Analytical geometry

MINISTRY OF EDUCATION AND SCIENCE RUSSIAN FEDERATION TOMSK STATE UNIVERSITY









E.E.Koryakina, A.V.Nikol’skii

ANALYTICAL GEOMETRY




Textbook












Tomsk
Publishing House Tomsk state university 2022

UDK 514.12


BBC В143

К 66








K66 Koryakina E.E., Nikol’skii A.V. Analytical geometry: Textbook.
- Tomsk: Tomsk state university, 2022. - 119 с.




     This textbook is intended for first-year students of physics and computer science studying analytical geometry in English. It covers course sections dealing with vector algebra, geometry of straight lines in the plane, and geometry of planes and straight lines in space.



UDK 514.12


BBC В143













© Koryakina E.E., Nikol’skii A.V., 2022


                 © Tomsk state university, 2022

1. Vector Algebra


1.1. Free Vectors. Linear Operations on Free Vectors. Coordinates of Vectors



    Definition. A fixed vector AB is a directed segment АВ. The modulus |ab| of the vector AB is the length of the segment АВ.

    A vector is called zero vector if its tail and head coincide: AA = 0.

    Definition. Two fixed vectors AB and CD are called collinear if the segments АВ and CD lie on the same straight line or on parallel straight lines.

    For collinear vectors, the concepts of codirectionality and contradirectionality are introduced.

    Definition. If vectors AB and CD lie on parallel straight lines, they are codirectional (AB TT CD) when the points В and D lie on the same side from the line АС and contradirectional (AB ^T CD) when the points В and D lie on opposite sides from the line АС.
    Definition. If vectors AB and CD lie on the same line, they are codirectional ( AB TT CD ) when the ray defined by one vector entirely contains the ray defined by the other vector. If not, they are contradirectional (AB ^T CD).


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    The zero vector 0 has no direction and is collinear to any vector.

    Definition. Two fixed vectors AB and CD are equal if the two following conditions are satisfied:
    1.  AB TT CD;
    2.  |AB| = \cd\ .
    Since this relation is a relation of equivalence (i.e., it is reflexive, symmetric, and transitive), one can introduce the following
    Definition. A free vector a is a class of equivalence in the set of fixed vectors.

    Let us introduce and consider linear operations on free vectors.
    1. Addition of free vectors.
    Definition. The sum of free vectors A and b is the free vector с = a + b which is obtained as follows:
    1. Vector AB = a is drawn from point А;
    2. Vector BC = b is drawn from point В;
          >     >    >     ^ⁱ
    3. AC = AB + BC = a + b .


    Properties of addition:

    1. a + b = b + a.
    2. (a + b)+ c = a + (b + c).
    3. EQ; Va : a + 0 = a .
    4. Vfi; E(-a): a + (-a) = 0.

4

    2. Multiplication of a vector by a number
    Definition. The product of a free vector <5 and a number a e □ is the free vector b = aa such that

   1.

2. a=0.

b=14 lcl,
b TT ci, a > 0; b ^T a, a < 0; b = 0,



    Properties of multiplication of a vector by a scalar:

   1.  a(P< )=(aP) a.
   2.  (a + P) a = aa + P<.
   3.  a(a + b) = aa +ab).
   4.  1 • 3 = c).


    Thus, the set of free vectors with the two considered operations is a (finite-dimensional) linear space over the set of real numbers □ .
    Definition. Coefficients of the resolution of a vector in terms of a basis are called coordinates of the vector in this basis.
    Addition of vectors is reduced to addition of corresponding coordinates; multiplication of a vector by a number is reduced to multiplication of every coordinate by this number.

5

PROBLEMS




    1. Let M, N, P, and Q be middles of sides AB, CD, BC, and DE of a pentagon ABCDE, and let K and L be middles of the segments MN and PQ. Prove that AE || KL

and

W - 4| AE

Solution.

From the tetragon MBCN, we have ---- ---- ---- --MN - MB + BC + CN


—- 1 —- —- 1 —-or MN Ab AB + BC + -CD.
2         2


It means that

--- 1 —- 1 —- 1 —- 1 —-MK --MN abab + -BC + -CD .
    2    4   2   4


    Similarly, from the tetragon PCDQ,


—- 1 —- —- 1 —-PQ bcBC + CD + -DE.
22


    Therefore,


—- 1 —- 1 —- 1 —-LQ--BC + -CD + -DE .
424


   From hexagon AMKLQE, we obtain --------------*- -*- --*- -*- --*- --*-AM + MK + KL + LQ + QE - AE;


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therefore,


         --* -* I -* --* --* I -*
KL = AE—AB -MK - LQ—DE. 2                  2

  Substituting MK and LQ obtained earlier, we obtain

  —* —*  1 —* 1 —* 1 —* 1 —* 1 —* 1 —*
  KL = AE--AB--AB--BC--CD--BC--CD 2    4   2    4    4   2



1 —* 1 —* —* 3 —* 3 —* 3 —* 3 —* — DE — DE = AE—AB—BC —CD—DE =
42       4444

   * 3 * * * *\ * 3* 1 *
  = AE - ³1 AB + BC + CD + DE ) = AE - ³ AE = ³ AE.
4\       /44


   Finally,


—* 1 —* KL = -AB.
4


  Thus


                       ----* ----* I--------* I 1 I-* I
                       KL || AB and |KL| = 4| AB|




     2.   Given a parallelepiped ----------------------*
  ABCDA1B1C1D1 in which AB = a, --* *•       ---*
  AD = b , and AA₁ = c*. Find
  d = A - b + *.
So/utzon.


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Since

---* ---* ---* AD + DB = AB,

we have
---*■   *
DB = 3 - b ,
---*  *
DD1 = AA1 = c .


Since

we have

----* ----* ----* DD1 + D1B1 = DB1,

                        -----* —*      —*
                        DB1 = d = a - b + c .



    3.       Given a vector <* such that \a| = 3 . Find vector b of the opposite direction if b| = 5 .
    Solution.


Since b ^T a, b = aa , where a < 0.



Hence |bj = |a| • \a | and |a|


—* b
a

5
3

                   5      *
Since a < 0, a = — and b =
                   3


	

5* -a.
3

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    4.      Decompose vector S = a + b + — in three noncomplanar vectors m , n, and p, if m = 3 + b - 2c, n = a - b , and p = 2b + 3c .
    Solution.


The decomposition of the vector in the general form is
S = xm + yn + zp . We have to find x, y, and z. Substituting the expressions from the condition, we obtain

S = x ( a + b

- 2c j + y (a - bj + z (2b— + 3c j =


          = a (x + y) + b (x - y + 2z) + — (-2x + 3z);


however, by condition, S = a + b + c . Therefore, by virtue of the uniqueness of the decomposition, we have

' x⁺y=¹
< x - y + 2 z = 1.
- 2 x + 3z = 1


    Solving the system, we obtain




2
x = 5’

3
y = 5’

3
z = —
5

       -  2    3  3
      1O  "J   J   J —*
   and S = — m + - n + — p .
          5    5  5


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    5.      Prove that vectors b = {2,4,5}, c = {1,0,2}, and d = {2,-1,3} are linearly independent and find coordinates of the vector a = {1,3,5} in the basis b , c , d.
    Solution. Calculate the determinant composed of coordinates of vectors b , cⁱ , and d :

2

4

5
2
3

4

5
3

2
2

4

d=1

-1

-1

-1

-2

-1

-17 + 20 = 3 * 0.

2

0

It means that rows of the determinant are linearly independent and three linearly independent vectors b , cⁱ , and d form a basis in the three-dimensional space.
a = xb + _yc + zd,
{1,3,5} = x{2,4,5}+ j{1,0,2}+ z{2,-1,3},
{1,3,5} = {2x + y + 2z, 4x - z, 5x + 2y + 3z}.

By virtue of uniqueness of the decomposition in terms of the basis, we have a system for finding coordinates of the vector aⁱ in the given basis:
2 x + y + 2 z = 1 <   4 x - z = 3  ;
5x + 2 y + 3z = 5

2
Л= 4

12
0 -1 = dT

= 3 ^ 0.

523

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