Called **cosine directors** of a vector ** A**, to the cosines of the angles that he with each with the coordinate axes. In a three-dimensional plane they are represented as:

Where: Cosα = A_{x} /|** A**| ;

_{ }Cosβ = A

_{y}/|

**| ;**

*A*_{ }Cosγ = A

_{z}/|

**|**

*A*_{ }

**Definition**

They are called cosine directors of a vector ** A** to the cosines of the angles that said vector forms with each one with the coordinate axes; these determine its direction along each axis. The number of direct cosines depends on the number of dimensions of the system, if it is two dimensions, there will be two direct cosines. If it is three-dimensional, there will be three.

**Cosine directors in two dimensions**

Be ** A** = xî + yĵ, then the principal cosines are given by:

Cosα = A_{x} /|** A**| ;

_{ }Cosβ = A

_{y}/|

**|**

*A*Where α and β are the angles that the vector forms ** A** with the x and y axes. And the sum of the squares of all the direct cosines is equal to one:

Cos^{2}α + Cos^{2}β = 1

**Cosine directors in space (3D)**

Be** A** = xî + yĵ + zk, then the direct cosines are given by:

Cosα = A_{x} /|** A**| ;

_{ }Cosβ = A

_{y}/|

**| ;**

*A*_{ }Cosγ = A

_{z}/|

**|**

*A*_{ }

Where α, β and γ are the angles that the vector forms with the x, y and z axes. And the sum of the squares of all the direct cosines is equal to one:

Cos^{2}α + Cos^{2}β + Cos^{2}γ = 1

**Exercises**

- Determine the cosines directors of the vector
= (5, 7, -3).*A*

How ** A** = (A

_{x}, TO

_{y}, TO

_{z}) = (5, 7, -3) then:

A_{x }= 5; TO_{y }= 7; TO_{z }= -3

Now, the cosine directors will be given by:

Cosα = A_{x} /|** A**| ;

_{ }Cosβ = A

_{y}/|

**| ;**

*A*_{ }Cosγ = A

_{z}/|

**|**

*A*_{ }

Let's calculate the modulus (or magnitude) of ** A** :

|** A**| = √ [(A

_{x})

^{2 }+ (A

_{y})

^{2}+ (A

_{z})

^{2}] = √ [(5)

^{2 }+ (7)

^{2}+ (-3)

^{2}] =

= √ (25 + 49 + 9) =

= √83

Thus:

Cosα = 5 / √83

_{ }Cosβ = 7 / √83

_{ }Cosγ = - 3 / √83 _{ }

- Determine the leading cosines of the vector
= (1, -2, 4) and check that the sum of the squares of all the leading cosines is equal to one.*A*

How ** A** = (A

_{x}, TO

_{y}, TO

_{z}) = (1, -2, 4) then:

A_{x }= 1; TO_{y }= -2; TO_{z }= 4

The cosine directors will be given by:

Cosα = A_{x} /|** A**| ;

_{ }Cosβ = A

_{y}/|

**| ;**

*A*_{ }Cosγ = A

_{z}/|

**|**

*A*_{ }

Let's calculate the modulus (or magnitude) of ** A**:

|** A**| = √ [(A

_{x})

^{2 }+ (A

_{y})

^{2}+ (A

_{z})

^{2}] = √ [(1)

^{2 }+ (-2)

^{2}+ (4)

^{2}] =

= √ (1 + 4 + 16) =

= √21

Thus:

Cosα = 1 / √21

_{ }Cosβ = - 2 / √21

_{ }Cosγ = 4 / √21 _{ }

Now, we must verify that:

Cos^{2}α + Cos^{2}β + Cos^{2}γ = 1

Thus:

(1 / √21)^{2 }+ (- 2 / √21)^{2 }+ (4 / √21 _{ })^{2 }= 1/21 + 4/21 + 16/21 = 21/21 = 1

Which is what we wanted to check.

- Using the principal cosines determine the angles α and β that the vector forms
= (-4, 3) with the x and y axes.*A*

How ** A** = (A

_{x}, TO

_{y}) = (-4, 3) then:

A_{x }= -4; TO_{y }= 3

The cosine directors will be given by:

Cosα = A_{x} /|** A**| ;

_{ }Cosβ = A

_{y}/|

**|**

*A*Let's calculate the modulus (or magnitude) of ** A**:

|** A**| = √ [(A

_{x})

^{2 }+ (A

_{y})

^{2}] = √ [(-4)

^{2 }+ (3)

^{2}] =

= √ (16 + 9) =

= √25 =

= 5

Thus:

Cosα = - 4/5 »α = arc cos (- 4/5) = 2,498 rad

Since 1 radian ≅ 57.296 °, then

α = 143.13 °

Analogously:

Cosβ = 3/5 »β = arc cos (3/5) = 0.927 rad = 53.13 °