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 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)
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
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 x-axis.
V.V is denoted as V2.
For P not 0, we have|
P2 > 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 = U2 +2.U.V + V2 (U-V)2 = U2 -2.U.V + V2 (U+V)(U-V) = U2 - V2
|U is orthogonal with V if and only if U.V = 0|
|sqrt(V2) = ||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 =(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.
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 = X2 + Y2 +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 <=> A2 = ||A||2 = |OA|2 = x2 + y2 So, if A(x,y) then |OA| = sqrt( x2 + y2 )
|A B| = sqrt( (x' - x)2 + (y' - y)2 )
The orthogonal projection of vector P on vector Q is (P.Q) Q ----------- Q2
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' Q2We have found a formula for the projection of P on Q as a function of P and Q.
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)
Let A = vector with image point A(x,y).|
Let B = vector with image point B(x',y').
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