Lesson Explainer: Direction Angles and Direction Cosines Mathematics

In this explainer, we will learn how to find direction angles and direction cosines for a given vector in space.

We know, in three-dimensional coordinate space, we have the 𝑥-, 𝑦-, and 𝑧-axes. These are perpendicular to one another as seen in the diagram below. The unit vectors 𝑖, 𝑗, and 𝑘 act in the 𝑥-, 𝑦-, and 𝑧-directions respectively.

Definition: The Direction Angles

Given a vector 𝐴=𝐴,𝐴,𝐴, the angles that this vector makes with the 𝑥-, 𝑦-, and 𝑧-axes, respectively, are 𝛼, 𝛽, and 𝛾. These are known as the direction angles and are written (𝛼,𝛽,𝛾).

These direction angles lead us to a definition for the direction cosines. We know, in right-angled trigonometry, the cosine of any angle 𝜃 is equal to the length of the side adjacent to the angle divided by the length of the hypotenuse: cosadjhyp𝜃=.

Definition: The Direction Cosines

The direction cosines are the cosines of the three direction angles, 𝛼, 𝛽, and 𝛾: coscoscos𝛼=𝐴𝐴,𝛽=𝐴𝐴,𝛾=𝐴𝐴.

𝐴, 𝐴, and 𝐴 are the 𝑥-, 𝑦-, and 𝑧-components of vector 𝐴, and 𝐴 is the magnitude or norm of this vector, where 𝐴=(𝐴)+𝐴+(𝐴).

We can rearrange the three equations such that 𝛼=𝐴𝐴,𝛽=𝐴𝐴,𝛾=𝐴𝐴.coscoscos

In our first example, we will demonstrate how to find a vector 𝐴 given its direction angles and magnitude.

Example 1: Finding a Vector given Its Norm and Direction Angles

Find vector 𝐴 whose norm is 41 and whose direction angles are (135,120,60).

Answer

The norm of a vector is its magnitude or length, and the direction angles 𝛼, 𝛽, and 𝛾 are the angles between the vector and the 𝑥-, 𝑦-, and 𝑧-axes respectively.

Using the formulae for the direction cosines, we know that, for a vector 𝐴=𝐴,𝐴,𝐴, coscoscos𝛼=𝐴𝐴,𝛽=𝐴𝐴,𝛾=𝐴𝐴.

Multiplying each of these equations by 𝐴, the components of 𝐴 are as follows: 𝐴=𝐴𝛼=41135=4122,𝐴=𝐴𝛽=41120=412,𝐴=𝐴𝛾=4160=412.coscoscoscoscoscos

Therefore, 𝐴=4122,412,412.

Before looking at our next example, we will consider a formula that links the direction cosines.

Consider the direction cosines as follows: coscoscos𝛼=𝐴𝐴,𝛽=𝐴𝐴,𝛾=𝐴𝐴.

Squaring both sides of these three equations gives us coscoscos𝛼=(𝐴)𝐴,𝛽=𝐴𝐴,𝛾=(𝐴)𝐴.

Adding these three equations, we get coscoscos𝛼+𝛽+𝛾=(𝐴)+𝐴+(𝐴)𝐴.

We also know that 𝐴=(𝐴)+𝐴+(𝐴).

So, 𝐴=(𝐴)+𝐴+(𝐴).

This means that the right-hand side of the equation is equal to 1.

Therefore, coscoscos𝛼+𝛽+𝛾=1.

Formula: Property of Direction Cosines for a Three-Dimensional Vector

If 𝛼, 𝛽, and 𝛾 are the three direction angles and cos𝛼 , cos𝛽, and cos𝛾 are their corresponding direction cosines, then coscoscos𝛼+𝛽+𝛾=1.

Example 2: Finding the Third Direction Angle of a Vector

Suppose that 31, 65, and 𝜃 are the direction angles of a vector. Which of the following, to the nearest hundredth, is 𝜃?

  1. 72.88
  2. 85.03
  3. 264.00
  4. 84.00

Answer

In order to answer this question, we will use the fact that if the three direction angles of a vector are 𝛼, 𝛽, and 𝛾, then coscoscos𝛼+𝛽+𝛾=1.

If we let 𝛼=31, 𝛽=65, and 𝛾=𝜃, we have, coscoscoscoscos31+65+𝜃=10.91334+𝜃1𝜃10.91334.

Taking the square root of both sides of our equation, cos𝜃0.29437.

Taking the inverse cosine of both sides, 𝜃72.8797.

So, the value of 𝜃, to the nearest hundredth, is 72.88.

In our next example, we will demonstrate how to calculate the direction cosines of a vector.

Example 3: Finding the Direction Cosines of a Vector

Find the direction cosines of the vector 𝐴=(5,2,8).

Answer

Given any vector 𝐴 with components 𝐴, 𝐴, and 𝐴, the direction cosines are coscoscos𝛼=𝐴𝐴,𝛽=𝐴𝐴,𝛾=𝐴𝐴, where 𝐴=(𝐴)+𝐴+(𝐴).

Substituting 𝐴=5, 𝐴=2, and 𝐴=8, we have 𝐴=(5)+(2)+(8)𝐴=93.

Therefore, coscoscos𝛼=593,𝛽=293,𝛾=893.

The direction cosines of the vector 𝐴 are 593,293,893.

Example 4: Finding the Direction Angles of a Vector

Find the direction angles of the vector 212,2122,212.

Answer

Given any vector 𝐴 with components 𝐴, 𝐴, and 𝐴, the direction cosines are coscoscos𝛼=𝐴𝐴,𝛽=𝐴𝐴,𝛾=𝐴𝐴, where 𝐴=(𝐴)+𝐴+(𝐴).

Substituting 𝐴=212, 𝐴=2122, and 𝐴=212, we have 𝐴=212+2122+212𝐴=441𝐴=21.

Therefore, coscoscos𝛼=21,𝛽=21,𝛾=21, which simplify to coscoscos𝛼=12,𝛽=22,𝛾=12.

Taking the inverse cosine of both sides of these three equations, 𝛼=60,𝛽=45,𝛾=60.

The direction angles of the vector 212,2122,212 are (60,45,60).

Example 5: Finding the Direction Angles of a Vector

Find the measure of the direction angles of the vector 𝐹, represented by the given figure, corrected to one decimal place.

Answer

We begin by writing the vector 𝐹 in terms of its three components, where we will define 1 unit to be 1 cm.

In the 𝑥-direction, we travel 8 cm, so the 𝑥-component is 8. In the 𝑦-direction, we travel 19 cm, so the 𝑦-component is 19. In the 𝑧-direction, we travel 9 cm, so the 𝑧-component is 9: 𝐹=(8,19,9).

The magnitude of the vector is 𝐹=(𝐹)+𝐹+(𝐹), where 𝐹, 𝐹, and 𝐹 are the 𝑥-, 𝑦-, and 𝑧-components of vector 𝐹: 𝐹=(8)+(19)+(9)𝐹=506.

Given any vector 𝐹 with components 𝐹, 𝐹, and 𝐹 and direction angles 𝜃, 𝜃, and 𝜃, 𝜃=𝐹𝐹𝜃=𝐹𝐹𝜃=𝐹𝐹.coscoscos

So, 𝜃=8506𝜃=69.2,𝜃=19506𝜃=32.4,𝜃=9506𝜃=66.4.coscoscos

Therefore, the direction angles of the vector 𝐹 are 𝜃=69.2, 𝜃=32.4, 𝜃=66.4.

We will finish this explainer by recapping some of the key points.

Key Points

  • The direction angles 𝛼, 𝛽, and 𝛾 are the angles between a vector and the 𝑥-, 𝑦-, and 𝑧-axes respectively.
  • The direction cosines are the cosines of the three direction angles, 𝛼, 𝛽, and 𝛾, such that coscoscos𝛼=𝐴𝐴,𝛽=𝐴𝐴,𝛾=𝐴𝐴.
  • This means that 𝛼, 𝛽, and 𝛾 are as follows: 𝛼=𝐴𝐴,𝛽=𝐴𝐴,𝛾=𝐴𝐴.coscoscos
  • The following formula links the three direction cosines: coscoscos𝛼+𝛽+𝛾=1.

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