Задача № 2
Даны координаты вершин пирамиды A1A2A3A4: A1(x1 ; y1 ; z1), A2(x2 ; y2 ; z2), A3(x3 ; y3 ; z3), A4(x4 ; y4 ; z4).
Требуется:
1) в декартовой прямоугольной системе
координат построить пирамиду A1A2A3A4 ;
2) записать векторы
,
,
в ортонормированном базисе и найти
модули этих векторов; 3) найти острый
угол между векторами
и
;
4) найти площадь треугольникаA1A2A3 ;
5) найти объем пирамиды A1A2A3A4 .
|
Вариант |
|
|
|
|
|
1 |
(2; 3; 2) |
(10; 7; 3) |
(6; 6; 3) |
(8; 9; 5) |
|
2 |
(3; 5; 2) |
(1; 7; 5) |
(5; 6; 8) |
(1; 6; 4) |
|
3 |
(6; 1; 4) |
(3;-3; 8) |
(5;-5; 8) |
(8; 3; 3) |
|
4 |
(2; 5; 4) |
(5; 3; 6) |
(8; 3; 5) |
(8; 2; 10) |
|
5 |
(3; 4; 3) |
(7;-4; 4) |
(6; 0; 4) |
(9; 10; 6) |
|
6 |
(1; 2; 3) |
(3; 4; 6) |
(-3; 1; 6) |
(3; 3; 5) |
|
7 |
(3; 5; 1) |
(0; 1; 5) |
(1; 0; 5) |
(7; 9;-1) |
|
8 |
(5;-2; 4) |
(7; 1; 6) |
(7; 4; 5) |
(8; 4; 10) |
|
9 |
(1; 2; 1) |
(9;-2; 2) |
(-3; 5; 0) |
(7; 8;-2) |
|
10 |
(4; 1; 3) |
(2; 3; 6) |
(5;-3; 6) |
(3; 3; 5) |
|
11 |
(3;-1; 2) |
(7; 2; 6) |
(9; 0; 6) |
(5; 1; 3) |
|
12 |
(3; 5; 4) |
(1; 8; 6) |
(-1; 2; 6) |
(9;-1; 1) |
|
13 |
(1; 1; 2) |
(-3; 9; 3) |
(-2; 5; 3) |
(7; 7;-1) |
|
14 |
(1; 4; 3) |
(-1; 6; 6) |
(6;-4; 0) |
(2; 2; 1) |
|
Вариант |
|
|
|
|
|
15 |
(2; 4; 1) |
(6; 7; 5) |
(7; 6; 5) |
(6; 8; 3) |
|
16 |
(1; 2; 2) |
(3; 5; 4) |
(5;-1; 4) |
(7; 8; 5) |
|
17 |
(2;-2; 1) |
(10; 2; 2) |
(6; 1; 2) |
(8; 4; 4) |
|
18 |
(3; 4;-1) |
(1; 6; 2) |
(5; 5; 5) |
(1; 5; 1) |
|
19 |
(2; 5; 3) |
(-1; 1; 7) |
(1;-1; 7) |
(4; 7; 2) |
|
20 |
(1; 4; 2) |
(4; 2; 4) |
(7; 2; 3) |
(7; 1; 8) |
|
21 |
(3; 1; 4) |
(7;-7; 5) |
(6;-3; 5) |
(9; 7; 7) |
|
22 |
(2; 4; 3) |
(4; 6; 6) |
(-2; 3; 6) |
(4; 5; 5) |
|
23 |
(5;-2;-1) |
(2;-6; 3) |
(3;-7; 3) |
(9; 2;-3) |
|
24 |
(5; 2; 1) |
(7; 5; 3) |
(7; 8; 2) |
(8; 8; 7) |
|
25 |
(2;-1; 7) |
(10;-5; 8) |
(-2; 2; 6) |
(8; 5; 4) |
|
26 |
(4; 7; 8) |
(2; 9; 11) |
(5; 3; 11) |
(3; 9; 10) |
|
27 |
(2; 1; 3) |
(6; 4; 7) |
(8; 2; 7) |
(4; 3; 4) |
|
28 |
(1; 5; 2) |
(-1; 8; 4) |
(-3; 2; 4) |
(7;-1;-1) |
|
29 |
(6; 1; 4) |
(2; 9; 5) |
(3; 5; 5) |
(12; 7; 1) |
|
30 |
(6; 5; 1) |
(4; 7; 4) |
(11;-3;-2) |
(7; 3;-1) |
Задача № 3
Построить кривую, заданную уравнением в полярной системе координат.
|
Вариант |
Уравнение кривой |
Вариант |
Уравнение кривой | |
|
1 |
r
=
|
16 |
r = 3 – 3sin 2 | |
|
2 |
r = 3 + sin 2 |
17 |
r
= | |
|
3 |
r
=
|
18 |
r = 3 – sin 2 | |
|
4 |
r = 2 – 2sin 3 |
19 |
r
=
| |
|
5 |
r
=
|
20 |
r = 2 + 2sin 3 | |
|
6 |
r = 4 + 2cos 3 |
21 |
r
=
| |
|
7 |
r
=
|
22 |
r = 4 – 2cos 3 | |
|
8 |
r = 3 – 3cos 2 |
23 |
r
=
| |
|
9 |
r
=
|
24 |
r = 3 + 3cos 2 | |
|
10 |
r = 3 + cos 2 |
25 |
r
=
| |
|
11 |
r
=
|
26 |
r = 3 – cos 2 | |
|
Вариант |
Уравнение кривой |
|
Уравнение кривой | |
|
12 |
r = 2 – 2cos 3 |
27 |
r
=
| |
|
13 |
r
=
|
28 |
r = 2 + 2cos 3 | |
|
14 |
r = 4 + 2sin 3 |
29 |
r
=
| |
|
15 |
r
=
|
30 |
r = 4 – 2sin 3 | |
+ 2cos
2
– 2cos
2
– 2cos
3
+ 2cos
3
+ 2sin
3
– 2sin
3
– 2sin
2
+2sin
2
+ 2sin
2
– 2sin
2
– 2sin
3
+ 2sin
3
+ 2cos
3
– 2cos
3
– 2cos
2