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ANALYTICAL GEOMETRY
Practical class №6
...

Decomposition of vector a (pic
...
1) by basis i , j , k has following form:

a  ax i  a y j  az k ,
where, ax  PrOx a , a y  PrOy a , az  PrOz a
...


0
k a

O
i

ax



write a  ax ; a y ; az
...

a

Point A  x A ; y A ; z A  is initial point of the vector,

x
Рис
...
1

point B  xB ; yB ; zB  is terminal point of the vector,

components of vector AB we find using following formula:

AB   xB  x A ; yB  y A ; zB  z A 
...
1)

Point C  xC ; yC ; zC  is the midpoint of segment AB , then

z  zB
x A  xB
y  yB
; yC  A
; zC  A

...
Then:

xM 

x A  xB  x C
3

; yM 

y A  yB  y C
3

; zM 

z A  zB  z C
3


...

Let a  ax ; a y ; az , b  bx ; by ; bz 









 a    ax ;  a y ;  az  ,   ,




a  b   ax  bx ; a y  by ; az  bz 
...

bx b y bz

(2
...
3)
27

Practical class №6
...

Problems
1
...

2
...
Find
a) vector AB ;

b) AB
...
The points A ( 2; 2) , B (2; 6) are given and point C is the midpoint of
AB
...

4
...
Find the unit vector for vector a and the
projection of a on Oy
...
The points A ( 2; p; 3) , B (1; 0; 2) and C (q; 1; 1) are given
...
The vector BC   1;  2; 3 and point B ( 3; 1;  2) are given
...

7
...

8
...


9
...
Verify that the points A (2; 1; 0) , B (0; 4;  3) , C (2; 3;  5) and
D(2;  3; 1) are vertices of a trapezoid
...

11
...
Find the coordinates of the intersection point of its
diagonals
...
The vectors AB  ;  ;  6  and AC  4; 2;  3 are given
...
The vectors a  3 ; 2 ; 0  , b  2 ; 0 ;  3 are given
...


28

14
...
Find the coordinates of the

triangle height base drawn from the vertex C
...
The points A  1; 1; 0  , B  1; 2; 2  and С  3; 2; 0  are the vertices of the
triangle
...

16
...
Find
components of the vector AB  AC
...
The points A  3; 7;  4  , B  2 ;  1; 1 and С  1; 3; 0  are the vertices of
the triangle
...

18
...


19
...
Find components of
vector x , if: a) a  x  b ; b) 3a  b  0,5 x
...
Which values of x are vectors a  1; 1; 2  and b x 2 ; x  2; x 2  12



collinear at?
21
...
Find
components of the vector BC
...
The vectors a  p; 0;  2  and b   2; p; 0  are given
...
Find components of the vector p , which is collinear and oppositely directed
to the vector q  3;  5;  2  , if p  3 38
...
BC and AD are the bases of the trapezoid ABCD and AB  7;4;5 ,

AC  3; 2;  1 , AD  20;  4;  12 
...

25
...


29

26
...
The points A 1;  1; 2  , B  3; 0; 2  and C  1; 2; 0  are
the vertices of the triangle
...

 Find coordinates of the point D which is midpoint of BC :
3   1
x x
y y
z z
02
20
xD  B C 
 1; y D  B C 
 1; z D  B C 
 1
...

Problem 2
...
Find the magnitude of vector

c  a  3b
...
Then

c  02   7    5  74
...
Vectors a  6; m;  2  and b  3; 2; n  are collinear
...

6 m 2
2 6
m 6
 If vectors a and b are collinear then  
 
 and

2 3
n
3
3 2 n

m  4 and n   1
...

bx 3

Task 4
...
Find the diagonal length DB
...
Find the components of the vectors CB и AB :
CB  2  1; 1  2; 0  0   CB 1;  1; 0  ;
AB  2  1; 1  0; 0  2   AB 1; 1;  2 
...

Answers
1
...
2
...
3
...


4
3
0
 ; 𝑃𝑟𝑜𝑗𝑂𝑦 𝑎⃗ = 0 5
...
6
...

5
5
7
...
8
...
9
...
10
...

4
...
 2,5; 9;  4,5
...
  8,   4
...


89
...
 4; 0,5;  2,5
...
2 2  2 5
...
 6; 5;  1
...
3
...
а) no; b) yes
...
а) 1;  2; 2  ;
b)  2;  4; 12
...
x  2
...
 7; 7;  6 
...
p  0; p  2
...
 9; 15; 6 
...
15;  3;  9 
...
а) a  b ; b) a  b ; c) a  b ; d) a  b
...
  6
...
Dot product of vectors
...
4)



 

where   a; b

is the angle between vectors a and b , 0    
...


a2  a  a  a
...


b



Let a  ax ; a y ; az , b  bx ; by ; bz the dot product is:

a  b  axbx  a y by  az bz
...

(2
...




Let a  ax ; a y ; az




(2
...
2
...
2
...
Cosines of direction angles we can find
31

using following formulae:

ax
;
a

ay

az

...

cos  

cos  

cos  

;

Property of direction cosines is cos2   cos2   cos2   1
...

Problems


 

1
...

2
...

3
...



4
...

5
...
Find the direction cosines of the vector a ( 6; 2; 3)
...
Find the cosine of the angle between vectors a  2; 1; 3 and b   1; 3; 2 
...
Find 𝑃𝑟𝑜𝑗𝑏⃗⃗ 𝑎⃗, if a   2; 2; 1 , b   3; 6; 2 
...
Find the work of the force F  2;  1; 4  , if the point of its application moves
from B  3; 5;  2  to C  2; 1; 0 
...
Find dot product of the vectors a  4b  a  b , if a  2 2 , b  0, 5 ,









 a; b   1350
...
The vectors a  p; 1; 1 and b  1;  p; 2  are given
...
The points A ( 3; 2; 1) and B (1; 2; 1) are given
...

13
...
Find the inner corners of this triangle
...
Find the angle between the diagonals of the parallelogram constructed on the
vectors a  2 i  j and b  2 j  k
...
The force F  3;5;  2  moves the material point from the point B  4; 2; 3
to the point C , lying on the axis Oy
...

16
...
Find the magnitude

of the vector AB , which is perpendicular to vector a , if the point B lies on
the axis Oz
...
Find dot product a  2b  2a  b , if a  2 2 , b  3 , a; b  1350
...
Can vector a form angles with coordinate axes:

  300 ;   600 ;   1500 ?
19
...
Find
component a z , if a  6
...
Calculate the angle between vectors a and b , if: a  2 ;  2 ; 0  ,

b  3 ; 0 ;  3
...
What value of  are vectors a   i  3 j  4 k and b  4 ;  ;  7 
perpendicular at?
22
...
Find the magnitude
of the vector AB , if the point B lies on the axis Oy and dot product

a  AB  1
...
The force F  2;  3; 1 moves the material point from B  3; 1;  2  to the
point C  4;  1; 3
...

⃗⃗⃗⃗⃗⃗
24
...


25
...
Verify that the quadrangle ABCD is square, if: A (3; 5; 6) ,
B (1; 8; 12) , C (5; 10; 9) , D (3; 7; 3)
...
What values of x and y is vector a  3 ; y ;  1 perpendicular to the
vector b  x i  5 j  2 k , and the vector b is perpendicular to the vector

c 1; 0,4 y ;  0,5 ?

28
...
Find dot product a  AB

if the point B lies on the axis Ox , and vectors AB and a are collinear
...
The vectors AB  3;  1;  2  , AD  3; 4; 8 BC  2; 5; 1 are given in the
quadrangle ABCD , m and n are its diagonals
...

30
...

The solution of typical problems







Problem 1
...



Using the definition of dot product and its properties, we obtain:
2a  b  a  2b  2 aa  b  a  4 a  b  2b  b 





 2 a 2  3 a  b  2b 2  2 a

 2

 
3

2

2

 3 a b cos1500  2 b

2



 3 3  2  cos 150 o  2  2 2  6  9  8  7
...
The vertices of the triangle A(1;  2; 4) , B(4;  2; 0) ,
C (3;  2; 1) are given
...

 а) To find angle at the vertex B we consider two vectors having a
beginning at point B : BA and BC
...

2

b) AB  2 AC  (3; 0;  4)  2(4; 0;  3)  (11; 0; 2)
...


Problem 3
...

 Using the condition of perpendicularity of vectors a and b :
a  b  0  k 1  ( 3)  2  2  ( k )  0  k   6
...
For what values of т the angle between vectors a (7;  1; 2 m)
and b  2; 4 m; 1 is sharp
...
Then m  7
...
6
...
–38
...
1
...
1
...
Yes
...
7
...


...
10
...
10
...
0;  2
...
45 0
...
A  45 , B  45 , C  90
...
90 0
...
C  0;4;0 
...
2 5
...
– 20
...
No
...
3
...
600
...
4
...
5
...
13
...
20
...
m 
...
x  1; y  1
...
–50
...
4
...


...
Cross product and scalar triple product of vectors
Cross product of two vectors a and b is vector c (pic
...
3), we write c  a  b ,
if:
1) c  a , c  b ;
2) vectors a , b , c form right-hand vector triple;




3) c  a  b  sin  ,   a; b



0     
...

To find cross product we use following formula:
a  ax ; a y ; az b  bx ; by ; bz :









i
a b  ax
bx

j
ay
by

k
az 
bz

 ay

 by


ax

az

bx

bz

az
bz

;

с

;

b


a

ax
bx

ay 

...
2
...
7)

The area of parallelogram formed by vector a and vector b is S  a  b
(pic
...
3)
...

M  AB F , where the moment M of a force F about point A
  
Scalar triple product of vecrors a , b , c is number denoted by a b c ,
where a b c  (a  b )  c
...

cz



(2
...
Also we can determine the volume
of triangular pyramid formed by these vectors (piс
...
4) :
1  
Vпир  a b c
...
9)

с

the condition of coplanarity of vectors: a b c  0
...

  
a b c  0  triple a , b , c is left-hand
...
2
...
Find a  b , if a  3, b  4, a  b
...
Find a  b , if a  1; 2; 3 ; b  2 i  j  3 k
...
Find at least one vector which is perpendicular to the vectors

a 1;  3; 4  ; b  5 i  2 j  k
...
Find scalar triple product a b c , if a 1; 2; 1 ,

b  3;  4; 2  , c  0; 1; 2 
...
Are the vectors a 1; 4; 3 ; b  0; 1; 2  ;

c  1; 2; 3 coplanar?

6
...
Find a  b  a  b , if a  3, b  4, a; b  1200
...
Find the area of parallelogram formed by vectors a  2;  2; 1 ,

b   i  j  4k
...
Find the area of triangle with the vertices in the points A 1;1; 1 ,

B  2; 3; 4  , C  4; 3; 2  and its height drawn from the vertex A
...
Find the magnitude and directional cosines of the moment of the force
F 1;  4; 2  , applied to the point C  2; 1; 1 relative to point D 1; 3;  1
...
Find the volume of a triangular pyramid with vertices at the points
A  0; 1; 2  , B 1; 2; 2  , C   1; 2; 2  , D  1;  1; 3 and its height lowered
onto the base ABC
...
Prove that points A 1; 2;  1 , B  0; 1; 5 , C   1; 2; 1 , D  2; 1; 3
lie in the same plane
...
The vectors a  2; 1;  1 , b  3; 0; 1
...
Find b  a  2a  b
...
Do the points A  2;  13; 3 , B 1; 4; 1 , C   1;  1;  4  ,

D  0; 0; 0  lie in the same plane?

16
...

17
...


18
...
What value of l a b c  a  c , if a  2;  1; l  ; b 1; 3;  3 and

c  i  2 j  k at?

20
...
Find the volume of a parallelepiped formed by the vectors a  4; 2; 3 ,

b 1; 3; 1 , c 1; 3; 7 
...
Do the vectors a  2; 3; 0  ; b 1;  1; 3 ; c 1;  1; 5 form a basis?
The solution of typical problems
Problem 1
...

 Let BH is the height of the parallelogram drawn from the vertex B
...

AD
37

i

j

k

AB  AD  4 5 2  i
2 1

2

5 2
1

2

j

4 2
2

2

k

4 5
2 1



  12 i  12 j  6 k
...
Then

AD : AD  4  1  4  3 and the height length of the parallelogram, drawn
from the vertex B is BH 

AB  AD



AD

18
 6
...
Find the volume of a triangular pyramid with vertices at the
points A 0; 0; 1 , B 2; 3; 5 , C 6; 2; 3 , D 3; 7; 2
...

1
1
Then Vпир   AB AC AD  120  20 (cubic units ) 
6
6
Problem 3
...
Are these vectors coplanar? In the case of their non-planarity to find
triple (right-hand or left-hand) they form, and calculate the volume of
parallelepiped formed by these vectors
...


2 4 3
Since the scalar triple product is non-zero, the vectors are non-planar and
the triple is left-hand (since the scalar triple product is negative)
...
12
...
3 i  3 j  3k
...
11i  21 j 13k
...
1
...
No
...
Left-hand
...
12 3
...
9 2
...
S  24, h  2 3
...
M  2 5; cos  
5

38

cos   0; cos   

1

1

...
V  , h  1
...
3 i  15 j  9k
...
Left
...
Yes
...
V  20, h  2 6
...
7
...
5
...
1
...
t  0,5
...
60
...

Yes
...
Equations of the straight line
Let L is a straight line, M 0  x0 ; y0  is a point on the line L , N is normal
vector of L , a is a direction vector, m is slope (pic
...
5)
...
2
...

y
M

y
N

y0
b

O

L

y  y0

M0
x  x0

a

x0

x

x

Рис
...
5

Table 2
...




A x  x 0   B  y  y 0   0

M0, N

M0 , m
M 0 , M1

Ax  B y  C  0
y  y0  m  x  x0 



Standard form of a line

N  A; B 

Slope-intercept form of a line

y  mxb

Equation of a line crossing
through two points M 0 and

x  x0
y  y0

x1  x0 y1  y0

M1
...
2
...

Distance from point M 0  x0 ; y0  to line L : Ax  By  C  0 we can find
using following formula:   M 0 , L  

Ax0  By0  C
A B
2

2


...
2
Angle
Lines equations



between L 1 and L 2

N 1  A1 ; B1  , N 2  A2 ; B 2 

L1 : A1 x  B1 y  C1  0

N1  N 2

cos  

L 2 : A 2 x  B2 y  C2  0

N1 N 2
(  is acute angle)

L1 :
L2 :

x  x0 y  y0

m1
n1
x  x1
m2



a1  m1 ; n1  , a2  m 2 ; n 2 

y  y1
n2

L1 : y  m1 x  b1

cos  

a1 a2

(  is acute angle)

tg  

L2 : y  m2 x  b2

a1  a2

m2  m1
1  m1 m2

Condition Condition for
for L 1 | | L2
L1  L2

N 1 || N 2
A1

B
 1
A 2 B2
 
a 1 || a 2

m1

n
 1
m 2 n2
m1  m2

N1  N 2
N1  N 2  0

a1  a 2
a1  a 2  0
m1  m2  1

Problems
1
...
Find the equation of a straight line, if
а) the straight line passes through the point M 0  2; 3 and form the angle with
axis Ох is equal to 450 ;
b) the straight line passes through the point M 0  2;  3 and it is parallel to the
vector a 1;  3 ;
c) the straight line passes through the point M 0  1; 4  and it is perpendicular to
the vector n  3;  2  ;

d) the straight line passes through the point M 0 1;  2  and it is perpendicular to
the line 3 x  2 y  5  0 ;
e) the straight line passes through the point M 0  3;  1 and it is parallel to the
straight line 2 x  3 y  4  0;
f) the straight line passes through the point M 0  5;2  and it is parallel to the
axis Оy;
g) the straight line passes through the points M 0 1;  2  and M1  2;3
...

3
...

40

4
...

5
...
What value of a are straight lines x  2 y  5  0 , ax  y  4  0 are
parallel at?
7
...
Find the angle between straight lines 3 x  y  5  0 and 2 x  y  3  0
...
The points А  7;  6  , B  7;  6  , C  9;  9  are the vertices of triangle
...

10
...

11
...

Material for individual work
12
...

13
...
Find slope of the straight line,
the value b and build this line
...
The equations AB : x  2 y  4  0 , AC : 2 x  5 y  10  0 ,
BC : 3 x  2 y  12  0 are the sides of the triangle
...

15
...
Find the equation of the middle line
of the triangle parallel to the side AC
...
The vertex C (2; 5) and the height equation (BH ) 4 x  y  12  0 of the
triangle are given
...

17
...

18
...
Find the equations of the incident and reflected
rays
...
Equations of two sides of a rectangle 2 x  3 y  5  0, 3 x  2 y  7  0 and
one of its vertices A 2;  3 are given
...

20
...

21
...

22
...


The solution of typical problems
Problem 1
...
Find the standard equation of the line: 1) the side AB ; 2) the
height BH
(pic
...
6)
...
The vector a  AB  (3;  3)
B
is the direction vector of the straight line
AB
...
2
...
Then

3
3

x  1  y  4  x  y  3  0 is the
standard equation of the line AB
...
The vector AC (1;  3) is the normal vector to the line BH
...

Problem 2
...

2
3

 Express y from this equation
...

3

Problem 3
...


 Coordinates of the intersection point we find from the system of
 7x  2 y  17  0,
equations: 
14 x  9 y  24  0
...

Problem 4
...
If L1  L 2  N 1  N 2 

N 1 N2  0 

 c  1 c  3   3  c  2c  3  0  c 2  7c  12  0  c1  3;

c2  4
...
а, b, c, d, g
...
а) x  y  5  0; b) 3x  y  3  0; c) 3x  2 y  11  0;
d) 2 x  3 y  8  0; e) 2 x  3 y  3  0; f) x  5  0; g) 5x  y  7  0; m  5
...
45
...
AB : y  6; CM : x  3 y  18  0; CH : x  9
...
y  5 x  5  0
...
(0; 4)
...
а) 3 x  5 y  21  0; b) y  2; c) 5x  y  9  0
...
(2; 1)
...
а) m   , b  0; b) m  0, b  7
...
No, obtuse
...
0,5
...
1
...
m   , b  6
...
A  0; 2  , B 8; 6 
...
3 x  2 y  9  0
...
x  4 y  18  0
...
y  2 x  6  0
...
y  2 x  6; y  2 x  6
...
2 x  3 y  13  0; 3 x  2 y  0
...
1; 2 
...
3
...
Quadratic curves (Curves of the second order)
Circle equation in standart form with radius R and center C  a; b  has
form:
(2
...

Ellipse (pic
...
7)
...
2
...
Points F1  c; 0  , F2  c; 0 
are foci of these curves ; r1  F1M , r2  F2 M
lies on the curve
...
2
...
3
...
2
...
3

Ellipse

Name

x2

Standard equation

a

2



y2
b

2

Hyperbola

x2

1

a

2



y2
b

2

1

а is transverse axis

Axes

а is major axis
b is minor axis (а>b)

Characteristic property

r1  r2  2a

r1  r2  2a

a 2  c 2  b 2 , a  c 
c
  1
a

c 2  a 2  b 2 , c  a 
c
  1
a
b
y x
a

Equation of parameters
Eccentricity
Asymptote

absent

b is conjugate axis

p
;
2

y 2  2 px is standard equation of parabola, where p > 0 , F 

of parabola; line x  


0  is focus


p
is directrix
...
2
...
2
...
2
...
Determine which lines in the plane define the equations:
1) x 2  4 y 2  4
2) x 2  y 2  4
3) 4 x 2  y  4
4) x 2  4 y 2  4
5) x  y  4
6) x 2  4 y 2  0
...
Do the points A (7;  6) , B (7;  6) and C (9;  9) lie on the circle

 x  1

2

 y 2  100 , inside the circle or outside the circle?

3
...
Find the center and
the radius of the circle
...
The circle is given by equation x 2  y 2  2 x  6 y  6  0
...

5
...
Find
the circle equation
...
Find the parabola equation, which vertex is in the origin, if:
а) the axis Oy is the axis of symmetry and the parabola passes though the point

B 1; 1 ;

b) the axis Oy is the axis of symmetry and the parabola passes though the point

D  4;  8 ;

c) the axis Ox is the axis of symmetry and its focus is in the point F  2; 0 
...
Find foci and
7
...

2
2
8
...
Find the equation
of hyperbola asymptotes
...
Find the circle equation that touches the axis Ox at the origin and passes
through the point A (0;  4)
...
Find standard equation of the ellipse, if:
а) a  10, c  8 ;
b) c  6;   0,6
...
Find standard equation of the hyperbola which passes through the points





M 1  6; 1 , M 2 8; 2 2
...
Build graphs of functions: а) y  2  x  1  1; b) y  x 2  4 x  2
...
Plot shapes on a plane bounded by lines: а) y  x 2  2 x, y  x  2;
b) y   3  x 2 , y  0; c) x 2  y 2  1, x  3
...
The circle is given by equation x  y  4 x  2 y  4  0
...

15
...

Find the equation of the line connecting the centers of these circles
...
Find srandard equation of the ellipse, if:
2

2









а) the ellipse passes through the points M 1 4;  3 , M 2 2 3; 3 ;
б) the ellipse passes through the point M





15;  1 , c  4
...
Find standard equation of the hyperbola , if:
3
а) c  10, b  4 ; b) c  6;  
...
Find the parabola equation, which vertex is in the origin, if:
а) the axis Oy is the axis of symmetry and the parabola passes though the point
A  9; 6  ;
б) the axis Oy is the axis of symmetry and the parabola passes though the point
C  1; 3
...
Find standard equation of the hyperbola, if:
а) the hyperbola is passes through the point M  2;  1 and y  

2
x are its
3

asymptotes;
b) the hyperbola is passes through the point M  5; 3 and   2 ;

x2 y2
c) the foci of the hyperbola coincide with the foci of the ellipse

1
25 9
and   2
...
Find points of intersection of the circle x  82   y  22  50 and the line
x  y  0
...
Find the equation that defines the geometric location of the plane points
equidistant from the center of the circles x  8 2   y  12 2  25 and

x  6 2   y  8 2  16
...
The circle is given by equation x 2  y 2  8 x  6 y  0
...

 Complete the squares in the equation:

x

2

 2  4  x  4 2   4 2   y 2  2  3 y  32   32  0 

  x  4   y  3  16  9   x  4    y  3  5 2
...

2

2

2

2

46

Problem 2
...


 The point C  1; 2  is the circle center
...
Then the point M lies on the circle
...
Find the equation of tangent
line passing through the point M :
3  x  2   4  y  2   0  3x  4 y  14  0
...
Find standard equation of the ellipse passing through the
16 

 12 
points M 1  4;  and M 2  3;  
...

5

 5

x2 y2
 The equation of ellipse is 2  2  1
...

 a 2 25 b 2
let

1
16

t
;
 z , then
a2
25 b 2

2

16 t  9 z  1,
t  1 25,
a  25,

 2

9
t

16
z

1
z

1
25



b  16
...
Then
25 16
c
3
c  a 2  b 2  25  16  3
...
Find standard equation of the hyperbola passing through the
point M  9; 8 , y  

2 2
x are its asymptotes
...
Let's substitute the coordinates of the point М
3
a
into the equation of the hyperbola
...

2
2
 
b  a
b 
a
 a


9
3
3
2
2
x
y

 1
...

The equation of hyperbola is
9
8
c
17
and F1  17; 0 , F2 17; 0
...
1) ellipse; 2) circle; 3) parabola; 4) hyperbola; 5) straight line; 6) two
intersecting lines
...
A on the circle, B inside the circle, C outside the circle
...
 3; 1 ; 10
...
8
...
 x  1   y  1  25
...
а) y  x ; b) y   ;
2
2

2

2

3
2
2
c) y 2  8 x
...
5; 3;  4; 0  ;  4; 0 
...
y   x
...
x  y  4 y  0
...
14
...



 1
...

 1; b)
10
...
3x  4 y  12  0
...
а)

 1
...
а) 
84 16
18 27
20 4
x2
y2
x2 y2
2
2

 1;

 1
...
а) y  4 x; b) y  9 x
...
а)
b)
16 20
74 79

x2 y2
x2 y2
b)

 1; c)

 1
...
1;  1 ;  9;  9 
...
x  2 y  27  0
...
Planes
We can write the equation of the plane passing through the point


M 0 x0 ; y0 ; z0  with normal vector N   A; B; C   0 :

Ax  x0   B y  y0   C z  z0   0
...


(2
...

Parallel planes, perpendicular planes and angle between planes are given in table
2
...

Table 2
...


(2
...
Does the point M 1; 2; 3 lie on the plane P : x  3 y  z  3  0 ?
2
...
Find а) the
coordinates of normal vector of this plane; b) the coordinates of the point on the
plane
...
Build the plane 3 x  2 y  6 z  12  0 and calculate the volume of the
pyramid bounded by the coordinate planes and the given plane
...
Find the equation of the plane Р passing through the point M  2;  1; 3
with normal vector N  3; 2; 5
...
Find the equation of the plane Р passing through the point M  2; 1; 3 and
is parallel to the plane P1 : 2 x  4 y  z  5  0
...
Find the equation of the plane Р passing through the point M (1;1;1) and is
parallel to the plane Oxy
...
What geometric image corresponds to the equation x  y  0
а) on a plane? b) in space?
8
...
Find the equation of the plane Р
...
Are the planes P1 :4 x  2 y  4 z  5  0 and P2 : 2 x  y  2 z  4  0
parallel?

49

10
...
Find the cosine of the angle between two planes P1 : 2 x  3 y  6 z  7  0
and P2 : 4 x  8 y  z  3  0
...
Find the equation of the plane passing through the point M  3;  4; 2 
and is parallel to the vectors a  3; 1;  1 and b 1;  2; 1
...
Find the equation of the plane passing through the three points:

M 1  3;  2; 1 , M 2  2; 1;  1 , M 3 1; 0; 1
...
Find the equation of the plane Р passing through the three points
O  0; 0; 0  , A  3;  2; 1 and B 1; 4; 0 
...
Find the equation of the plane passing through the point A  2; 3; 1 and is
perpendicular to the planes P1 : x  y  z  0, P2 : 2 y  x
...
Find the equation of the plane passing through the point K 1; 2;  3 and
the axis Oz
...
Which value of " m " do the plane 2 x  m y  3z  3m  9  0 pass
through the origin at?
18
...

19
...


20
...
Find the equation of the plane P
...
Are the planes P1 : x  3 z  2  0 and P2 : 2 x  6 z  7  0 parallel ?
22
...


23
...


24
...


25
...


50

26
...


27
...


28
...


29
...

The solution of typical problems
Problem 1
...


 Vector N (2; 3; 1) is normal vector of the plane, then
 2( х  1)  3( у  1)  1( z  0)  0 or 2 х  3 у  z  5  0
...
Find the equation of the plane Р passing through the three
points М1 (0; 1; 5) , М 2 (3; 0; 1) and М 3 (1; 1; 2)
...
Then components of the normal vector of the plane Р is N
 М1М 2  М 1М 3 :
i

j

N 3

k

1 4  3i  13 j  k
...

Problem 3
...


 The normal vectors of Р1 and Р2 are: N1 (0;  3; 1) , N 2 (0; 2; 1)
...
Then   45 о
...
Find distance from point М 0 (1;  1; 1) to plane
Р : 2 х  3 у  6 z  25  0
...

7

Answers
1
...
2
...
8
...
3x  2 y  5 z  7  0
...
2 x  4 y  z  3  0
...
z  1
...
а) a line; б) a plane
...
x  4 y  4 z  15  0
...
yes
...
yes
...
cos  
...
x  4 y  7 z  1  0
...
x  y  z  2  0
...
4 x  y  14 z  0
...
2 x  y  z  0
...
2 x  y  0
...
3
...
x  1  0
...
3 x  2 y  z  10  0
...
2 x  5 y  6 z  25  0
...
yes
...
cos  
...
6 x  7 y  4 z  7  0
...
13 x  8 y  6 z  23  0
...
1
...
2
...
x  5 y  32  2  0
...
x  y  z  0
...
5 x  7 y  9 z  20  0
...
Lines in space and planes
...
2
...

(2
...

0

Equations for the line of intersection of two planes are:

 A x  B1 y  C1 z  D1  0,
L: 1
 A2 x  B2 y  C2 z  D2  0
...
5
...
2
...
5

L1 | | L 2

L1  L 2

a1 a2
m1 n1 p1


m2 n2 p2

a1  a2
m1  m2  n1  n2  p1  p2  0

between L1 and

L2

cos  

a1  a2
a1  a2

Let P : А х  В у  С z  D  0 is the plane with normal vector

N   A; B; C 
...
6
...
6

Problems
1
...



2
1
3

2
...


3
...
Find а) symmetric
equations of the straight line; b) parametric equations of the straight line; c)
equations of the line of intersection of two planes
...
Find symmetric equations of the line passing through the point A  1; 2; 3
perpendicular to the plane P : 2 x  y  z  1  0
...
Find the equation of the plane passing through the point M  2;  1; 4  ,
perpendicular to the line

x y 1 z  2

...
Find parametric equations of the straight line

x 1 y 1

 z  2
...

7
...
Find the angle between two lines L1 :

L2 :

x4 y 3 z 4

...
Find the angle between the line L :

P : 2 x  2 y  1  0
...
Find the equation of the plane passing through the point A 1;  2; 1

x 1 y  2 z  3
x y z

...
Find the equation of the plane passing through the point A  3; 1;  2  and
x4 y2
z
line L :

...
Find the coordinates of the intersection point of the plane

x 5 y 4 z 5



...
Find the sine of the angle between the line 
and the plane
3
x

2
y

2
z

0

3x  2 y  z  4  0
...
Find symmetric equations of the line passing through the point
M  1; 2;  3 parallel to the line L: x  2 t  1, y  t  3; z  4 t  2, t 
...

x

y

2
z

0


15
...
Find symmetric equations of the line passing through the points
A  1; 3; 2  and B 1; 2;  3
...
Find symmetric equations of the line passing through the point
A  2;  3 ;1 perpendicular to the plane P : x  y  z  3  0
...
Which value of m is the line L: x  t  1, y  2 mt  1, z   t  1,

t  R , parallel to the plane P : 3 x  y  z  1  0 at?
x 1 y 1 z  3
19
...

20
...
Find the projection of the point M  2;  3; 6  onto plane
а) L 1 :


...

The solution of typical problems
Problem 1
...


x  3 y  2 z 1

...
From the

 Find symmetric equations of the line L:

symmetric equations it is easy to obtain the equations for the line of intersection
x 3 y 2
 2  1 ,
 x  2 y  7  0,
of two planes: 

 
y

2
z

1
5
y

z

9

0
...
Find the angle between the plane P : 2 x  2 y  2 z  1  0
and the line L :

 N



x 1 y  3 z


...
Then

sin  

2 22
2  2  4 1 2 1



1
2 1
 ;   arcsin
...
Find symmetric equations of the line
2 x  3 y  z  8  0,

 x  y  z  1  0
...

1

To find the coordinates of M 0  L we assume that z0  0
...

x  y 1  0
Since M 0  1; 2; 0  and a  2; 3; 5 , we obtain the symmetric equations

of the line L :

x 1 y  2 z


...
Find the angle between the lines L1 :

x 1 y 1 z

 and
2
2
6

L 2 : x  2 t  1, y  t  2 , z  t  1, t 
...
a1  2;  2; 6  and a2  2; 1; 1 are direction vectors of the lines
...


Problem 5
...



3
1
2
 x  3t  5,

 Find parametric equations of the straight line L :  y  t  4,

...


Let's denote by t 1 the parameter value corresponding to the point
...

1
 1

must satisfy the equation of the plane P : 2  3t1  5  3 t1  4   5  2t1  5  1  0
...
Then
M1 ( 1; 2; 1)
...

1
...




...


2
...
3
...
2 x  3 y  1  0
...
x  2t  1; y  1; z  t  2; 1; 1;  2  ;  3; 1;  1
...
3
...
450
...
450
...
8 x  3 y  14 z  0
...
4 x  6 y  7 z  4  0
...
14
...
15
...

2
1
4
42
x  2 y  3 z 1
x 1 y  3 z  2



...
m  2
...
900
...




...

1
1
1
2
1
5
22
1
20
...
21
...

3
7
12
...
13

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