Vectors in a plane

Vectors

Translations and Vectors

Take in the plane a fixed origin O.

A translation t in the plane, is now completely determined by the image point P of O by this translation. The translation is determined by point P or also by any couple points (A,B) so that t(A)=B .

With a that translation t, we associate a free vector. The free vector is the set of couples of points (A,B) so that t(A)=B .

There is a one one correspondence between the set of all translations and the set of all free vectors in the plane.

We can represent a free vector by means of an arrow with origin in O and with terminus t(O)=P. Point P is the image point of the vector. The couple (O,P) is a representation of the vector, and the couple (A, t(A)) is another representation.

A free vector is also called a vector

We note the vector with image point P as OP or P

Vectors AB and CD are equal if and only if there is a translation t such that t(A)=B and t(C)=D.

The vector OO corresponds with the identical translation and is called the zero vector.

Addition

Take two translations t and t' in the plane, with t(O)=A and t'(A)=B, then combine the translations t' after t. Let t" be this new translation. Then t"(O)=B. With translation t corresponds vector OA, with translation t' corresponds vector AB and with translation t" corresponds vector OB. By definition we say that OA + AB = OB . So, the addition of vectors is associated with the composition of the corresponding translations.

It is easy to see that the set off all vectors forms a commutative group for the addition. The opposite vector of vector AB is vector BA corresponding with the inverse translation. The opposite vector of vector A is noted -A.

The difference of two vectors is defined by A - B = A + (-B).

For any vector AB we have :
OA + AB = OB <=> AB = B - A

Real multiple of a vector

Take the vector P and fix the origin of an x-axis in O so that P is on that axis. With point P corresponds one and only one real number u. We say : point P is at x = u and we call u the abscissa of point P. We write abs(P)= u . Now with any real number r, take on that axis the point Q with abs(Q) = r.u . Then the vector Q is defined as r times the vector P. We write Q = r.P .

It can be proved that for each r and s in R and each vector P and Q

If O,P,Q are on the same axis with origin in O, then we define abs(P)=abs(P). It can be proved that abs(P+Q)=abs(P)+abs(Q) and abs(r.P)= r. abs(P)

Orthogonal projection of a vector on an axis through O.

Take an x-axis with origin in O and any vector P.
Let point P' be the orthogonal projection of point P on x. Then vector P' is called the vector-projection of P on x. We write proj(P)=P'.
For all vectors P and Q and for any real number r
  • proj(P+Q)=proj(P)+proj(Q)
  • proj(r.P)=r. proj(P)

Dot product of two vectors

Definition

Let x be an axis with origin O and through point P.
The dot product of two vectors P and Q is defined by 
 
        P.Q = abs(P).abs(proj(Q))

From this, it follows immediate that the dot product is a real number!!

The dot product is independent of the orientation of the x-axis.

V.V is denoted as V2.

Properties

  1. For any vector P not 0
    P2 = abs(P).abs(proj(P)) = abs(P).abs(P) > 0
    So,
    For P not 0, we have
    P2 > 0

  2. For any vector P and Q and any r,s in R 
     
(r P).(s Q)     = abs(r P).abs(proj(s Q))
                = abs(r P).abs(s.proj(Q))
                = r.abs(P).s.abs(proj(Q))
                = r.s.abs(P).abs(proj(Q))
                = r.s.(P.Q)
    So,
    (r P).(s Q) = r.s.(P.Q)

  3. It can be proved that
    P.Q = Q.P

  4. For any vector U,V and W 
     
U.(V+W)         = abs(U).abs(proj(V+W))
                = abs(U).abs(proj(V)+proj(W))
                = abs(U).(abs(proj(V))+abs(proj(W)))
                = abs(U).abs(proj(V) + abs(U).abs(proj(W))
                = U.V + U.W
    The dot product of two vectors is distributive relative to the addition of vectors.

  5. For any Vector U,V and W and any r,s in R 
     
U.(r.V+s.W)     = U.(r.V) + U.(s.W)
                = r.(U.V) + s.(U.W)
  6. It is now easy to show that 
     

(U+V)2 = U2 +2.U.V + V2


(U-V)2 = U2 -2.U.V + V2


(U+V)(U-V) = U2  - V2

Orthogonality

U is orthogonal with V if and only if U.V = 0

Norm of V

The norm of V is defined by the distance from O to the image point V.
We write the norm of V as ||V||.
From this if follows that ||V|| = | abs(V) |.

Properties of norm

  1. sqrt(V2) = sqrt(abs(V).abs(V)) = |abs(V)| = ||V||

    sqrt(V2) = ||V||

  2. ||r.V|| = |abs(r.V)| = |r|.|abs(V)| = |r|.||V||

    ||r.V|| = |r|.||V||

Unit vectors

A unit vector is defined by a vector with norm = 1.
The image points of all the vectors are on a circle with radius = 1.

Dividing a vector by its own norm results in a unit vector with the same direction and with the same sense. 

 
  V
 ------ = U = a unit vector
 ||V||

Dot product and cosine

Let E and U be two unit vectors. Take OE as x-axis, with abs(E)=1.
Say t is the angle EOU.
Then, E.U = abs(E).abs(proj(U)) = 1.cos(t)
Now, take any two vectors P and Q, then there are unit vectors E and U such that P = ||P||.E and Q = ||Q||.U
We have : 
 
P.Q = ||P||.E.||Q||.U
      = ||P||.||Q||.E.U
P.Q = ||P||.||Q||. cos(t)
P.Q = ||P||.||Q||. cos(t)

The last formula has many applications in physics.

Cosine rule

Take a vector A and B. The distance |AB| is the norm of vector AB. This norm is ||AB|| . 
 

||AB||2  = AB.AB

=(B-A)(B-A)

= B.B + A.A - 2.A.B


=||B||2 +||A||2 -2.||A||.||B||.cos(t)
Here t is the angle from A to B. Now, let us view the triangle OAB. Then it follows that 
 

|AB|2 =|OB|2 +|OA| -2.|OB|.|OA|.cos(t)
If we translate triangle OAB in triangle CAB, we have the cosine rule in any triangle. 
 
|AB|2 =|CB|2 +|CA|2 -2.|CB|.|CA|.cos(t)
with t the angle in point C of the triangle.

Unit vectors and dot product

Say E and U are unit vectors, then E.U = 1.1.cos(t); With t the angle from E to U.
So, E.U is always in [-1,1].

Inequality of Cauchy-Schwarz

Let X and Y Be two vectors, then 
 
  X            Y
-----  and  ------  are unit vectors and then
||X||        ||Y||

 X      Y
-----.----- is  always in [-1,1].
||X|| ||Y||


So X.Y is always in [  -||X||.||Y|| , ||X||.||Y||  ]

Inequality of Minkowski

 

||X+Y||2 = (X + Y)2

        = X2 + Y2  +2.X.Y

       =< ||X||2 + ||Y||2 + 2.||X||.||Y||

       =< (||X|| + ||Y||)2

So,
||X+Y|| =< (||X|| + ||Y||)

Orthonormal basis

Take two orthogonal unit vectors E and V.
Fix an x-axis and an y-axis, such that E(1,0) and U(0,1).
We say that (E,U) forms an orthonormal basis. Let A = vector with image point A(x,y).
Then, A = xE + yU .

Properties of coordinates

Let (E,U) form an orthonormal basis.
Let A = vector with image point A(x,y).
Let B = vector with image point B(x',y').
Let r any real number.
Then 
 
        A = x E + y U
        B = x'E + y'U
        A + B = x.E + y.U + x'E + y'U
        A + B = (x+x')E + (y+y')U
        r.A    = r(x.E + y.U)
        r.A    = rx.E + ry.U
So, we have
        co(A + B) = (x+x',y+y')
        co(r.A)    = (rx,ry)

Dot product and coordinates

Let (E,U) form an orthonormal basis.
Let A = vector with image point A(x,y).
Let B = vector with image point B(x',y').
Then 
 
        A = x E + y U
        B = x'E + y'U
        A.B = (x.E + y.U)(x'E + y'U)
            = x.x'.E.E' + x.y'.E.U + y.x'.U.E + y.y'.U.U'
            = x.x' + y.y'
So, we have the formula
        A.B = x.x' + y.y'
Remark : 
 
        A.A = x.x + y.y
<=>
        A2 = ||A||2 = |OA|2 = x2 + y2

So, if A(x,y) then |OA| = sqrt( x2 + y2 )

Distance |A B|

Let (E,U) form an orthonormal basis.
Let A = vector with image point A(x,y).
Let B = vector with image point B(x',y').
Then |A B| = ||AB||
The vector AB = B - A has coordinates (x'-x,y'-y). 
 

|A B| = sqrt( (x' - x)2  + (y' - y)2  )

Projection of vector P on vector Q

Theorem 
 
    The orthogonal projection of vector P on vector Q is

           (P.Q) Q
         -----------
             Q2

Proof:

We take an axis u with O as origin and through the imige point Q of Q.
Choose a unit vector E such that E = Q/||Q||.
Point P' is the orthogonal projection of point P on the u-axis.
The vector P' is called the orthogonal projection of P on Q.
Choose D such that D + P' = P.
 
        P = P' + D

=>    E.P = E.P' + E.D

=>    E.P = E.P'

=>   (E.P) E = (E.P') E

=>   (E.P) E = ||P'|| E

        Q          Q
=>   (----- .P)  ----- = P'
      ||Q||      ||Q||

        Q.P        Q
=>   (--------)  ----- = P'
       ||Q||     ||Q||

           (Q.P) Q
=>     --------------- = P'
           ||Q||2

           (Q.P) Q
=>     --------------- = P'
             Q2

           (P.Q) Q
=>     --------------- = P'
             Q2
We have found a formula for the projection of P on Q as a function of P and Q.

Example:

Relative to an orthonormal basis we take two vectors A(2,7) and B(5,-1). We calculate the orthogonal projection A' of A on B. 

 
          (A.B) B
    A' = ----------
            B2

          3 B
       = -------
           26

   A' has coordinates (3/26) (5, -1) = (15/26 , -3/26)

Formulas

Let A = vector with image point A(x,y).
Let B = vector with image point B(x',y').
Then 
 

        A.B = x.x' + y.y'

        A2 = x2 + y2  = ||A||2

        ||A|| = |OA| = sqrt( x2 + y2 )

        |A B| = sqrt( (x'-x)2  + (y'-y)2  )

        A/ ||A|| is a unit vector

     The orthogonal projection of vector P on vector Q is

           (P.Q) Q
         -----------
             Q2




Topics and Problems

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