Sean ** A** = (A

_{x}, TO

_{y}, TO

_{z}) Y

**= (B**

*B*_{x}, B

_{y}, B

_{z}); he

**Scalar product**(also called

**Product point**or

**Internal product**) of two vectors is defined as:

** A** ∙

**= |**

*B***| |**

*A***| cosθ**

*B*Where θ is the angle between both vectors. Also, it can be expressed as:

** A** ·

**= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z }

The dot product is always a **real number**, is commutative and distributive, from it arises the theorem of the cosine. Also, when the dot product of two vectors ** A** y

**is null (zero) means that they are perpendicular to each other.**

*B***Definition**

It is called **dot product (or dot product or internal product) of two vectors A** y

**yet**

*B***climb**whose value will be equal to the product of their modules multiplied by the cosine of the angle that they form:

** A** ∙

**= |**

*B***| |**

*A***| cosθ**

*B*The dot product represents the projection of the vector ** A** about the vector

**and equivalently to the projection of**

*B***on**

*B***(Figure I). Another way to express the dot product is:**

*A*** A** ·

**= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z }

**Properties**

- The scalar product of a vector with itself is always positive:

ζ**_{A}** =

**∙**

*A***= |**

*A***|**

*A*^{2}≥ 0

And it will only be null if ** A** it is a null vector. Thus:

|** A**| = √ (

**∙**

*A*

*A) =***ζ**

*√*

_{A}- The scalar product is commutative:

** A** ∙

**=**

*B***∙**

*B*

*A*Since the angle between the vectors is the same and the multiplication between scalars is commutative.

- The scalar product is distributive:

** A** ∙ (

**+**

*B***) =**

*C***∙**

*A***+**

*B***∙**

*A*

*C*- Multiplication by a scalar:

β ∙ (** A** ∙

**) = | β ||**

*B***||**

*A***| cosθ**

*B*(β** A**) ∙ (β

**) = | β**

*B***||**

*A***| cosθ = |**

*B***|| β**

*A***| cosθ**

*B*- From the scalar product arises the Cosine Theorem:

** C **=

**+**

*A*

*B*** C **·

*= (*

**C****+**

*A***) · (**

*B***+**

*A***)**

*B*|** C**|

^{2}= |

**|**

*A*^{2 }+ |

**|**

*B*^{2 }+ 2|

**||**

*A***| cosθ**

*B*Which is nothing other than the cosine theorem.

- We will say that two vectors, not null, are orthogonal (
**perpendicular**) if your dot product is null (zero):

** A** ⊥

**→ θ = π / 2 →**

*B***∙**

*A***= |**

*B***||**

*A***| cosθ = 0**

*B***Exercises**

- Calculate the dot product of the vectors
= (2, 4, 5) and*A*= (- 2, 3, 7).*B*

From the formula of the scalar product we have:

** A** ·

**= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z }

Thus:

** A** ·

**= (2)(-2) + (4)(3) + (5)(7) =**

*B*= – 4 + 12 + 35 =

= 43 _{ }

- Calculate the dot product of the vectors
= (2, 3) and*A*= (-1, 1), considering that the angle between the two is θ = 30⁰.*B*

From the formula of the scalar product we have:

** A** ∙

**= |**

*B***| |**

*A***| cosθ**

*B*We calculate the modulus of both vectors:

|** A**| = √ [(A

_{x})

^{2 }+ (A

_{y})

^{2 }] = √ [(2)

^{2 }+(3)

^{2 }] = √ (4 + 9) = √13

|** B**| = √ [(B

_{x})

^{2 }+ (B

_{y})

^{2 }] = √ [(-1)

^{2 }+(1)

^{2 }] = √ (1 + 1) = √2

Thus:

** A** ∙

**= √13 √2 cos30 ⁰ =**

*B*= (√26 √3) / 2 =

= √78 / 2

- Determine if the vectors
= (2, - 3) and*A*= (-5, -10/3) are perpendicular.*B*

Two vectors ** A** y

**they are perpendicular if the dot product of both is zero, that is:**

*B*** A** ·

**= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z }= 0

Thus:

** A** ·

**= (2)(-5) + (-3)(-10/3) =**

*B*= – 10 + 10 =

= 0

Both vectors are perpendicular.

- Given the vectors
= (2, a) and*A*= (3, -2), calculate a so that both vectors are perpendicular.*B*

For both vectors to be perpendicular, the scalar product of both must be zero, that is:

** A** ·

**= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z }= 0

Thus:

** A** ·

**= (2) (3) + a (-2) = 0**

*B*6 - 2a = 0

6 = 2a

6/2 = a

a = 3