β,γ\alpha, \beta, \gamma with the axes. Show sin⁡2α+sin⁡2β+sin⁡2γ=2\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2.
  1. Each sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta.
  2. Sum: (1−cos⁡2α)+(1−cos⁡2β)+(1−cos⁡2γ)=3−(cos⁡2α+cos⁡2β+cos⁡2γ)(1-\cos^2\alpha) + (1-\cos^2\beta) + (1-\cos^2\gamma) = 3 - (\cos^2\alpha+\cos^2\beta+\cos^2\gamma).
  3. The bracket is the direction-cosine identity =1= 1.
  4. So the sum =3−1=2= 3 - 1 = 2.
Answer:sin⁡2α+sin⁡2β+sin⁡2γ=2\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 73D GeometryHARD
If a line in 3 dimensions makes angles α,β\alpha, \beta and γ\gamma with the positive directions of the coordinate axes, then what is cos⁡(α+β)cos⁡(α−β)\cos(\alpha+\beta)\cos(\alpha-\beta) equal to?

[Q61 · Apr · 2025]

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

If α=β=γ\alpha=\beta=\gamma, then 3cos⁡2α=13\cos^2\alpha = 1, so cos⁡α=13\cos\alpha = \tfrac{1}{\sqrt3} (≈54.7°\approx 54.7°), NOT 45°45° or 60°60°. Options offering 45°/60° for the equal-angle case are distractors.
Drill 1 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

    l2+m2+n2=1,l=cos⁡α, m=cos⁡β, n=cos⁡γl^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⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12 a22+b22+c22\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

    a1a2+b1b2+c1c2=0a_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⟩

    proj=(x2−x1) l+(y2−y1) m+(z2−z1) n\text{proj} = (x_2-x_1)\,l + (y_2-y_1)\,m + (z_2-z_1)\,n
  • Direction-angle identities

    Core identities

    ∑cos⁡2 ⁣θ=1,∑sin⁡2 ⁣θ=2,∑cos⁡2θ=−1\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
LineDirection cosinesNote
x-axis⟨1,0,0⟩\langle 1, 0, 0 \ranglemakes 0° with x, 90° with y and z
y-axis⟨0,1,0⟩\langle 0, 1, 0 \rangleDCs ⟨0,1,0⟩\langle 0,1,0\rangle; DRs e.g. ⟨0,4,0⟩\langle 0,4,0\rangle
z-axis⟨0,0,1⟩\langle 0, 0, 1 \rangleperpendicular to the whole XY-plane
⊥ to z-axisn=0n = 0, e.g. ⟨5,6,0⟩\langle 5, 6, 0\ranglelies 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)

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