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Elementary Mathematics
Geometrical Vectors
This is an article from my home page: www.olewitthansen.dk
Ole Witt-Hansen 2011
Contents
1. Parallel displacements in the plane. Vectors............................................................................1 2. Sum of two vectors, and difference between two vectors.......................................................2 3. Multiplication of a vector with a number.................................................................................4
4. Resolution of a vector after two given directions ....................................................................5 5. Coordinates of a vector ............................................................................................................6 6. Position vector. The length of a vector ....................................................................................7 7. Projection of a line-segment on a line. Rotation angle ............................................................9 8. The Scalar product of two vectors .........................................................................................10 9. Projection of a vector on a vector ..........................................................................................14 10. The scalar product expressed in coordinates........................................................................14 11. Cross vector..........................................................................................................................16 12. Determinant for a pair of vectors .........................................................................................18 13. The equation for the line. The distance from a point to a line .............................................21 14. Intersection of two lines. Two linear equation with two unknowns ....................................23 14. Coordinate transformations..................................................................................................24
Geometrical vectors 1
1. Parallel displacements in the plane. Vectors A parallel displacement is a translation, where all objects are moved the same distance in the same direction. The parallel displacement that moves the point A to the point B is written A → B. A parallel displacement can therefore most suitable be represented by an oriented line segment or a stylized arrow. Obviously two such arrows represent the same parallel displacement, and the set of all such arrows having the same length and orientation is defined as a vector. Strictly speaking a single arrow should be designated as a representative of a vector, but in praxis it is always referred to as a vector.
A vector, which connects two points A and B in the plane, is written
AB , but a vector can also be
written as a single fat letter such as a or b or a low letter with an vector arrow, such as bora
. The following three ways to write a vector are equivalent.
a =
ABa
The length of a vector is written with two vertical parentheses: |a| =
|||| ABa .
Among the vectors there is one special vector, which has zero length. It is called the nil-vector, and is written o. The nil-vector has no direction.
A vector, which is not the nil-vector, is called a proper vector. A vector which has the length 1 is called a unit-vector. Unit vectors are mostly designated with
the letters e, i, j and k. Two vectors are said to be parallel (usually written a || b) if there is a representative for a that is
parallel to a representative for b. Two proper vectors are said to be unidirectional or opposite directed, when their representatives
are either unidirectional or opposite directed Two proper vectors a and b are said to be orthogonal (perpendicular to each other) (usually
written a b) , if there is a representative for a which is perpendicular to a representative for b. By a directional vector for a line we understand a vector which is parallel to the line. By a normal vector to a line, we understand a vector which is perpendicular to the line.
Geometrical vectors 2
2. Sum of two vectors, and difference between two vectors Let us assume that we make a displacement corresponding to the vector a, and subsequently a displacement corresponding to the vector b. If the starting point of b is placed at the end point of a, then the resulting displacement is a vector from the starting point of a to the end point of b. (Shown in the figure below). Since this displacement is the result of the two displacements then vector c which represent this displacement is called the sum of the two vectors, and it is written:
c = a + b Notice that |a + b| is the length of a side in a triangle with the two other sides |a| and |b|. According to the triangle inequality, then a side in a triangle is always less than the two others sides.
|a + b| < |a| + |b| and |a + b| = |a| + |b| only if a and b are unidirectional. The addition of vectors is commutative. (2.1) a + b = b + a If we in the same figure would construct a + b, by placing the starting point of a in the end point of b, then a and b will span a parallelogram, since opposite sides are parallel and they have the same length, so both a + b and b + a are the diagonal in the parallelogram and therefore a + b = b + a. From the figure above we can also see, that we may construct the sum of the two vectors in a different way. Namely if we place the starting point of a and b in the same point and construct the parallelogram that is spanned by these two vectors, then the diagonal will be a+b. I physics this is called the parallelogram of the forces, since if a body is affected by two (non
parallel) forces: 21 FandF
then the resulting force on the body will be: 21 FFF
The addition of vectors is also associative:
a + (b + c) = (a + b) + c In the figure above are constructed the two vectors: a + (b + c) and ( a + b) + c. It is seen from the figure that the sum of the three vectors in both cases are the vector that connects the starting point of a with the endpoint of c.
For vectors apply the insertion rule: If A, B and C are point in the plane, then (without restriction):
Geometrical vectors 3
(2.3)
CBACAB
This follows immediately from the definition of a vector sum. Specially:
BAABAA0 By the opposite vector to a vector, we understand a vector having the same length, but reversed. The opposite vector to a is written –a, and it is often read as ”minus a” (Although it has nothing to do with a negative number). The opposite vector to the nil-vector is the nil-vector. In all cases we have:
a + (-a) = o. If we consider the vector equation: x + b = a, the solution obviously is the vector when added to b gives a. This vector is called the difference-vector between a and b, and is written (2.4) a – b The construction of a – b is shown above. The two vectors a and b are set from the same point. It is then seen from the figure (to the left) that the vector which connects the end point of b with the end point of a is the difference vector a – b. From the next figure it is seen that the difference vector can be obtained as the sum of a and (-b). This is illustrated in the third figure, since the rectangle with the sides b and (-b) is a parallelogram, such that the two other sides a – b and a + (– b) are identical vectors. Thus we have: (2.5) a – b = a + (– b) (Quite analogous to what we have for real numbers a and b, where: a – b = a + (-b) ). One useful consequence of this rule, is that you may move a vector b from one side of an equation to the other, by replacing it with (-b) or equivalently to “change sign” of the vector. This is explained below: c = a + b <=> c + (-b) = a + b + (-b) <=> c – b = a
Geometrical vectors 4
3. Multiplication of a vector with a number By t·a, where a is a vector and t is a real number, we understand the following:
If t = 0, then t·a = o (The nil-vector) If t > 0, then t·a is a vector having length t|a| and unidirectional with a. If t < 0, then t·a is a vector having length –t |a|, and opposite directed to a.
So we have in all cases: |t·a| = |t||a|.
3.1 Example Below is illustrated some cases of multiplication of a vector by a number. Multiplication of a vector with a number is subject to some simple rules, they are, however, a bit circumstantial to prove. If s and t are real numbers and a and b are vectors, then the following apply: (3.2) I: (s·t)·a = s·(t·a) II: (s+t)·a = s·a +t·a III: t·(a + b) = t·a + t·b Proof for the first multiplication rule: If s = 0 or t = 0 Then the vector on both sides is the nil-vector. If neither s nor t is 0, then the two vectors (s·t)·a and s·(t·a) must have the same direction or opposite directed to a, whether s·t > 0 or s·t < 0. Furthermore:
|(s·t)·a| = |(s·t)|·|a| = |s|·|t|·|a| = |s|·|t·a| = |s·(t·a)| The second rule is a bit more circumstantial to prove. If s and t has the same sign then (s+t)·a, s·a, t·a and s·a +t·a have the same direction. As shown below they also have the same length, so it is the same vector. | s·a +t·a| = |s·a| +|t·a| = |s|·|a| + |t|·|a| = |s+t|·|a| = |(s+t)·a| If s and t have opposite signs and s + t 0, then s + t has either the same sign as s or t. If we assume that s + t has the same sign as s, then s + t and –t have the same sign. Then we find according to the rule above.
s·a = ((s+t) + (-t))·a = (s+t)·a +(-t))·a =(s+t)·a - t·a => (s+t)·a = s·a +t·a If s + t = 0 then t = -s and then: s·a +t·a = s·a +(-s)·a = s·a -s·a = o
If a and b are proper, non parallel vectors then rule III can be proven if we use the theorem (or axiom) about singled angled triangles. If the figure to the left is shown the two singled angled triangles that are formed from the vectors a, b and a+b, and from the vectors t·a, t·b og t·(a+b). Since the triangles are singled angled, then is the side which correspond to a+b must be t·(a+b). According to the construction, however, t·(a+b) is also equal to t·a + t·b. Consequently:
t·(a+b) = t·a + t·b
Geometrical vectors 5
If a and b are proper parallel vectors then the two vectors t·(a+b) and t·a + t·b are also parallel and unidirectional, and the rule becomes trivial. If either a or b is the nil-vector, then the rule is also trivial. We shall then prove the following small but important rule: (3.3) If a and b are two vectors where a o, and b either is the nil-vector or parallel to a, then
there exists only one number t, such that: b = t·a.
Proof: If b = t·a, then |b| = |t|·|a| and thus ||
||||
a
bt . Furthermore if b has the same direction as a then t > 0, whereas, if
t < 0, if b is opposite directed to a, and t = 0, if b = o. In any case there is only one number which satisfy b = t·a. That t determined in this manner actually satisfy b = t·a is then obvious, since the two vectors have the same length and the same direction. From (3.3) one may as a special case determine a unit vector, having the same direction as a proper vector .
|| b
be
Obviously |e| = 1 and e has the same direction as b.
4. Resolution of a vector after two given directions
Let there be given two non parallel lines l and m
together with a vector c =
AB . We draw two lines one parallel to l through A and one parallel to m through B. The two lines intersect in the point C. According to the insertion rule, we have:
CBACAB
If we put cl =
AC and cm =
CB , then we have:
(4.1) c = cl + cm Thus we have obtained to write the vector c as a sum of two vectors parallel to l and m respectively. This resolution is unique. Assuming that c also can be written as a sum of two other vectors bl and bm, parallel to l and m, then we would have: cl + cm = bl + bm <=> cl - bl = bm - cm On the left side we have a vector parallel to l and on the right side a vector parallel to m. Since l and m are assumed to be non parallel, they must both be equal to the nil-vector, and consequently: cl = bl and bm = cm. The formula (4.1) is called the resolution of a vector after two directions.
Geometrical vectors 6
If c = cl + cm , and a and b are two vectors parallel to l and m, respectively, then we may, according to (3.3) determine two numbers s and t, such that: cl = s∙a and cm = t∙b For two non parallel directions given by the lines l and m and for two proper vectors a and b, then c can uniquely be written as a linear combination of a and b. This is also called to resolve c after a and b. (4.2) c = s∙a + t∙b
5. Coordinates of a vector We now introduce an ordinary right angled coordinate system, having origin O. The two points E (1,0) and F (0,1) are the unit points on the axes We let i and j denote the two orthogonal unit vectors:
(5.1) i =
OE and j =
OF These two vectors are called the basis vectors of the coordinate system. Since the coordinate system is uniquely specified by O and the two basis vectors, we designate it (O, i, j).
According to (4.2) any vector a, can uniquely be written as a linear combination of the two basis vectors. (5.2) a = x∙i + y∙j
(x,y) are denoted the coordinates to the vector a. There is a (very sensible) tradition to designate the coordinates to a vector, with the same letter as the vector, having index (1) and (2), and writing the coordinate set vertically (without a separation comma), and furthermore to use the equal sign between the vector and the coordinate set, as shown below.
(5.3) a = a1∙i + a2∙j =
2
1
a
a i =
0
1 j =
1
0
If a = a1∙i + a2∙j and b = b1∙i + b2∙j then it follows from the calculation rules for addition of vectors, and the rule for multiplication of a vector with a number: a + b = a1∙i + a2∙j + b1∙i + b2∙j = (a1 + b1) ∙i + (a2 + b2) ∙j a - b = a1∙i + a2∙j - (b1∙i + b2∙j) = (a1 - b1) ∙i + (a2 - b2) ∙j k∙a = k∙ (a1∙i + a2∙j) = (k∙a1)∙i + (k∙a2)∙j From this follows (not surprisingly) some rules for operating with coordinates.
Geometrical vectors 7
(5.4)
2
1
a
a+
2
1
b
b =
2
1
2
1
b
b
a
a
2
1
a
a-
2
1
b
b =
2
1
2
1
b
b
a
a k
2
1
a
a =
2
1
ka
ka
6. Position vector. The length of a vector Let there be given a coordinate system (O, i, j) in the plane. For an arbitrary point P = (x,y) in the plane, the
vector
OP is called the position vector P. The projection of the point P on the axes are Q (x,0) and R (0, y). According to (3.3), we have:
OQ x∙i and
OR y∙j and thus
(6.1)
OP
OQ +
OR x∙i + y∙j =
y
x
(6.1) shows that the coordinates to a point are the same as the coordinates to the position vector of the point.
A vector a =
2
1
a
a is position vector for the point A (a1, a2), such that a =
OA .
According to the distance formula, we therefore have:
(6.2) | a | = |
OA | =|OA| = 22
21 aa
The length of a vector is the square root of the sum of the squares of the coordinates. If A and B are arbitrary points in the plane, we have according to the insertion rule.
(6.3)
ABOAOB and thus
OAOBAB
If A (a1 , a2) and B (b1 , b2) we have:
2
1
a
aOA and
2
1
b
bOB
When this applied in (6.3) we find:
(6.4)
22
11
ab
abAB
Thus, we have shown the very important rule: The coordinates to the vector which connects two arbitrary points A and B, are found as the coordinates to the endpoints minus the coordinates to the starting point.
Geometrical vectors 8
For the length of
AB , one finds according to (6.2) a formula identical to the distance formula.
(6.5) 222
211 )()(|| ababAB
6.6 Example.
Many minor geometrical theorems may be proven elegantly by operating with vectors. First we shall prove: The coordinates to the midpoint M of a line segment, which connects A and B are the mean of the coordinates of the endpoints. Let A (a1 ,a2) and B (b1 ,b2) be points in the plane. According to the insertion rule, we have:
)(21
21
OAOBOAABOAAMOAOM
(6.7) )(21
OAOBOM
When written out in coordinates:
(6.7)
22
1121
ba
baOM
6.8 Example. We shall determine the intersection point S of the three medians in a triangle A, B, C .
Let A = (a1 ,a2), B = (b1 ,b2) and C = (c1 ,c2). From the geometry we know, that the median connects a corner with the midpoint M of the opposite side, and that the intersection point divides the length of the median in the ratio 2:1, such that:
AMAS3
2
According to the insertion rule:
)(3
2
3
2
OAOMOAAMOAASOAOS
According to (6.7): )(21
OCOBOM so we get:
)))(2
1(
3
2)(
3
2
OAOBOCOAOAOMOAOS
This may be reduced to:
(6.9) )(3
1
OBOCOAOS
or written out in coordinates:
(6.9)
222
11131
cba
cbaOS
The coordinates to the intersection point of the medians is the mean of the coordinates to the three corners of the triangle. Since the expression for the intersection point is symmetric in A, B and C, it is irrelevant, which of the corners we start with, and we may therefore also conclude that the three medians intersect in the same point.
Geometrical vectors 9
7. Projection of a line-segment on a line. Rotation angle To introduce the scalar product of two vectors, it is necessary to clarify some statements from the geometry. 7.1 example. Signed line segment.
On an oriented number line (a coordinate axis), it is sometimes advantageous to have signed line segments. The line-segment which connects the two points A and B is written AB. The length of the line segment is as usual written |AB|. AB is counted positive if B lies in the positive direction of the oriented line, otherwise negative. Precisely: If AB = |AB|, then BA = –|AB|. When line-segments are signed, then it apply for three points A, B and C on a oriented number line, and irrespectively of their mutual location, we have the insertion rule:
(7.1) AB = AC + CB
If B lies to the right of A and C lies between A and B, then (7.1) follows immediately from |AB| = |AC| + |CB|. If B lies to the right of A and C lies to the right of B, then we have:
|AC| = |AB| + |BC| <=> |AB| = |AC| – |BC| <=> AB = AC + CB If AB = – |AB|, that is, B lies to the left of A, and C lies between B and A, then we have in the same way: |BA| = |BC| + |CA| <=> –AB = –CB –AC <=> AB = AC + CB The other 3 cases of locations of A, B, C can be verified in the same manner. 7.2 Example. Rotation angle If we have two vectors or two oriented line-segments, we define the angle between them as the numerical least rotation which takes the one vector (or oriented line-segment) over into the other vector (or oriented line-segment). Using this definition, the angle between two vectors (or oriented line-segments) always between 0 and 1800 .
Earlier we have introduced the concept of directional angle for a line, as the rotation which takes the positive direction of the 1. axis over in to the same direction as a semi-line l. If v is a directional angle for a line l, then all directional lines can be written as: v + p∙3600 and v +1800 + p∙3600, together v + p∙1800 , where p is an integral number. In the figure is shown a coordinate system and two semi-lines l and m, having directional angles u and v. We wish to find the angle w between l and m. In the same manner as for the insertion rule for points in an oriented line we may establish an insertion rule for signed directional angles. From the figure is seen (with the shown location of l and m) that:
v = u + w => w = v – u Generally we may say that the rotation that takes l over in m consists of two rotations. One is the angle –u which takes l into the x-axis plus a rotation an angle v, which takes the x-axis into m. Together a rotation an angle w = –u + v = v – u. Using the insertion rule, this applies independently of the sign of the angles and magnitude of u and v. We may therefore always (using the procedure described above) find an angle between two vectors (or two oriented line segments) having directional angles u and v, as v - u. To establish the angle between to vectors (as defined above), it may be necessary to change the sign or add an integral number of 3600. However it is important to note that cos(v-u) is unchanged by such operations.
Geometrical vectors 10
7.3 Example. Projection of a line-segment on a oriented line. The projection of a point on a line l is the intersection between the line l and the perpendicular line through the point. A dashed line perpendicular to l is drawn (constructed) through the point. The projection Pl of P on the line l is the intersection between the two lines. See the figure below. The projection of a line segment AB on the line l is the line segment AlBl , connecting the projections of A and B on l. If AB is perpendicular to l, then the projection will degenerate into a point. If we introduce an orientation on the line l, and let v, be the rotation angle from l to the line-segment AB, between the two lines positive orientations, then we have, (if the line segments are signed):
AlBl = AB cos v.
Proof. If AB is parallel displaced perpendicular to the line l, such that A now lies on l , then the projection of AB is unchanged. See the figure above. If v < 900, it is seen from the right angled triangle AlB’Bl , that |AlBl| = |AB’| cos v and thus: AlBl = AB cos v. If 900 < v < 1800 , we have: |AlBl| = |AB| cos(1800-v). Since AlBl = –|AlBl|, we get as before: AlBl = AB cos v. If we have a broken line, AC and CB, as show on the third figure above, we have that the sum of the signed projections of AC and CB will be equal to the projection of the line segment AB. If the rotation angles to AC and CB both are less than 900, (or both are between 900 and 1800 ), then it follows directly from the figure. In other cases the rule follows, by applying the insertion rule on Al , Cl and Bl, which are the projections of A, C and B on l. This can be generalized to the following rule: (7.4) The projections of a broken line on an oriented line are equal to the projection of the line which connects
the end point of the broken line.
8. The Scalar product of two vectors
The results from the examples (7.2) and (7.3), can directly be applied for vectors since an oriented line-segment can be perceived as a representative for a vector. The angle between two vectors and the projection of a vector on a vector has the same meaning as for their representatives. This leads to the rule (8.1): The projection of the sum of the two vectors a + b on the vector c is the sum of the projections of the two vectors a and b on the vector c. ac + bc = (a + b)c
We are now ready to introduce the scalar product of two vectors.
Geometrical vectors 11
A scalar means a number, in contrast to a vector. Often the scalar product is introduced using coordinates, but its significance is more transparent, if it is introduced geometrically, that is, without having a coordinate system, although it requires somewhat more classical geometry, such as the projection rules, stated above. By the scalar product of two proper vectors we understand the product of their lengths times the cosine to the angle between them. If one of the vectors is the nil-vector, then the scalar product is zero. The scalar product is written with a dot ”∙” which cannot be omitted as it can when writing the product of two variables. The scalar product is often referred to as the dot product.
(8.2) a∙b = |a|∙|b|∙cos v where v = / (a, b) is the angle between a og b. It is important to notice that |b|∙cos v is the signed projection of a representative for b on a line segment having the same orientation as a. correspondingly |a|∙cosv is the signed projection of a representative for a on a line segment having the same orientation as b. For the operation with scalar products the following important rules apply:
(8.3)
abba (commutative law)
(8.4)
2|| aaa
(8.5)
bcacbac )( (distributive law)
(8.6) )()()(
abkakbbka
(8.7)
baoboaba 0 (8.3) Is an immidiate consequence of the definition since: / (b, a) = – / (a, b) and cos(v) = cos(-v).
(8.4) follows from:
2||0cos|||| aaaaa
If we put: u = ),(
ca , v = ),(
cb and w = ),(
cba , then we can show (8.5) by the following rewriting:
vbcuacbcac cos||||cos||||
)cos||cos||(|| vbuac
)(cos||||)(||)(||
bacwbacbacbac ccc
Geometrical vectors 12
Her we have used that ac , bc and (a + b)c are the signed projections of a representative for a , b and a + b on a line segment oriented as c, and then applied (7.4). The rule (8.6) follows, since each of the 3 expressions can be interpreted as the length of a times the constant k times the projection of b on a. The rule (8.7) follows directly from the definition of the scalar product and the zero-rule for products:
0900
0cos0||0||0cos||||0
voboaba
vbavbaba
From which follows the important rule: Two proper vectors are orthogonal if and only if their scalar product is zero.
baoboaba 0 Following the algebraic rules for the scalar product, we retrieve most of the rules known from algebra with numbers, but where addition is replaced with the sum of vectors and multiplication is replaced with the scalar product of two vectors. Firstly there is a tradition to put a∙a = a2. Then we shall show the formulas for the square of a two term quantity, by applying (8.2) to (8.7). (a + b)2 = (a + b) ∙(a + b) = a ∙(a + b) + b∙(a + b) = a∙a + a∙b + b∙a + b∙b
(a + b)2 = a2 + b2 + 2a∙b Correspondingly
(a - b)2 = a2 + b2 - 2a∙b In the same manner we find: (a - b)∙(a + b) = a ∙(a + b) - b∙(a + b) = a∙a + a∙b - b∙a - b∙b
(a - b)∙(a + b)= a2 - b2
Finally we notice that: |a∙b| = ||a|∙|b|∙cos v| |a|∙|b| , Since numeric cosine is always less or equal to 1. 8.9 Example. For two vectors a and b we have: |a| = 2, |b| = 3 and |a + b| = 3. Determine the angle between a and b. First we determine the scalar product of the two vectors |a + b| = 3 => |a + b|2 = 9 => (a + b)2 = 9 <=> a2 + b2 + 2a∙b = 9 <=>
Geometrical vectors 13
|a|2 + |b|2 + 2a∙b = 9 <=> a∙b = -2
047.1096
2
||||cos
vba
bav
8.10 Example We shall then show, using vectors, a theorem from geometry that in a triangle the heights intersect in the same point,
In the figure is shown a triangle ABC, where the heights from A and B, intersect in the point P.
The vector c =
PC , and we will show that c
AB .
The vector a =
PA and the vector b =
PB , so we have:
AB =
PB -
PA = b – a Correspondingly
BC =
PC -
PB = c – b and
AC =
PC -
PA = c – a According to the construction: a (c – b) and b (c – a), from which it follows a∙(c – b) = 0 and b∙(c – a) = 0 <=> a∙c = a∙b and b∙c = b∙a =>
a∙c = b∙c <=> a∙c - b∙c = 0 <=> (a – b)∙c = 0 <=> c (a – b) <=> c
AB This demonstrates that CP is the height from C intersecting P, so the three heights intersects in the same point Notice that the proof in (8.10) is purely geometric, without reference to any specific coordinate system. We shall continue to derive the cosine relation, using vector relation only. 8.11 Example. The cosine-relation.
In the figure is shown the triangle ABC.
We set: c =
AB , b =
AC and thus
a =
BC =
ABAC = b – c Clearly the lengths of the sides in the triangle is: c = |c| , b = |b| and a = |b - c|. We then evaluate a2.
a2 = |b - c|2 = (b – c)2 = b2 + c2 – 2 b∙c = |b|2 + |c|2 – 2 |b|∙|c| cos A (8.12) a2 = b2 + c2 – 2∙b∙c cos A
The formula (8.12) is the familiar cosine relation for the side a. The two other cosine relations are derived similarly.
Geometrical vectors 14
9. Projection of a vector on a vector We have earlier shown that the scalar product of two
vectors: a∙b = |a|∙|b|∙cos v , where v = ),(
ba , can be interpreted as |a| times ba, where ba is the signed projection of a representative of b on a line unidirectional to a. Multiplying this projection with a unit vector that is unidirectional with a, then we find the projection ba of b on a.
||||
)cos|(|a
a
a
baevbb aa
(9.1)
aa
baba
2||
For the length of the projection |ba|, we get by taking the length of both sides of (9.1).
(9.2)
||
||||
a
baba
10. The scalar product expressed in coordinates We now invent a coordinate system (O, i, j) in the plane. Since i and j are orthogonal unit vectors, we have: (10.1) i2 = i∙i = 1 and j2 = j∙j = 1 and i∙j = j∙i = 0
Let a and b have the coordinates
2
1a
aa and
2
1b
bb . We may resolve a and b after i and j.
a = a1i + a2j og b = b1i + b2j We then calculate the scalar product a∙b, using the rules for the scalar product (10.1) a∙b = (a1i + a2j) ∙( b1i + b2j) = a1b1i
2 + a1b2i∙j + a2b1j∙i + a2b2j2 = a1b1 + a2b2
We have the found an expression for the scalar product expressed in coordinates. (10.2) a∙b = a1b1 + a2b2
Geometrical vectors 15
10.3 Example
We shall then determine an expression for the projection of 5
2
b on
3
4
a , together with the angle between the
two vectors. According to (9.1):
25
2125
28
3
4
25
7
3
4
2)2)3(24(
)3(5422
||
a
a
baba
For the angle, we find according to the definition of the scalar product:
0
222207.105
295
7
52)3(4
5342
||||cos
vba
bav
10.4 Example. The addition formulas. The addition formulas are the generic name for the calculation of cos(u-v), sin(u-v), cos(u+v) og sin(u+v). Let eu = (cos u, sin u) and ev = (cos v, sin v) be two unit vectors, corresponding to the directional angles u and v. The angle between them is u-v (apart from a sign and an integral number of 3600, where both let cosine unchanged). So we have: eu ∙ev = |eu |∙|ev| cos(u-v) = 1∙1∙ cos(u-v) = cos(u-v) On the other hand, if we evaluate the scalar product in coordinates, we find: eu ∙ev = cos u ∙cos v + sin u ∙sin v Then we get the first of the additions formulas. (10.5) cos(u-v) = cos u ∙cos v + sin u ∙sin v If we replace v by –v we get:
cos(u+v) = cos u ∙cos(-v) + sin u ∙sin(-v) = cos u ∙cos v - sin u ∙sin v (10.6) cos(u+v) = cos u ∙cos v - sin u ∙sin v If we in (10.6) replace u by 90-u, we get: cos(90-(u-v)) = cos(90 – u) ∙cos v – sin(90 – u) ∙sin v which gives: (10.7) sin(u-v) = sin u ∙cos v – cos u ∙sin v Finally if we replace v by –v in (10.7) , we get: (10.8) sin(u+v) = sin u ∙cos v + cos u ∙sin v It should be noticed how elegantly the addition formulas are derived using vectors. The trigonometric derivation is much more circumstantial. 10.9 Example. Cosine and sine to the double and the half angle. If we in (10.6) and (10.8) set u = v = x, then we get the relations: cos(x+x) = cos(2x) = cos x ∙cos x - sin x ∙sin x = cos2 x - sin2 x => cos(2x) = cos2 x - sin2 x sin(x+x) = sin(2x) = sin x ∙cos x + cos x ∙sin x = 2sin x ∙cos x => sin(2x) =2sin x ∙cos x
Geometrical vectors 16
Applying the basic relation: cos2 x + sin2 x = 1, and doing the rewriting cos2 x = 1 - sin2 x and sin2 x = 1 - cos2 x on the formula for cos(2x), we get the following formulas: (10.10) cos(2x) = cos2 x - sin2 x = 2cos2 x – 1 = 1 - 2sin2 x Replacing x with ½x i (10.10) we get formulas for cos(½x) and sin(½x).
cos(2x) = 2cos2 x – 1 => cos x = 2 cos2 (½x) – 1 If we solve the last equation for cos(½x) we find:
(10.12) 2
cos1
2cos
xx (“+” should be applied, if ½x is the 1. or 4. quadrant, otherwise “-“)
Similarly, from cos(2x) = 1 - 2sin2 x, we get.
2
cos1
2sin
xx (“+” should be applied, if ½x is the 1. or 2. quadrant, otherwise “-“)
10.13 Example- The logarithmic formulas for the addition of sine and cosine. By adding the addition formulas (10.5) and (10.6) we have
cos(u-v) + cos(u+v) = cos u ∙cos v + sin u ∙sin v + cos u ∙cos v - sin u ∙sin v = 2cos u∙cos v. Introducing x = u - v and y = u + v and solving with respect to u and v, we find: u =½(x + y) and v = ½ (x - y) Inserting this above, we find the first of the logarithmic formulas.
(10.14) 2
cos2
cos2coscosyxyx
yx
The formulas are called logarithmic, because a multiplication is replaced by an addition. In a similar way, we get the other logarithmic formulas:
(10.15) 2
sin2
sin2coscosyxyx
yx
(10.16) 2
cos2
sin2sinsinyxyx
yx
(10.17) 2
sin2
cos2sinsinyxyx
yx
11. Cross vector
For an arbitrary vector
a , is defined the cross vector a to
a , as the vector, which comes about by
rotating the vector
a an angle 900 in the positive circular direction. From the definition, we have
|||ˆ|
aa and
aa
Furthermore it is seen that:
0 =
0 ,
i =
j ,
j = -
i
There are some smaller rules, when doing operations with cross vectors.
Geometrical vectors 17
(11.1)
akka and
baba
From the figure below, it is apparent that the vectors
ka and
a are unidirectional or opposite
directed, dependent of whether
ak and
a are unidirectional or opposite directed.
Furthermore the vectors
ka and
ak have the same length. So we have:
||||||||||||||^^
akakakkaka
From which follows:
akka .
If
a and
b are vectors, then we consider the triangle OAB where
aOA ,
bAB and
baOB . By a rotation +900 around O, A is passed into A1, B is passed into i B1. So, we have:
aOA1 ,
bBA 11 and
baOB1 . Since
1OB =
111 BAOA , we have:
baba The cross vector of the sum of two vectors is the sum of the individual cross vectors.
Let
2
1
a
aa be a vector in a coordinate system (O, ),
ji , so that:
jaiaa 21 .
By applying the rules for operating with cross vectors, we have:
iajajaiajaiajaiaa 21212121 From which we see that the coordinates to the cross vector is:
(11.2)
1
2
a
aa
11.3 Example
In a coordinate system is given the vector
3
2a . About the vector b is informed:
4
ba and 19
ba
Geometrical vectors 18
Determine the coordinates to
b .
We put
2
1
b
bb
which implies
1
2ˆb
bb and write the two scalar products in coordinates.
2∙b1 - 3∙b2 = 4 and 2∙(-b2) - 3∙b1 = 19 <=> 2∙b1 - 3∙b2 = 4 and 3∙b1 + 2∙b2 = -19
By solving these two equations with respect to b1 and b2 , we find
2
5b
12. Determinant for a pair of vectors For two vectors a and b, we define the determinant det(a, b) for the pair of vectors as:
(12.1) det(a, b) =
ba
If
2
1
a
aa ,
1
2
a
aa and
2
1
b
bb and if the determinant is evaluated in coordinates:
det(a, b) =
ba = (-a2)∙b1+a1∙b2 = a1∙b2 - a2∙b1
(12.2) det(a, b) = a1∙b2 - a2∙b1 From the coordinate expression, we can see that: det(b, a) = - det(a, b), since det(b, a) = (-b2)∙a1+b1∙a2 = -( a1∙b2 - a2∙b1) = - det(a, b) The determinant is often written with a determinant symbol, which is two vertical line-segments, in which the coordinates are placed vertically.
(12.3) det(a, b) = 22
11
ba
ba
As a calculation shows the coordinates may as well be written horizontally.
det(a, b) = 22
11
ba
ba =
21
21
bb
aa = a1∙b2 - a2∙b1
For proper vectors a and b applies that a is parallel to b, if and only if the determinant det(a, b) =0. (12.3) a || b <=> det(a, b) = 0 This follows from:
Geometrical vectors 19
det(a, b) = 0
bababa ||0 The following geometrical theorem applies: For proper vectors a and b applies that |det(a, b)| is the area of the parallelogram spanned by the vectors a and b.
Let two proper vectors a and b be given together with three points A and B so that
OAa and
OBb . We remind you of the formula for the length of the projection of a vector on a vector.
||||||||
||||
aa babaa
bab
The formula expresses that the numerical value of the scalar product is equal to the length of the one vector times the length of the projection of the other vector on the first. Applied on the vector pair (a, b) we have:
(12.4) |||||||||||)det(|
aa babababa
|
a | is the length of
a , and || ˆ
ab is the projection of
b on
a . As it appears from the figure above,
then || ˆ
ab is the height in the parallelogram, which has
a , as the one set of parallel sides and
b as
the other sides. This is true if either || ˆ
ab is unidirectional with
a or opposite directed to
a .
From this follows:
(12.4) |||),det(|
baba
(Equal to the area of the parallelogram spanned by
a and
b ). 12.5 Example: The area of a triangle, spanned by two vectors. Let there be given a triangle A(-3,2), B(4,5) and C(7,0).
Geometrical vectors 20
The area of the triangle is half the area of the parallelogram, which is spanned by the vectors AB and
AC .
3
7
2
3
5
4AB
2
10
2
3
0
7AC
The area of the triangle is therefore: 22|103)2(7|½|23
107|½|),det(|½
ACABT
12.6 Example: The area of a triangle expressed by the coordinates of the three corners.
Let )(),(),(212121
ccCbbBaaA . ),,(,),(22112211
acacACababAB
2222
111121 ) |,det(|
acab
acabABABT
We define the direction of rotation for the vector pair (
a ,
b ), as positive, if the least rotation which
takes
a over in
b is in the positive rotation direction.
In the figure above we can see that '
b (the projection of
b on
a ), is either unidirectional with
a ,
when the rotational direction for the vector pair (
a ,
b ) is positive, or opposite directed to
a , when
the rotational direction for the vector pair (
a ,
b ) is negative.
From this we see that the sign for the rotational direction for the vector pair (
a ,
b ) is the same as
the sign of the determinant ),det(
ba when expressed as
ba . By the directional angle for a vector a, we understand the directional angle for a semi-line which is
parallel to a. If P is the directional point on the unit circle for an angle v, then we have v
vOPe
sin
cos
.
A vector a having directional angle v can therefore be written:
v
vaa
sin
cos|| .
It is then possible to write an expression for the determinant: |det(a, b)|, which looks like the one, we have for the scalar product. The scalar product is defined purely geometrically, so it is independent of the choice of coordinate system. We then choose the 1. axis, so it is unidirectional to a. If the angle between a and b is v, then v is the directional angle for b. Consequently:
Geometrical vectors 21
0
|| aa and
v
vbb
sin
cos||
And thus:
(12.6) ),det(
ba = vba
vb
vba sin||||
sin||0
cos||||
=> ),det(
ba vba sin||||
12.7 Example. The sine-relations.
Let us consider a triangle ABC spanned by the two vectors
CBa and
CAb , then |a| is the side a, |b| is the side b and the intermediate angle is C. According to the formula (12.6) we may find the area of the triangle as: T = ½ det(a, b) = ½∙a∙b∙sin C In quite the same manner, we find: T = ½∙b∙c∙sinA and T = ½∙a∙c∙sinB. If we put
½∙b∙c∙sin A = ½∙a∙c∙sin B = ½∙a∙b∙sin C
And dividing by ½∙a∙b∙c we retrieve the sine-relations
c
C
b
B
a
A sinsinsin
3. The equation for the line. The distance from a point to a line
A line l is determined by a point P0 (x0,y0) on the line and a normal vector to the line
b
an .
We may then establish the following:
The point P(x,y) lies on the line, if and only if the two vectors
n and
PP0 are orthogonal:
n ·
PP0 = 0
Geometrical vectors 22
a(x - x0) + b(y - y0) = 0
ax + by – ax0 - by0 = 0
(13.1) ax + by +c = 0 Where we have set: c = -ax0 - by0 .
We notice that the equation also is fulfilled for P = P0 , since
00PP = o.
The distance d = dist(P1,l) from a point P1 to the line l , may then be found by expressing that d is
the length of the projection of the vector 10
PP on
n .
For the projection ba
of a vector
a on a vector
b , and for the length of the projection we have previously derived the two expressions.
(13.2)
b
b
baa b 2
||
and
||
||||
b
baa b
From which we find:
(13.3) 22
11
22
010110 )()(
),(ba
cbyax
ba
yybxxa
n
PPnPldistd
This last expression is the same as the one we found in the analytical geometry 4.4 Example: 1) A line l has the normal vector n = (3,4) and the line passes through P(-2,2). a) We shall determine an equation for the line.
We insert n = (3,4) and P(-2,2) in a(x - x0) + b(y - y0) = 0 and get: 3(x + 2) + 4(y – 2) = 0 <=> 3x + 4y – 2 = 0 b) Determine the distance from the line to A(4, -5).
When we insert in (4,3) we get: d = 22423
25443
2) A line having a direction vector r = (1,-2) passes through (-4, 2). Determine an equation for the line.
The cross vector to r, will be a normal vector n to the line. n = (2, 1). The line is then established according to the formula. 2(x + 4) +1(y – 2) = 0 <=> 2x + y + 6 = 0
Geometrical vectors 23
14. Intersection of two lines. Two linear equation with two unknowns When you must solve to linear equations with two unknowns, then traditionally it is done with the following line up. a1x + b1y = c1 a2x + b2y = c2 This may also be perceived as the equations for two lines,(where we have moved the constant to the right side). A solution is a set of coordinates (x, y), which satisfy both equations, and therefore is the point of intersection of the two lines.
The system of equations may, however, also be perceived as a vector
2
1
c
cc , written as a linear
combination of the vectors:
2
1
a
aa and
2
1
b
bb .
When viewed as such, it can be written as:
(13.1)
cbyax The point is then to solve this vector equation by operations with vectors. Namely if we take the
scalar product of the vector equation with the cross vector to
a , we have:
cabayaax
aa = 0, since they are orthogonal vectors.
If 0),det(0
baba , then the lines
a and
b are not parallel (the lines intersect in one point), so the equation has only one solution y.
(13.2) ),det(
),det(
ba
ca
ba
cay
If quite the same manner we find, if we take the scalar product with the cross vector to
b .
cbbbyabx
),det(
),det(
),det(
),det(
),det(
),det(
ba
bc
ba
bc
ab
cb
ab
cbx
D=det(a, b) is called the determinant for the system of equations. If the determinant is non zero, then the equations have one and only one solution. If we write the solution in coordinates: a1x + b1y = c1
Geometrical vectors 24
a2x + b2y = c2 The determinant written with coordinates becomes:
D =22
11
ba
ba.
If D 0, there is only one solution the equations.
22
11
22
11
22
11
22
11
ba
ba
ca
ca
y
ba
ba
bc
bc
x
To solve two linear equations with two unknowns in this manner is called the determinant method. Example. Two linear equations with two unknowns. Let the two equations be: 2 x -3y =-4
-x +5y =7
The determinant is: D = 0751
32
, so the equations have one solution:
7
10
7
71
42
7
1
7
57
34
yx
14. Coordinate transformations We shall look at coordinate transformations induced by rotations in the plane. Below is illustrated two coordinate systems having the same origin, but where the one system is rotated an angle with respect to the other. The basis vectors of the two systems are apparent form the figure. y y1
x1
j1 j i1 x i Since i1 is a unit vector having directional angle , then it has the coordinates i1 = (cos, sin). j1 is the cross vector to i1, so j1 = (-sin, cos). Thus we have:
Geometrical vectors 25
jij
jii
cossin
sincos
1
1
In the normal coordinate system, the point P and also
OP has the coordinates (x,y), while P in the rotated coordinate system has the coordinates (x1,y1). Therefore:
1111 jyixOPandjyixOP
If we insert the expression for
1i and
1j in the second equation above, collecting the coefficients
with
i and
j we find:
jyxiyxyxOP )cossin()sincos(),( 1111
From which we collect x and y: cossinsincos 1111 yxyandyxx If the equations are solved with respect to x1 and y1 ,then we finally get the wanted transformation formulas: (14.1) cossinsincos 11 yxyandyxx There is a tradition to write the transformation formulas on matrix form. The first coordinate in the column to the left is obtained by multiplying the first row in the matrix by the column to the right, and son on.
(14.2)
y
x
y
x
cossin
sincos
1
1
