What are Direction Cosines of a Line Segment?

We use vector algebra to three-dimensional geometry. This approach to 3-dimensional geometry is to make the study simple. The cosines of the angles made by a directed line segment with the coordinate axes are termed as the direction cosines of that line.

What are Direction Cosines of a Line Segment? If a directed line L passing through the origin makes angles α, β and γ with x, y, and z-axes, respectively, called direction angles, then the cosine of these angles, cos α, cos β, and cos γ are known as direction cosines of the directed line L. If we change the direction of L, then the direction angles are replaced by their supplements, i.e., π-α, π-β and π-γ. So, the signs of the direction cosines are reversed.

Remember that a given line in space can be extended in two opposite directions and so it has two sets of direction cosines. To have a unique set of direction cosines for a given line in space, we should take the given line as a directed line. These are denoted by l, m, and n.

If the given line in space does not pass through the origin, then, to find its direction cosines, draw a line through the origin and parallel to the given line. Then take one of the directed lines from the origin. Find its direction cosines as two parallel lines have the same set of direction cosines.

Relation between Direction Cosines of a Line

Direction Cosines of a Line Passing Through Two Points

The direction cosines of the line segment joining the points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) are

Here,

Solved Examples

Example 1:

If a line makes an angle 90°, 60°, and 30° with the positive direction of the x, y and z-axis, respectively, find its direction cosines.

Solution:

Let l, m, n be direction cosines of line.

Then l = cos 90° = 0

m = cos 60° = ½

n = cos 30° = √3/2

Example 2:

Find the direction cosines of a line joining the points (3, -4, 6) and (5, 2, 5).

Solution:

Given points are A(3, -4, 6) and B(5, 2, 5).

Direction cosines are given by (x₂ – x₁)/AB, (y₂ – y₁)/AB, and (z₂ – z₁)/AB.

x₂ – x₁ = 5 – 3 = 2

y₂ – y₁ = 2 + 4 = 6

z₂ – z₁ = 5 – 6 = -1

AB = √[(x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)²] = √(2² + 6² + (-1)²) = √41

So direction cosines of the line = 2/√41, 6/√41, -1/√41.

Example 3: The plane passing through the points (1, 2, 1), (2, 1, 2) and parallel to the line, 2x = 3y, z = 1 also passes through the point:

(a) (0, -6, 2)

(b) (0, 6, -2)

(c) (-2, 0, 1)

(d) (2, 0, -1)

Solution: 

Given that plane passes through (2, 1, 2) is a(x – 2) + b(y – 1) + c(z – 2) = 0.

It also passes through the point (1, 2, 1).

So -a + b – c = 0

⇒ a – b + c = 0…(i)

Given line x/3 = y/2 = (z – 1)/0 is parallel to (1)

So 3a + 2b + 0c = 0

⇒ a/(0 – 2) = b/(3 – 0) = c/(2 + 3)

⇒ a/2 = -b/3 = -c/5

So the plane is 2x – 4 – 3y + 3 – 5z + 10 = 0

⇒ 2x – 3y – 5z + 9 = 0

Substitute options in the above equations and check.

Hence, option (c) is the answer.

Example 4: The direction cosines of the normal to the plane 2x + 3y – 6z = 5 are 

(a) 2, 3, -6

(b) 2/7, 3/7, -6/7

(c) 2/5, 3/5, -6/5

(d) None of these.

Solution:

Direction ratios of normal to the plane are (2, 3, -6).

So direction cosines = (2/√49, 3/√49, -6/√49)

= (2/7, 3/7, -6/7).

Hence, option (b) is the answer. 

Also, Read
Types of coordinate systems

3D geometry