Vectors in a plane
- Vectors
- Dot product of two vectors
- Definition
- Properties
- Orthogonality
- Magnitude of V
- Properties of magnitude
- Unity vectors
- Dot product and cosine
- Cosine rule
- Unity vectors and dot product
- Inequality of Cauchy-Schwarz
- Inequality of Minkowski
- Orthonormal basis
- Properties of coordinates
- Dot product and coordinates
- Distance |A B|
- Formulas
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 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 vectors in the plane.
We can represent a 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 arrow OP is a representation of the vector, but the arrow from A to the point t(A), is a representation too.
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 and 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 abscis 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
- r.(s.P)=(r.s)P
- (r+s)P = rP + sP
- r(P+Q)= rP + rQ
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
Take an x-axis with origin in O and any vector P. Say P' is the orthogonal projection of P on x. Then vector P' is called the vector-projection of P on x. We write proj(P)=P'.
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For all vectors P and Q and for any real number r
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Dot product of two vectors
Definition
Let x be an axis with P on x.
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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!!
Properties
- For any vector P not 0
P.P = abs(P).abs(proj(P)) = abs(P).abs(P) > 0
So,
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For P not 0, we hawe |
- For any vector P and Q and any r,s in R
(r P).(s Q) = abs(r P).abs(proj(s P))
= abs(r P).abs(s.proj(P))
= r.abs(P).s.abs(proj(P))
= r.s.abs(P).abs(proj(P))
= r.s.(P.Q)
So,
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(r P).(s Q) = r.s.(P.Q) |
- It can be proved that
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P.Q = Q.P |
- 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
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The dot product of two vectors is distributive with respect to the addition of vectors. |
- 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)
- 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
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Orthogonality
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U is orthogonal with V if and only if U.V = 0 |
Magnitude of V
The
magnitude of V is defined by the distance from O to the image point V.
We write the magnutude of V as ||V||.
From this if follows that ||V|| = | abs(V) |.
Properties of magnitude
- sqrt(V.V) = sqrt(abs(V).abs(V)) = |abs(V)| = ||V||
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sqrt(V.V) = ||V|| |
2. ||r.V|| = |abs(r.V)| = |r|.|abs(V)| = |r|.||V||
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||r.V|| = |r|.||V|| |
Unity vectors
A
unity vector is defined by a vector with magnitude = 1.
The image points of all the vectors are on a circle with radius = 1.
Dividing a vector U by its own magnitude ||U|| results in a unity
vector with the same direction as U and with the same sense.
Dot product and cosine
Let
E and U be two unity 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 Unity vectors E
and U such that P = ||P||.E and Q = ||Q||.U
We have :
P.Q = ||P||.E.||Q||.U
= ||P||.||Q||.E.UP.Q = ||P||.||Q||. cos(t)
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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 magnutude of vector AB. This magnitude 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.
Unity vectors and dot product
Say
E and U are unity 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 Unity vectors and then
||X|| ||Y|| X Y-----.----- is always in [-1,1].||X|| ||Y||
So X.Y is always in [ -||X||.||Y|| , ||X||.||Y|| ]
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Inequality of Minkowski
||X+Y||2 = (X + Y)2
= X2 + Y2 +2.X.Y
=< ||X||2 + ||Y||2 + 2.||X||.||Y|| =< (||X|| + ||Y||)2So,
||X+Y|| =< (||X|| + ||Y||)
Orthonormal basis
Take
two orthogonal unity 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')Ur.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)
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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 )
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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 )
Formulas
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Let A = vector with image point A(x,y). A.B = x.x' + y.y' A2 = x2 + y2 ||A|| = |OA| = sqrt( x2 + y2 )|A B| = sqrt( (x'-x)2 - (y'-y)2 ) |