# Vector multiplication Multiplying two vectors A y B  It is done in two ways:

• How Scalar product, whose result is a number:

A · B = C; Where C ∈ R.

• How vector product, whose result is another vector.

A × B = C

## Scalar product Be A = (Ax, TOy, TOz) Y B = (Bx, By, Bz), the scalar product (also called dot product or internal product) of two vectors is defined as:

A · B = AxBx + AyBy + AzBz

Now, another way to express the dot product is:

AB = |A| |B| cosθ

Where |A| and |B| are the modules of A y B, and θ is the angle between both vectors.

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

Example 1: Determine the dot product of A = (2, 4, 6) and B = (-2, 3, 8).

We see that for the vector A , 2 is the component "x", 4 is "y" and 6 is "z". For vector B, -2 is the component "x", 3 "y" and 8 is "z". The dot product will be:

A · B = AxBx + AyBy + AzBz = (2)(-2) + (4)(3) + (6)(8) = – 4 + 12 + 48 = 56

Example 2: Determine the dot product of A = (5, 7) and B = (- 1, -3), considering that the angle between the two is θ = 60 ⁰.

We see that for the vector A, 5 is the component "x" and 7 is  "y. For vector B, -1 is the component  "x ”y - 3 is“ y ”. The dot product will be:

AB = |A| |B| cosθ

• Modulus calculation A:

|A| = √ [(Ax)2 + (Ay)2  ] = √ [(5)2 +(7)2 ] = √ (25 + 49) = √74

• Modulus calculation B:

|B| = √ [(Bx)2 + (By)2  ] = √ [(-1)2 +(-3)2 ] = √ (1 + 9) = √10

Thus:

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

## Vector product Be A = (Ax, TOy, TOz) Y B = (Bx, By, Bz),, the vector product (also called the cross product) of two vectors is defined as:

A × B = (AyBz - TOzBy) î + (AxBz - TOzBx) ĵ + (AxBy   AyBx) k

Now if we multiply the magnitudes of A y B  and we multiply them by the sine of the angle formed by both vectors (<180 ⁰), the magnitude The vector product is:

A × B = |A| |B| sinθ

Where |A| and |B| are the modules of A y B, and θ is the angle between both vectors.

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 |B| are the modules of A toward B, by the shortest path, the stretched thumb points the direction and direction of the vector product vector  A × B.

Example: Determine the cross product of A = (6, 8, 10) and B = (-2, 3, 8):

We see that for the vector A , 6 is the component "x", 8 is "y" and 10 is "z". Now for the vector B, -2 is the component "x", 3 "y" and 8 is "z". The cross product will be:

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

= (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 A = (1, -2, 1) and B = (-1, 3, 1):

We see that for the vector A , 1 is the component "x", .2 is "y" and 1 is "z". Now for the vector B, -1 is the component "x", 3 "y" and 1 is "z". The cross product will be: = [(-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