Задачи по аналитической геометрии. Микенберг М.А. - 71 стр.

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Составители: 

cos(ϕ) =
4
21
x2
6
=
y+3
3
=
z+5
5
2x 3y + 4z 1 = 0
x + 2y + 3z = 0
a = 3 b = 4, 5
4
2 3, 5 6, 5
8
(0; 0; 2) (0; 0; 6
4
13
)
23x y 4z 24 = 0
x y z 1 = 0
x + y + 2z = 0
3x 2y + 6z + 21 = 0 189x + 28y + 48z 591 = 0
2x 3y 6z + 19 = 0 6x 2y 3z + 18 = 0
4x 3y + 6z 12 = 0 12x 49y + 38z + 84 = 0
(3; 2; 4)
(2; 3; 2)
(4; 1; 3)
(1; 4; 7)
(5; 1; 0)
(2; 2; 5)
21 6 15
25
x 8y 13z + 9 = 0
13 3 7
2x
2
+ 2y
2
z
2
= 0
18y
2
+ 50z
2
+ 75xz + 225x 450 = 0
3x
2
+ 123y
2
+ 23z
2
18xy 22xz + 50yz + 18x 54y 66z + 27 = 0
(y 5x)
2
10(x
2
+ y
2
+ z
2
) = 0
xy + xz + yz = 0
40(x 2)
2
9y
2
9z
2
= 0
(x 2, 5z)
2
+ (y 1, 5z)
2
= 25
(x y)
2
+ 3z
2
8(x y) 8z 26 = 0
(2x + z)
2
10(2x + z) + 25y
2
= 0
3[(x z)
2
+ (y z)
2
1] (x + y 2z)
2
= 0
x
2
+ 2y
2
4x = 0
(2; 3; 0) (0; 0; 2) (4; 2; 9)