Multiplying two vectors ** A** y

**It is done in two ways:**

*B*- How
**Scalar product**, whose result is a**number**:

** A** ·

**= C; Where C ∈ R.**

*B*- How
**vector product**, whose result is another**vector**.

** A** ×

**=**

*B*

*C***Scalar product**

Be ** A** = (A

_{x}, TO

_{y}, TO

_{z}) Y

**= (B**

*B*_{x}, B

_{y}, B

_{z}), the scalar product (also called dot product or internal product) of two vectors is defined as:

** A** ·

**= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z }

Now, another way to express the dot product is:

** A** ∙

**= |**

*B***| |**

*A***| cosθ**

*B*Where |** A**| and |

**| are the modules of**

*B***y**

*A***, and θ is the angle between both vectors.**

*B**The Scalar product of two vectors results in a number real.*

**Example 1: **Determine the dot product of ** A** = (2, 4, 6) and

**= (-2, 3, 8).**

*B*We see that for the vector ** A** , 2 is the component "x", 4 is "y" and 6 is "z". For vector

**, -2 is the component "x", 3 "y" and 8 is "z". The dot product will be:**

*B*** A** ·

**= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z }= (2)(-2) + (4)(3) + (6)(8) = – 4 + 12 + 48 = 56

**Example 2: **Determine the dot product of ** A** = (5, 7) and

**= (- 1, -3), considering that the angle between the two is θ = 60 ⁰.**

*B*We see that for the vector ** A**, 5 is the component "x" and 7 is

**"**y

**”**. For vector

**, -1 is the component**

*B***"**x ”y - 3 is“ y ”. The dot product will be:

** A** ∙

**= |**

*B***| |**

*A***| cosθ**

*B*- Modulus calculation
:*A*

|** A**| = √ [(A

_{x})

^{2 }+ (A

_{y})

^{2 }] = √ [(5)

^{2 }+(7)

^{2 }] = √ (25 + 49) = √74

- Modulus calculation
:*B*

|** B**| = √ [(B

_{x})

^{2 }+ (B

_{y})

^{2 }] = √ [(-1)

^{2 }+(-3)

^{2 }] = √ (1 + 9) = √10

Thus:

** A** ∙

**= √74 √10 cos60 ⁰ = (√74 √10) / 2 = √740 / 2 = 13.60**

*B***Vector product**

Be ** A** = (A

_{x}, TO

_{y}, TO

_{z}) Y

**= (B**

*B*_{x}, B

_{y}, B

_{z}),, the vector product (also called the cross product) of two vectors is defined as:

** A** ×

**= (A**

*B*_{y}B

_{z }- TO

_{z}B

_{y}) î + (A

_{x}B

_{z }- TO

_{z}B

_{x}) ĵ + (A

_{x}B

_{y }–

_{ }A

_{y}B

_{x}) k

Now if we multiply the magnitudes of ** A** y

**and we multiply them by the sine of the angle formed by both vectors (<180 ⁰), the**

*B***magnitude**The vector product is:

** A** ×

**= |**

*B***| |**

*A***| sinθ**

*B*Where |** A**| and |

**| are the modules of**

*B***y**

*A***, and θ is the angle between both vectors.**

*B*The direction of the vector of the vector product is determined by the right-hand rule.

**Right hand rule**

If we place the right hand so that the fingers point in the direction of rotation of Where |** A**| and |

**| are the modules of**

*B***toward**

*A***, by the shortest path, the stretched thumb points the direction and direction of the vector product vector**

*B***×**

*A***.**

*B***Example: **Determine the cross product of ** A** = (6, 8, 10) and

**= (-2, 3, 8):**

*B*We see that for the vector ** A** , 6 is the component "x", 8 is "y" and 10 is "z". Now for the vector

**, -2 is the component "x", 3 "y" and 8 is "z". The cross product will be:**

*B*** A** ×

**= (8 · 8 - 10 · 3) î + [6 · 8 - 10 · (-2)] ĵ + [6 · 3 - 8 · (-2)] k =**

*B*= (64 - 30) î + (48 + 120) ĵ + (18 + 16) k =

= 34 î + 68 ĵ + 34 k

**Determinant of the vector product**

The vector product is represented compactly by means of a determinant that for the 3 × 3 dimension is:

** Example:** Determine the cross product of

**= (1, -2, 1) and**

*A***= (-1, 3, 1):**

*B*We see that for the vector ** A** , 1 is the component "x", .2 is "y" and 1 is "z". Now for the vector

**, -1 is the component "x", 3 "y" and 1 is "z". The cross product will be:**

*B*= [(-2) · 1 - 1 · 3)] î + [1 · 1 - 1 · (-1)] ĵ + [1 · 3 - (-2) · (-1)] k =

= (-2 - 3) î + (1 + 1) ĵ + (3 + 2) k =

= -5 î + 2 ĵ + k