9 min readConcept sheet1-mark MCQ or assertion-reason; 2-mark questions; used again in three dimensional geometry

Direction cosines and direction ratios of a vector

Direction cosines are the cosines of the angles a vector makes with the axes, so l² + m² + n² = 1. Direction ratios are numbers proportional to them.

Direction cosinesl = cos α, m = cos β, n = cos γ, where α, β, γ are the angles with the positive x, y and z axes

For r⃗ = aî + bĵ + ck̂l = a/|r⃗|, m = b/|r⃗|, n = c/|r⃗|, so r̂ = lî + mĵ + nk̂

Identityl² + m² + n² = 1

Direction ratiosa, b, c (the components) or any multiple λa, λb, λc; they need not have squares adding to 1

Direction angles and cosines

Let r⃗ = OP⃗ = xî + yĵ + zk̂ with |r⃗| = r. Suppose OP⃗ makes angles α, β, γ with the positive x, y and z axes. These are its direction angles, each between 0 and π. Drop a perpendicular from P to the x-axis at A. Triangle OAP has a right angle at A, so cos α = x/r. In the same way cos β = y/r and cos γ = z/r. The numbers l = cos α, m = cos β, n = cos γ are the direction cosines of r⃗.

Why l² + m² + n² = 1

1
l = x/r, m = y/r, n = z/r, where r = |r⃗|.
2
l² + m² + n² = (x² + y² + z²)/r².
3
But r² = x² + y² + z².
4
So, l² + m² + n² = 1.

Direction cosines vs direction ratios

Direction cosines (l, m, n)Fixed for a given vector. They are the components of the unit vector r̂. Always l² + m² + n² = 1.

Direction ratios (a, b, c)Not unique: any numbers proportional to l, m, n. The components of r⃗ are one set. Usually a² + b² + c² ≠ 1.

From direction ratios a, b, c, get the direction cosines by dividing each by √(a² + b² + c²).

Example 1: direction cosines

Problem

Find the direction cosines of a⃗ = 2î − 3ĵ + 6k̂ and the angle it makes with the x-axis.

Working

|a⃗| = √(4 + 9 + 36) = √49 = 7
l = 2/7, m = −3/7, n = 6/7
Check: 4/49 + 9/49 + 36/49 = 49/49 = 1.
cos α = l = 2/7, so α = cos⁻¹(2/7).

Answer

Direction cosines 2/7, −3/7, 6/7; angle with the x-axis cos⁻¹(2/7).

Example 2: board-style, equal angles

Problem

A vector makes equal acute angles with all three axes. Find its direction cosines. Find the vector of magnitude 5√3 in this direction.

Working

Equal angles give l = m = n.
l² + m² + n² = 1, so 3l² = 1 and l = 1/√3 (positive, since the angles are acute).
Direction cosines: 1/√3, 1/√3, 1/√3.
Unit vector = (1/√3)(î + ĵ + k̂).
Required vector = 5√3 × (1/√3)(î + ĵ + k̂) = 5(î + ĵ + k̂).

Answer

Direction cosines 1/√3, 1/√3, 1/√3; vector 5î + 5ĵ + 5k̂.

Exam tips

Direction angles lie between 0 and π, so a direction cosine can be negative.
No vector makes 45° with all three axes, because 3 × (1/2) ≠ 1.
The direction cosines of î, ĵ, k̂ are (1, 0, 0), (0, 1, 0) and (0, 0, 1).
The direction cosines of −r⃗ are −l, −m, −n.

In real life

Spacecraft engineers, including those at ISRO, describe how a satellite is turned using a 'direction cosine matrix'. It is built from exactly these cosines of angles with fixed axes.

Exam angle

1 marks: “Can 1/2, 1/2, 1/2 be the direction cosines of a vector? Give a reason.” No. (1/2)² + (1/2)² + (1/2)² = 3/4, which is not 1.

2 marks: “Find the direction cosines of the vector joining A(2, −1, 3) to B(4, 3, −1).” AB⃗ = 2î + 4ĵ − 4k̂, |AB⃗| = √(4 + 16 + 16) = 6. Direction cosines: 1/3, 2/3, −2/3.

2 marks: “A vector r⃗ has magnitude 6 and direction ratios 2, −1, 2. Find r⃗.” √(4 + 1 + 4) = 3, so the direction cosines are ±(2/3, −1/3, 2/3). r⃗ = 6(lî + mĵ + nk̂) = ±(4î − 2ĵ + 4k̂).

Remember it

Squares of the cosines add to one. Ratios: any multiple. Cosines: divide by the length.