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.
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
Take the vector P and fix the origin of an xaxis 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)
Take an xaxis 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 vectorprojection of P on x.
We write proj(P)=P'.
For all vectors P and Q and for any real number r

The dot product of two vectors P and Q is defined by
P.Q = abs(P).abs(proj(Q)) 
The dot product is independent of the orientation of the xaxis.
V.V is denoted as V^{2}.
For P not 0, we have P^{2} > 0 
(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) 
P.Q = Q.P 
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. 
U.(r.V+s.W) = U.(r.V) + U.(s.W) = r.(U.V) + s.(U.W)
(U+V)^{2} = U^{2} +2.U.V + V^{2} (UV)^{2} = U^{2} 2.U.V + V^{2} (U+V)(UV) = U^{2}  V^{2}
U is orthogonal with V if and only if U.V = 0 
sqrt(V^{2}) = V 
r.V = r.V 
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
Then, E.U = abs(E).abs(proj(U)) = 1.cos(t)
P.Q = P.E.Q.U = P.Q.E.U P.Q = P.Q. cos(t)
P.Q = P.Q. cos(t) 
AB^{2} = AB.AB =(BA)(BA) = 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.
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 ]
X+Y^{2} = (X + Y)^{2} = X^{2} + Y^{2} +2.X.Y =< X^{2} + Y^{2} + 2.X.Y =< (X + Y)^{2} So, X+Y =< (X + Y)
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)
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 <=> A^{2} = A^{2} = OA^{2} = x^{2} + y^{2} So, if A(x,y) then OA = sqrt( x^{2} + y^{2} )
A B = sqrt( (x'  x)^{2} + (y'  y)^{2} )
The orthogonal projection of vector P on vector Q is (P.Q) Q  Q^{2} 
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 uaxis.
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' Q^{2} (P.Q) Q =>  = P' Q^{2}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' =  B^{2} 3 B =  26 A' has coordinates (3/26) (5, 1) = (15/26 , 3/26)
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' A^{2} = x^{2} + y^{2} = A^{2} A = OA = sqrt( x^{2} + y^{2} ) 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  Q^{2} 