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Analytic geometry. Textbook

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4.3. Analytic geometry in the space

131

•the line has no common points with the plane;

•the line has one common point with the plane and in this case nd this point.

№

Straight line

Plane

x 1

y + 3

z + 2

4x + 3y z + 3 = 0

x 1

z 2

1 = 0

x + 2

y 5

z + 15 = 0

170. Find equation of the plane passing through the point (3; 1; 2)

and the straight line

x 4

y + 3

171. Find equation of the plane passing through the straight line

x 2

y 3

z + 1

and orthogonal to the plane x + 4y 3z + 7 = 0.

172. Check that the straight lines

x 3

y + 1

z 2

and

x 8

y 1

z 6

intersect and nd the equation of the plane passing through them.

173. Find equation of the plane passing through the following parallel

straight lines:

y + 2

z 1

and

x 1

y 3

z 2

174. Find equation of the plane passing through the point (4; 3; 1)

and parallel to the straight lines

;

and

x + 1

y 3

z 4

132

Chapter 4. Problems

175. Find equation of the plane passing through the straight line

x 3

y + 4

z 2

and parallel to the straight line

x + 5

y 2

z 1

176. Find equation of the plane passing through the straight line

x + 5 = y 2 = z 3 1 4

and parallel to the plane x + y z + 15 = 0.

177. Find equation of the plane that is orthogonal to the plane x

3y +2z 5 = 0 and intersects it by the straight line lying on the plane xOz.

178.Find equation of the plane that is perpendicular to the plane x + 4y 3z + 7 = 0 and passes through the straight line

x 2	=	y 3	=	z + 1	:
5
	1		2

179. Find equations of the straight line passing through the points (2; 3; 1) and (1; 0; 5).

180. Find intersection point of the straight line

x 7	=	y 4	=	z 5
5	1		4

and the plane 3x y + 2z 5 = 0.

181. Check whether the straight lines

x 1	=	y 7	=	z 5	and	x 6	=	y + 1	=	z
2						3		2
	1		4						1

intersect, and if they intersect, nd the intersection point.

182. Find equations of the straight line passing through the point (3; 2; 1) and orthogonal to the plane 4x + y 6z + 11 = 0.

4.3. Analytic geometry in the space

133

183. Find equations of the straight line passing through the point (2; 1; 1) and parallel to the line 2x + y + z 12 = 0, x 3y + z 21 = 0.

184. Find the distance of the point (7; 9; 7) from the straight line

x 2 = y 1 = z :

4 3 2

185. Find the distance of the point (7; 9; 7) from the straight line

x 2 = y 1 = z :

4 3 2

186. Find the distance between the skew straight lines

x 9

y + 2

;

y + 7

z 2

187. Find equation of the plane passing through the point M0(1; 2; 3) and orthogonal to each of the planes 2x y+5z+3 = 0 and x+3y z 7 = 0.

Chapter 5

Answers and solutions

We remind that in some gures below, we place the coordinate system on a grid with squares of unit length, without specifying in some cases the coordinates of points.

5.1Straight lines on the plane

1.a) The distance between the points (2; 2) and (5; 6) equals

pp p

(5 2)2 + (6 2)2 = 32 + 42 ] 25 = 5;

b) 10; c) 13; d) 10.

2. We have the following equation:

(7 3)2 + (k 3)2 = 5;

squaring the both sides we get that 16 + (k 3)2 = 25, (k 3)2 = 9,

k 3 = 3, therefore, k = 6 or k = 0.

3. We equate the distances between the given pairs of points:

(3 4)2 + (k 2)2 = (k 4)2 + ( 1 2)2 ;

(k 2)2 + 1 = (k 4)2 + 9; k2 4k + 5 = k2 8k + 25;

4k = 20; k = 5:

134

5.1. Straight lines on the plane

135

1) + 64 = 81 +p

4. (3 ( 5)2 + (k 1)2) =

(4 ( 5))2 + ( 3 1)2 ,

16; (k

1)2 = 33,

k 1 = 33 ; k = 1 33 .

5.x + 2y + 1 = 0.

6.2x y 6 = 0.

7.x + 5y 24 = 0, x + 2y 6 = 0.

8.a) x = 5; b) x = 1; c) x = a; d) x = a.

9.a) y = 3; b) y = 4; c) y = p; d) y = q.

10.k = 3.

11.We can take the normal vector f2; 1g of the given line as the normal vector of the desired line. Therefore this line has the equation

2(x + 2) (y 3) = 0; or 2x y + 7 = 0:

12. The line passing though the points (4; 1) and (3; 2) has the slope k = 23 14 = 1. Equation of the desired line takes the form y 2 = (x 3), or x + y 5 = 0.

13. The normal vector fA; Bg of the desired line is orthogonal to the normal vector of the given line, therefore 2A B = 0. We can take A = 1,

B = 2 (other possible values of A and B that do not vanish simultaneously are proportional to these and de ne a collinear vector). Therefore this line has equation (x 3) + 2(y + 1) = 0, or x + 2y 1 = 0.

14.a) 3; b) 4; c) 523; d) 0; e) 215.

15.The distance equals a) 2 units, b) 3 units.

16.The desired straight line has the equation

y = 2x + b; or 2x y + b = 0:

The distance of the origin from this line equals

j2 0 0 + bj	=	jbj	:

p22 + ( 1)2		p5

136						Chapter 5. Answers and solutions
	jbj			p		p

We have the equality	p		= 3, jbj = 3		5 , b = 3 5 .
		5
	p
Answer: y = 2x 3			5 .

17. We have the following general form of equations of straight lines that are perpendicular to the given straight line: 4x + 3y + C = 0. We

obtain the following relation

j4 2 + 3 3 + Cj

p = 2; 42 + 32

therefore jC + 17j = 10, C + 17 = 10, C = 7 or C = 27, the equations 4x + 3y = 0 and 4x + 3y 20 = 0.

18. The line passes through the point (2; 1), therefore if can be de ned by the equation A(x 2) + B(y 1) = 0. Equating the distance of the

origin to this line and the given value we get that

2) + B(		1)
					j2A + Bj = 2pA2 + B2 :
jA(p			j	= 2;
	A2 + B2

Now we square the both sides:

4A2 + 4AB + B2 = 4(A2 + B2); 4AB 3B2 = 0; B(4A 3B) = 0:

If B = 0, we let A = 1, equation of the line takes the form x 2 = 0.

Suppose that 4A 3B = 0. Let A = 3, B = 4, we get the equation 3(x 2) + 4(y 1) = 0, or 3x + 4y 10 = 0.

Answer: x 2 = 0, 3x + 4y 10 = 0.

19. We have the following general form of equations of straight lines

passing through the given point:

A(x + 1) + B(y + 4) = 0; jAj + jBj =6 0:

Equating the distance of the point (3; 2) from this line the the given value d = 6 we get

4A + 6B	j4A + 6Bj = 6pA2 + B2 :
pj A2 + B2j = 6;

5.1. Straight lines on the plane

137

Canceling by 2 and squaring the both sides we obtain:

j2A + 3Bj = 3 A2 + B2 ; 4A2 + 12AB + 9B2 = 9A2 + 9B2; 5A2 + 12AB = 0; A(5A 12B) = 0:

If A = 0 we let B = 1 and obtain the equation y + 4 = 0. If 5A 12B = 0 we let A = 12, B = 5 and get the equation 12(x + 1) + 5(y + 4) = 0, or 12x + 5y + 32 = 0.

Answer: y + 4 = 0, 12x + 5y + 32 = 0.

20.y + 1 = 2(x 3), or 2x y 7 = 0.

21.45 ( =4)

22.y + 1 = 3(x + 1), or 3x + y + 4 = 0.

23.This line passes through the points (5; 6) and (2; 0). Equation of

the line takes the form
	x 5	=	y 6	;
	5 2		6 0

or 2x y 4 = 0.

Answer: 2x y 4 = 0 or in the intercept form x2 + y4 = 1.

24.C a s e s a ) { c ) . We use equations of straight lines in the in-

tercept form Ax + By = 1. The line passes through the point ( 3; 4), so

A3 + B4 = 1,
4A 3B = AB:	(1)

a) A = 10, 40 3B = 10B, B = 4013, the equation takes the form 10x + 1340y = 1, or 4x + 13y 40 = 0.

b)A + B = 12, we substitute B = 12 A into (1). After simpli cation we get A2 5A 36 = 0, A = 9 or A = 4. The desired equations take the form x + 3y 9 = 0, 4x y + 16 = 0.

c)AB = 50, we substitute B = 50A into (1). after simpli cation we get

4A2 50A 150 = 0, A = 15, B = 103 or A = 52, B = 20. The equations take the form 2x + 9y 30 = 0, 8x + y + 20 = 0.

138 Chapter 5. Answers and solutions

C a s e s d ) , e ) . We use equations of the desired straight lines in the form A(x + 3) + B(y 4) = 0, or Ax + By + (3A 4B) = 0.

d) From the condition we get that

j3A 4Bj

p = 3; A2 + B2

after simpli cation there is obtained that B(24A 7B) = 0. If B = 0 we let A = 1 and get the equation x + 3 = 0. If 24A 7B = 0 we let A = 7,

B = 24, and get the equation 7x + 24y 75 = 0. e) From the condition we get that

j15A + 5Bj

p = 0; A2 + B2

or after simpli cation A(4A + 3B) = 0. If A = 0 we let B = 1 and get the equation y 4 = 0. If 4A + 3B = 0 we let A = 3, B = 4, and get the equation 3x 4y + 25 = 0.

Answer: a) 4x + 13y 40 = 0; b) x + 3y 9 = 0, 4x y + 16 = 0;

c) 2x + 9y 30 = 0, 8x + y + 20 = 0; d) x + 3 = 0, 7x + 24y 75 = 0; e) y 4 = 0, 3x 4y + 25 = 0.

25. The lines with the slope 4=3 may be de ned by the equations of the form 4x + 3y + C = 0.

a)The line passes through the point (6; 0), therefore C = 24, the desired equation is 4x + 3y 24 = 0.

b)In this case

p44 + 32			j		j
	jCj	= 5;		C		= 30; C = 30;
		= 5;		C		= 30; C = 30;
we obtain equations 4x + 3y 30 = 0.
c) We have
4 8 + 3 2 + C			= 4;					C + 38 = 20;
			= 4;					C + 38 = 20;
	p44 + 32						j	j

C + 38 = 20; C = 18; or C = 58;

5.1. Straight lines on the plane

139

we obtain equations 4x + 3y 18 = 0; 4x + 3y 58 = 0.

d) Equating the distances of the given points from the line we get

j4 2 + 3 6 + Cj	=	j4 3 + 3 ( 2) + Cj	;

p44 + 32		p44 + 32
jC + 26j = jC 12j;		C + 26 = (C 12):

Equation C + 26 = C 12 has no solution. If C + 26 = (C 12), we get that C = 7, the equation has the from 4x + 3y 7 = 0.

26. a) Equation of the straight line has the general form x + y A = 0. If we let x = 4, y = 2 we obtain A = 2, x + y 2 = 0.

Ax + Ay = 1, or the answer is

b)The point of the given line with x = 1 has the coordinates (1; 3). The desired line can be de ned by the equation (x 1) + 4(y + 3) = 0, or x + 4y + 11 = 0.

c)The desired straight line may be de ned by the equation x+y A = 0 (see the solution a)). The distance of the center of the circle equals the radius of the circle,

	j2 1 Aj		= 5; A 1 = 5p			;
	j2 1 Aj				2
					2
	p12 + 12			j j
			p	p
A 1 = 1 5 2 ; A = 1 5 2 ;
		p
the answer is x + y 1 5 2 = 0.

d) The desired straight line may be de ned by the equation x+2y +C =

0, the distance from the center O(0; 0) equals p

, therefore

p12 + 22

jCj

= p5 ;

= 5; C =

the answer is x + 2y 5 = 0.

Answer: a) x + y 2 = 0; b) x + 4y + 11 = 0; c) x + y 1 5 2 = 0;

d)x + 2y 5 = 0.

27.We write equation of the pencil with the center at the intersection

140 Chapter 5. Answers and solutions

point of these lines

(3x 2y 13) + (x + y 6) = 0; j j + j j 6= 0:

We put here x = 2, y = 3 and obtain that 7 = 0. Taking = 1 we obtain that = 7. The desired straight line has the equation 7(3x

2y 13) (x + y 6) = 0, or 20x + 15y + 25 = 0. Canceling by 5 and changing the sign we get the equation 4x 3y 17 = 0.

28. We write the equation of the pencil de ned by the given straight lines:

(4x + y 7) + (3x 2y 10);

(4 + 3 )x + ( 2 )y (4 + 10 ) = 0; j j + j j =6 0:

The desired line is parallel to the line x 3y 6 = 0, therefore

4 + 3

;

3(4 + 3 ) =

2 ;

13 =

7 :

We let = 7, = 13, so we get the line 11x 33y 81 = 0.

29.

Here is equation of the pencil de ned by the given straight lines:

(3x + 5y 13) + (x + y 1) = 0;

(3 + )x + (5 + )y (13 + ) = 0;

j j + j j 6= 0:

The desired line is orthogonal to the line 7x 5y 10 = 0, therefore

7(3 + ) 5(5 + ) = 0;

4 + 2 = 0;

= 2 :

We can take = 1, then = 2, and we get the answer 5x + 7y 15 = 0.

y 4p

= 0.

30.

x +

31.

53x 54y 10 = 0, 54x + 53y 10 = 0.

32.

The slope k of the straight line passing through two di erent points

; y

) and M

; y

) is found by the formula k =

y2 y1

if x

= x

x2 x1

6 1

In this case the angle of inclination is acute if k > 0 and obtuse if k < 0. If

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