Scalar product

Scalar product
Figure 2. Parametric equations of the curve

Sean A = (Ax, TOy, TOz) Y B = (Bx, By, Bz); he Scalar product (also called Product point or Internal product) of two vectors is defined as:

AB = |A| |B| cosθ

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

A · B = AxBx + AyBy + AzBz

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

Definition

It is called dot product (or dot product or internal product) of two vectors A y B  yet climb whose value will be equal to the product of their modules multiplied by the cosine of the angle that they form:

AB = |A| |B| cosθ

The dot product represents the projection of the vector A about the vector B and equivalently to the projection of B on A (Figure I). Another way to express the dot product is:

A · B = AxBx + AyBy + AzBz

Properties

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

ζA = AA = |A|2 ≥ 0

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

|A| = √ ( AA) = ζA

  1. The scalar product is commutative:

AB = BA

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

  1. The scalar product is distributive:

A ∙ (B + C) = AB + AC

  1. Multiplication by a scalar:

β ∙ (AB) = | β ||A||B| cosθ

A) ∙ (βB) = | βA||B| cosθ = |A|| βB| cosθ

  1. 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||B| cosθ

Which is nothing other than the cosine theorem.

Cosine theorem

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

AB           → θ = π / 2 →          AB = |A||B| cosθ = 0

Exercises

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

From the formula of the scalar product we have:

A · B = AxBx + AyBy + AzBz

Thus:

A · B = (2)(-2) + (4)(3) + (5)(7) =

= – 4 + 12 + 35 =

= 43  

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

From the formula of the scalar product we have:

AB = |A| |B| cosθ

We calculate the modulus of both vectors:

|A| = √ [(Ax)2 + (Ay)2  ] = √ [(2)2 +(3)2 ] = √ (4 + 9) = √13

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

Thus:

AB = √13 √2 cos30 ⁰ =

= (√26 √3) / 2 =

= √78 / 2

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

Two vectors A y B  they are perpendicular if the dot product of both is zero, that is:

A · B = AxBx + AyBy + AzBz = 0

Thus:

A · B = (2)(-5) + (-3)(-10/3) =

= – 10 + 10 =

= 0

Both vectors are perpendicular.

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

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

A · B = AxBx + AyBy + AzBz = 0

Thus:

A · B = (2) (3) + a (-2) = 0

6 - 2a = 0

6 = 2a

6/2 = a

a = 3