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Definition 1.30. Projection of a vector a on an axis l is called a positive number | A1B1 | , if the axis l and the vector A1 B1 are equally directed (Fig.

1.3, а), and a negative number − | A1B1 | , if the axis l and the vector A1 B1 are opposite directed (Fig. 1.3, b).

Projection of a vector a to an axis l is designated as: Prl a .

If φ is an angle between a direction of the axis l and a direction of the vector a , then

Prl a =

cos ϕ.

А1

В1

А1

В1

Fig. 1.3

Thus Pr

a > 0 , if an angle 0 ≤ ϕ < π ,

Pr a < 0 if

π < ϕ ≤ π .

4.4. Linear dependence and independence of vectors. Basis

Definition 1.31. An expression of such a kind as

x1a1 + x2 a2 + ... + xn an

is called a linear combination of vectors a1 , a2 , ... , an .

Definition 1.32. Vectors a1 , a2 , ... , an are called linearly dependent if there are exist such numbers c1 , c2 ,..., cn not all equal to zero, that their linear combination c1a1 + c2 a2 + ... + cn an = 0 , and linearly independent if this equality is carried out only if all numbers c1 , c2 , ..., cn equal to zero.

A set of linearly independent vectors a1 , a2 , ... , an is called a basis of space

Rn	if for each vector b in		Rn	there exist such real numbers	x ,	x , ...,	x ,
					1	2	n
that	b = x1a1 + x2 a2 + ... + xn an .
	This	equality is called	an	expansion of the vector b	in	the basis
a1 , a2 ,		... , an .

	Definition 1.33. Arbitrary nonzero vector on a straight line is called						a
basis on this straight line.

If a vector a is a basis on a straight line then there is a unique expansion of a

vector b such that b = λa , where λ is the coordinate of the vector b in the basis a .

Definition 1.34. An arbitrary ordered pair of noncollinear vectors is called a basis on a plane.

Definition 1.35. An arbitrary ordered triple of noncomplanar vectors is called a basis in space.

If vectors a , b and c are basis in space and a vector d is d = αa + βb + γc, then α, β, γ are coordinates of the vector d in the given basis.

4.5. Cartesian coordinate system

Any coordinate system in space is given by a point O and three non-coplanar ordered vectors e1, e2 , e3 (basis).

The point O is called an origin of coordinate system and straight lines passing through the origin in a direction of basic vectors are called axes of coordinates.

They are usually labeled as x-axis, y-axis, z-axis.

We consider basis i , j, k such that: | i |= 1, | j |= 1, | k |= 1 and

i j, j k , i k . Such basis is called orthonormal basis.

Coordinate system in space with orthonormal basis is called the Cartesian coordinate system.

If to connect any point M in space with the origin O we can consider a vector

r = OM called a radius-vector of the point M relative to the point O. Then there are three unique numbers (x, y, z) such that

r = xi + yj + zk.

Coordinates x, y, z of a radius-vector OM are called coordinates of a point M and are designated as M (x, y, z) (Fig. 1.4).

z		М(х, у, z)
k		у
	j	у
i	j	у
i		у
х
х
	Fig. 1.4

z
аz
		aG
	γ	ау
α	β	ау
α		у
ах		у
х

Fig. 1.5

4.6. Vectors in CCS

Let a vector

be given in CCS (Fig. 1.5). In the basis i , j,

the vector a

may be decomposed as

a = axi + ay j + azk

. It is denoted as

a = {a x ,

a y , a z }.

Here

ax , ay ,

are coordinates of the vector

in this

basis.

These

coordinates are projections of the vector a on coordinate axes, i.e.

ax = Prx a =

cosα,

ay = Pry a =

cosβ,

where α,

β, γ

az = Prz a =

cos γ,

are angles formed by the vector

with

coordinate axes

Οx ,Οy ,Οz respectively.

The length (module) of the vector a can be found by the formula

| a |= ax2 + a2y + az2 .

Then

cos α = |aax | , cos β = |aay | , cos γ = |aaz | .

Definition 1.36. cos α , cos β , cos γ are called directing cosines of the vector a . They define the direction of the vector a and satisfy the condition

cos2 α + cos2 β + cos2 γ = 1.

If A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ) , then

AB = (x2 − x1, y2 − y1, z2 − z1).

The length of the vector AB is written down as:

\| AB \|= (x − x )2		+ ( y	2	− y )2	+ (z	2	− z )2 .
2	1		2	1		2	1

Let vectors be given by their coordinates, i.e.

a = (xa , ya , za ) , b = (xb , yb , zb ),

then

λ a = (λ x a , λ y a , λ z a )

a + b = (xa + xb , ya + yb , za + zb ).

Definition 1.37. Vectors a and b are equal if their coordinates are equal:

xa = xb , ya = yb , za = zb .

Vectors a and b are collinear if their coordinates are proportional:

	xa		=	ya	=	za
	x			y		z
	b			b		b
4.7. Division of line segment according to a given ratio
Let a line segment АВ be		defined				by the points A(x1, y1, z1 ) and

B (x2 , y2 , z2 ) . Then coordinates of a point M(x, y, z) dividing this segment in the ratio | AM |:| MB |= λ may be found by the formulas:

x =	x1 + λx2	y =	y1 + λy2	z =	z1 + λz2

	1+ λ		1+ λ		1+ λ

Coordinates of the point dividing a segment in half

(

λ = 1 , are:

)

x =

x1 + x2

y =

y1 + y2

z =

z1 + z2

Micromodule 4

EXAMPLES OF PROBLEMS SOLUTION

Example 1. The points M1 (3 ; 3; −2)

M 2 (0 ; 1; 4)

are given. Find

а) coordinates, the length, directing cosines and ort of the vector M1M 2 ;

b) coordinates

of a the point

M dividing

a segment

M1M 2 in

the

ratio

| M1M |:| MM2 |= 2 : 3

Solution. а) M1M2

= (0 − 3 ; 1− 3; 4 − (−2)) = (−3;

−2;

6) ;

| M1M 2 |=

9 + 4 + 36 = 7 ;

cos α =

−3

cos β =

−2

cos γ =

An ort of the vector M1M 2

is the following:

e = {cos α, cos β,

cos γ} = {−3 / 7; −2 / 7;

6 / 7} ;

b) λ =

, then

3 +

3 + 2 1

−2 +

xM =

, yM =

, zM =

1+ 2

Example 2. Find a vector

a = {ax ; ay ; az }

forming identical angles with

coordinate axes provided

= 2

Solution. Taking into account the equalities

=| a | cos α , ay

=| a | cos β ,

az =| a | cos γ

and a condition α = β = γ

we write down that

cos2 α + cos2 α + cos2 α = 1.

We get cos2 α =

cos α = ±

= ay = az

= 2

ax =

= ay = az = −2 . The final answer is a = {2;

2; 2} or a = {−2;

−2;

−2} .

Example 3. Given

vectors

a =

{

}

and b =

{

4; 2;

}

−2; 3

−1

. Define

whether the vectors c1 = 2a − 5b

and c2

= a − 2b

are collinear.

Solution. We get

{

}

{

}

{

}

= 2a − 5b =

−4;

20; 10;

−18;

−14;

6 −

−5

{

}

−

{

8; 4;

}

−7;

}

= a − 2b = 1; −2; 3

−2

−6; 5 .

As the coordinates of the vectors c1

and c2

are not proportional, vectors are

not collinear.

Micromodule 4

CLASS AND HOME ASSIGNMENTS

1. The points M1(−4; 5;

−6), M 2 (5;

−7; 2)

are given. Find:

a) coordinates, length, directing cosines and an ort of the vector M1M 2 ;

b) coordinates of a point M which divides a segment

M1M 2 in the ratio

| M1M |:| MM 2 | = 3 : 5 .

2. Find the vector a = {ax ; ay ; az }

if it forms with axes Ox and Oy angles

α =

and β =

respectively, and

= 6 .

3. Define whether the vectors c1 = −a + 4b and c2

= 3a − 2b , constructed on

vectors a =

{

}

and b =

{

3; 1;

}

−2; 3

−1 are collinear.

Answers

а) M M

= {9; − 12; 8};

cos α =

, cos β = −

, cos γ =

;

b) M (−5 / 8;

0,5;

−3). 2.

a = {3 2; 3; ± 3} . 3. No.

Micromodule 4

SELF–TEST ASSIGNMENTS

4.1. Two points M1 and M2 are given. Find:

а) coordinates, the length, directing cosines, an ort of a vector M1M 2 ;

b)coordinates of the point M, if M1M : MM 2 = m : n ;

c)coordinates of the point M2 , if M1M3 = λM1M2 .

4.1.1.	M1 (1; 2; − 1) , M 2 (3; 4; − 2) , m : n = 2 : 5 , λ = 3 .
4.1.2.	M1 (−2; 0; − 4),		M 2 (−4; 1; − 2) , m : n = 3 :1,		λ = 2 .
4.1.3.	M1	(−5;1; 4), M2 (1; 3; 1) , m : n = 3 : 2 , λ = 4 .
4.1.4.	M1 (5; − 1; − 4),		M 2 (11; 1; − 1) ,	m : n = 2 :1,	λ = −2 .
4.1.5.	M1	(−3; − 1; 8),	M2 (−7; − 5; 6) ,	m : n = 1: 4,	λ = −3 .
4.1.6.	M1	(15; − 2; − 14), M2 (11; 0; 10) , m : n = 2 : 3, λ = 4 .

4.1.7.M1 (−8; − 12; 3), M2 (0; − 3; 15) , m : n=1:5, λ = –2.

4.1.8.M1 (10; − 5; − 4), M2 (1; 7; 5) , m : n=3:5, λ = –3.

4.1.9.М1(5; 2; –6), М2(25; –10; 3), m : n = 4:5, λ = 3.

4.1.10.М1(–3; –2; 16), М2(9; 18; 7), m : n = 2:3, λ = –2.

4.1.11.М1(–1; 8; 26), М2(23; 0; 20), m : n = 3:2, λ = –4.

4.1.12.М1(–7; 7; 15), М2(–1; –1; –9), m : n = 2:7, λ =2.

4.1.13.М1(–4; 5; 22), М2(4; –1; –2), m : n = 6:5, λ = 4.

4.1.14.М1(1; –8; 12), М2(25; –2; 4), m : n = 1:2, λ = –2.

4.1.15.М1(4; 9; 14), М2(–2; –15; 22), m : n = 1:3, λ = –3.

4.1.16.М1(–5; 17; 21), М2(4; 5; 1), m : n = 4:3, λ = –5.

4.1.17.М1(2; 11; 33), М2(22; –1; 24), m : n = 4:1, λ = 4.

4.1.18.М1(–7; 4; 13), М2(1; –5; 1), m : n=5:3, λ = –6.

4.1.19.М1(3; –8; 14), М2(–9; 1; 6), m : n = 5:2, λ = 5.

4.1.20.М1(–9; 3; 5), М2(0; 15; 13), m : n = 4:7, λ = –1.

4.1.21.М1(–1; 4; 12), М2(3; 0; 10), m : n = 6:7, λ = –2.

4.1.22.М1(2; 6; 4), М2(6; 4; 8), m : n = 2:1, λ = –3.

4.1.23.М1(–11; 16; 1), М2(–5; 10; 4), m : n = 3:4, λ = 2.

4.1.24.М1(–14; –3; 2), М2(–8; 3; –1), m : n = 4:5, λ = –3.

4.1.25.М1(2; 4; 7), М2(4; 7; 1), m : n = 2:3, λ = 4.

4.1.26.М1(–11; 18; 36), М2(1; 14; 30), m : n = 4:5, λ = –2.

4.1.27.М1(–4; –3; 0), М2(2; 1; 12), m : n = 1:6, λ = –4.

4.1.28.М1(9; 4; 16), М2(49; 28; –2), m : n = 4:3, λ = 2.

4.1.29.М1(0; 5; 21), М2(18; 11; 12), m : n = 6:5, λ = 3.

4.1.30.М1(–3; 5; 20), М2(3; 14; 2), m : n = 1:4, λ = –5.

4.2. Define whether the vectors c1 and c2 , constructed on vectors a and b are collinear.

4.2.1.

a =

{

−2;

}

, b

{

}

= 2a + 4b ,

= 3b − a .

{

−1

4.2.2.

a =

1; 0;

}

, b =

{

−2;

}

= 3a − b .

{

= a + 2b , c

4.2.3.

a =

−2;

}

, b

{

−2;

}

= 5a + 3b ,

= 2a − b .

{

4.2.4.

a =

}

, b

−1;

}

= 4a + 3b

= 8a − b .

{

−3

−1

4.2.5.

a =

}

, b =

{

}

= 3a − 2b .

{

= −2a + b , c

4.2.6.

a =

}

, b

{

}

= 4a + 2b .

{

−2

{

−1

, c

= a + b , c

4.2.7.

a =

−2;

}

, b

−1;

}

= 4a − 2b ,

= b − 2a .

{

4.2.8.

a =

}

, b

{

−1;

}

= 6a − 3b ,

= b − 2a .

{

−1

4.2.9.

a =

−2;

−3;

}

, b =

{

}

= 3a − 9b

= −3b − a .

−2

, c

4.2.10. a =

{

−1;

}

, b

{

−2;

}

= 2a − b ,

= 3b − 6a .

{

4.2.11. a =

}

, b =

}

= 3b + 6a .

{

−1

{

= 2a − b , c

4.2.12. a =

}

, b

−2;

}

= 5a − 2b

= 5b + 3a .

{

−2

4.2.13. a =

−2;

}

, b =

{

−3;

}

{

−1

c = 2a + 3b ,

= 2b + 3a .

4.2.14. a =

}

, b

{

−3;

}

= 4a − 2b ,

= b − 2a .

{

4.2.15. a =

}

, b

−3;

}

= 4a − 2b ,

= b + 2a .

{

4.2.16. a =

−2;

}

, b

{

}

= 4b − a .

{

−4

= 2a + 3b , c

4.2.17. a =

}

, b

{

−2;

}

= a − 2b ,

= 4b − 3a .

{

−5

4.2.18. a =

−3;

}

, b =

{

−2;

}

{

−1

= 4a + 3b ,

= 3a − b .

4.2.19. a =

}

, b =

{

}

= 4a + 3b

= 6a − b .

−3

−1

4.2.20. a =

{

}

, b

{

}

= −2a + 3b ,

= 3a − 2b .

{

4.2.21. a =

}

, b

{

}

= 5a + 2b .

{

−3

−1

= a + b , c

4.2.22. a =

−2;

}

, b =

{

−1;

}

= 4a − b ,

= b − 3a .

{

4.2.23. a =

}

, b

{

−1;

}

= 5a − 3a

= b − 4a .

{

−1

{

4.2.24. a =

}

, b

}

= 3a − 7b ,

= −2b − a .

{

−2

{

, c

4.2.25. a =

−1;

}

, b =

−2;

}

= 2a − b ,

= 3b − 4a .

{

4.2.26. a =

}

, b =

}

= 3b + 2a .

−1

= 2a − b , c

4.2.27.

a =

{

}

, b =

{

−2;

}

= 5a − 2b

= 4b + 3a .

{

−2

4.2.28.

a =

}

{

−3;

}

{

−2; 4; −1 , b =

c = 2a + 3b

c = 2b + a .

4.2.29.

a =

}

, b =

{

−3;

}

= 4a − 2b ,

= b − 2a .

{

4.2.30.

a =

}

, b =

−3;

}

= 4a − 2b ,

= 2b + a .

Micromodule 5

BASIC THEORETICAL INFORMATION.

DOT PRODUCT OF TWO VECTORS

Definition of dot product of two vectors, its properties and the coordinate form. Condition of perpendicularity of two vectors.

Literature: [1, chapter 4], [4, section 3, item 3.2], [6, chapter 2, § 4], [7, chapter 1, § 3], [10, chapter 1, § 2], [11, chapter 1, § 2].

5.1. Dot product of two vectors

Definition 1.38. Dot (scalar) product of two vectors a and b is the number

a b (or (a, b) ) equal to the product of lengths of these two vectors and a cosine of the angle between them:

a b =| a | | b | cos ϕ.

If one of vectors a or b is zero, then according to the definition

a b = 0.

If equalities | a | cos ϕ = Prb a , | b | cos ϕ = Pra b are true then a b =| b | Prb a =| a | Pra b.

The geometrical significance of the scalar product is next. The scalar product of two vectors is equal to the product of the length of one vector by a projection on it of the other vector.

Then

Prb a = a| bb| .

5.2. Properties of dot product

Algebraic properties of a dot product are

1)	a b = b a ;	2) (λa) b = λ(a b) ;
3)	a(b + c) = a b + a c .
Geometrical properties of a dot product are next.
1)	If a ≠ 0 and b ≠ 0, then	a b > 0 if the angle ϕ < 900, and a b < 0 if

the angle ϕ ≥ 900.

2) A dot product of two nonzero vectors is equal to zero if and only if these vectors are perpendicular.

3) The scalar square of a vector is equal to a square of its length, i.e. a a = a2 =| a |2 .

Thus,

| a |= a a .

Condition of perpendicularity of two vectors

Two nonzero vectors a and b are perpendicular if and only if their scalar product is equal to zero:

a b a b = 0.

5.3. Representation of a dot product in coordinate form. An angle between two vectors

Let vectors a and b be a set determined by coordinates a = (ax , ay , az ) , b = (bx , by , bz ) .

Then	a b = ax bx + ay by			+ az bz .
Conclusions from this formula are the following:
1) condition of perpendicularity of two vectors a and b :
	ax bx + ay by + az bz = 0 ;
2) the length of the vector a is \| a \|= ax				2 +ay	2 +az	2	;
3) a cosine of the angle between the vectors a and b is
cos ϕ =	a b	=	axb x +ayby + azbz					.

	\| a \|\| b \|		ax2 + ay2 + az2 bx2 + by2 + bz2

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