Lesson Formula for Dot-product of vectors in a plane via the vectors components

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Formula for Dot-product of vectors in a plane via the vectors components


In this lesson you will learn how to derive the formula for Dot-product of vectors via the vectors components.

Reminder  (see the lessons  Dot-product of vectors and the angle between two vectors  in this site).

Let  u  and  v  be two vectors in a plane.
Then the  dot-product  (or the  scalar product)  of vectors  u  and  v  is the real number  |u|*|v|*cos%28alpha%29,
 where  |u|  and  |v|  are the magnitudes  (the lengths)  of the vectors  u  and  v,  and  alpha  is the angle between the vectors.

Deriving the formula for Dot-product of vectors via the vectors components


  Theorem   (the formula for the dot-product via the vectors components).

  If  u = (a,b)  and  v = (c,d)   is the component form of the vectors  u  and  v  in a coordinate plane then   (u,v)  =  a*c + b*d   


Proof
Let  u  and  v  be two vectors in a plane and  alpha   be an angle between these vectors  (Figure 1).

The vectors  u  and  v  have the common initial point  P =  P(x1,y1).
The vector  u  has the terminal point  Q =  Q(x2,y2),  and
the vector  v  has the terminal point  R =  R(x3,y3).

The vector  u  has the components  (a,b),  u = (a,b),  a = x2-x1,  b = y2-y1.
The vector  v  has the components  (c,d),  v = (c,d),  c = x3-x1,  d = y3-y1.

Finally, let us introduce the vector  w  with the initial point  Q  and the terminal point  R.              
Denote the component of this vector as  w = (e,f),  so e = x3-x2,  f = y3-y2.

Now, when all the designations are made, we can start the proof.

We will use the formula for the length of the third side of a triangle

abs%28w%29%5E2  =  abs%28u%29%5E2  +  abs%28v%29%5E2  -  2abs%28u%29abs%28v%29cos%28alpha%29.          (1)

              Figure 1.  To the  Theorem
This formula is called the  Law of cosines.  For the proof of this formula see the lesson  Proof of the Law of Cosines revisited  under the topic  Trigonometry  of the section  Algebra-II  in this site.  The proof of the  Law of cosines  I am referencing to is based on the  "first principles",  i.e. uses the  Pythagorean Theorem  only.  So, this reference does not make a wrong logical loop.  For now,  let us return to the proof of the formula for dot-product.

We have the following expressions for the lengths of the vectors:

abs%28u%29%5E2 =%28x2-x1%29%5E2+%2B+%28y2-y1%29%5E2 =x2%5E2+-2x1x2+%2Bx1%5E2+%2B+y2%5E2+-2y1y2+%2By1%5E2,   abs%28v%29%5E2 =%28x3-x1%29%5E2+%2B+%28y3-y1%29%5E2 =x3%5E2+-2x1x3+%2Bx1%5E2+%2B+y3%5E2+-2y1y3+%2By1%5E2,   abs%28w%29%5E2 =%28x3-x2%29%5E2+%2B+%28y3-y2%29%5E2 =x3%5E2+-2x2x3+%2Bx2%5E2+%2B+y3%5E2+-2y2y3+%2By2%5E2.

Substitute this to the expression   2abs%28u%29abs%28v%29cos%28alpha%29 = abs%28u%29%5E2 + abs%28v%29%5E2 - abs%28w%29%5E2  obtained from  (1).  You get

2abs%28u%29abs%28v%29cos%28alpha%29 = x2%5E2+-2x1x2+%2Bx1%5E2+%2B+y2%5E2+-2y1y2+%2By1%5E2 + x3%5E2+-2x1x3+%2Bx1%5E2+%2B+y3%5E2+-2y1y3+%2By1%5E2 - ( x3%5E2+-2x2x3+%2Bx2%5E2+%2B+y3%5E2+-2y2y3+%2By2%5E2 ).

Now,  combine and cancel the like terms.  You get

2abs%28u%29abs%28v%29cos%28alpha%29 = 2*( x1%5E2 - x1x2 + y1%5E2 - y1y2 - x1x3 - y1y3 + x2x3 + y2y3 ).

You can check that the right side is exactly equal to

2*(ac + bd) = 2*( %28x2-x1%29%2A%28x3-x1%29 + %28y2-y1%29%2A%28y3-y1%29 ).

It implies that  abs%28u%29abs%28v%29cos%28alpha%29  =  ac + bd,

exactly as the  Theorem  states.  The proof is completed.



Example 1
Calculate the scalar product of the vectors  u = ( 1, sqrt%283%29 )  and  v = ( 2, sqrt%283%29 ).

Solution
According to the formula for the dot-product via the vectors components

(u,v) = 1*2 + sqrt%283%29*sqrt%283%29 = 2 + 3 = 5.



Example 2
Calculate the scalar product of the vectors  u = ( 1, sqrt%283%29 )  and  v = ( 3, -sqrt%283%29 ).

Solution
According to the formula for the dot-product via the vectors components

(u,v) = 1*3 - sqrt%283%29*sqrt%283%29 = 3 - 3 = 0.



Example 3
Prove that the vectors  u = ( 1, sqrt%283%29 )  and  v = ( 3, -sqrt%283%29 ) are orthogonal.

Solution
According to the formula for the dot-product via the vectors components

(u,v) = 1*3 - sqrt%283%29*sqrt%283%29 = 3 - 3 = 0.

Since  (u,v) = |u|*|v|*cos%28alpha%29,  it implies that cos%28alpha%29 = 0.
This means that the vectors  u  and  v  are orthogonal.


My lessons on Dot-product in this site are
    - Introduction to dot-product
    - Formula for Dot-product of vectors in a plane via the vectors components                               (this lesson)
    - Dot-product of vectors in a coordinate plane and the angle between two vectors
    - Perpendicular vectors in a coordinate plane
    - Solved problems on Dot-product of vectors and the angle between two vectors
    - Properties of Dot-product of vectors in a coordinate plane
    - The formula for the angle between two vectors and the formula for cosines of the difference of two angles

There are short lessons of the  "HOW TO . . . "  type on Dot-product:
    - HOW TO find the length of the vector in a coordinate plane
    - HOW TO find dot-product of two vectors in a plane
    - HOW TO find scalar product of two vectors in a coordinate plane
    - HOW TO find the angle between two vectors in a coordinate plane
    - HOW TO prove that two vectors in a coordinate plane are perpendicular
    - HOW TO prove that a triangle in a coordinate plane is a right triangle
    - HOW TO check if a quadrilateral in a coordinate plane is a parallelogram
    - HOW TO check if a quadrilateral in a coordinate plane is a rectangle
    - HOW TO check if a quadrilateral in a coordinate plane is a rhombus
    - HOW TO check if a quadrilateral in a coordinate plane is a square

For the full list of my lessons on dot-product with short annotations see the file  OVERVIEW of lessons on Dot-product.

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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