Chapter 8 Lines and Planes Chapter 8 Prerequisite …8...MHR • Calculus and Vectors 12 Solutions 819 Chapter 8 Lines and Planes Chapter 8 Prerequisite Skills Chapter 8 Prerequisite

MHR • Calculus and Vectors 12 Solutions 819

Chapter 8 Lines and Planes Chapter 8 Prerequisite Skills Chapter 8 Prerequisite Skills Question 1 Page 426 a) 4 8y x= !

b) 2 3y x= ! +

c)

3x ! 5y = 15

y =3

5x ! 3

x y 1 –4 2 0 3 4 0 –8

–1 –12 –2 –16

x y 1 1 2 –1 3 –3 0 3

–1 5 –2 7

x y 5 0

10 3 15 6 0 –3

–5 –6 -10 –9


d)

5x + 6y = 20

y = !5

6x +

20

6

Chapter 8 Prerequisite Skills Question 2 Page 426 To find the x-intercept, let y = 0 and solve for x. To find the y-intercept, let x = 0 and solve for y. a) 0 3 7

7

3

x

x

= +

= !

y = 3(0)+ 7

y = 7

b) 0 5 10

2

x

x

= !

=

y = 5(0)!10

y = !10

c)

2x ! 9(0) = 18

x = 9

2(0)! 9y = 18

y = !2

x y

1 15

6

2 10

6

3 5

6

0 20

6

–1 25

6

–2 5


d)

4x + 8(0) = 9

x =9

4

= 2.25

4(0)+ 8y = 9

y =9

8

= 1.125

Chapter 8 Prerequisite Skills Question 3 Page 426 a) Plot y-intercept. Use the slope to find other points, such as (1, 3) and (2, 1).

b) Plot x-intercept. Use the slope to find other points, such as (5, –5) and (7, –4)

c) Plot point (1, –3). Use the slope to plot other points. Move 5 right and 3 up to point (6, 0). Again move

5 right and 3 up to point (11, 3).


d) Plot the point (–5, 6). Use the slope to plot other points. Move 3 right and 8 down to point (–2, –2). Again, move 3 right and 8 down to point (1, –10).

e)

2x ! 6 = 0

x = 3

All points on graph have x = 3. It is a vertical line.

f)

y + 4 = 0

y = !4

All points on the graph have y = - 4. It is a horizontal line.


Chapter 8 Prerequisite Skills Question 4 Page 426 a) By observation, the point of intersection is (7, 2). b) By observation, the point of intersection is (–2, 2). c) By observation, the point of intersection is not obvious. One line passes through (0, 2) and (2, 1). Slope:

1! 2

2 ! 0= !

1

2

y-intercept is 2.

The equation is y = !

1

2x + 2 or 2 4x y+ = .

The other line passes through (0, –1) and (1, 1). Slope:

1! (–1)

1! 0= 2

y-intercept is –1. The equation is 2 1y x= ! or 2 1x y! = . Solve the system of equations using substitution.

x + 2(2x !1) = 4

5x = 6

x = 1.2

y = 2(1.2)!1

y = 1.4

The intersection point is (1.2, 1.4). Chapter 8 Prerequisite Skills Question 5 Page 426 a) Use elimination.

y = 3x + 2 !

y = !x ! 2 "

0 = 4x + 4 !!"

x = !1

Substitute x = –1 into equation .

y = 3(–1)+ 2

y = !1

The point of intersection is (–1, –1).


b) Use elimination.

x + 2y = 11 !

x + 3y = 16 "

0x ! y = !5 !!"

y = 5

Substitute y = 5 into equation .

x + 2(5) = 11

x = 1

The point of intersection is (1, 5) c) Use elimination.

4x + 3y = !20 !

5x ! 2y = 21 "

8x + 6y = !40 2!

15x ! 6y = 63 3"

23x = 23 2!+3"

x = 1

Substitute x = 1 into equation .

4(1)+ 3y = !20

3y = !24

y = !8

The point of intersection is (1, –8).


d) Use elimination.

2x + 4y = 15 !

4x ! 6y = !15 "

4x + 8y = 30 2!

!4x + 6y = 15 !"

14y = 45 2!!"

y =45

14

Substitute 45

14y = into equation .

2x + 445

14

!

"#$

%&= 15

28x +180 = 210

28x = 30

x =15

14

The point of intersection is 15 45,

14 14

! "# $% &

.

Chapter 8 Prerequisite Skills Question 6 Page 426 a) Parallel lines have equal slopes. The line 3 5y x= + has slope 3. The x-intercept 10 corresponds to the point (10, 0). Use the point-slope form of the equation of a line.

y ! 0

x !10= 3

y = 3x ! 30

The equation of the line is 3 30y x= ! .

b) Parallel lines have equal slopes. The line 4 5 7x y+ = has slope 4

5! .

The slope and a point on the line are known. Use the point-slope form of the equation of a line.

y ! 6

x ! (!2)=!4

5

5y ! 30 = !4x ! 8

5y = !4x + 22

y = !4

5x +

22

5

The equation of the line is 4 22

5 5y x= ! + .


c) Perpendicular lines have negative reciprocal slopes. The line 36

2y x= ! + has slope 3

2! .

The required line will have slope 23

.

The x-intercept of 5 2 20x y! = is 4 (let y = 0). Therefore the point (4, 0) is on the required line. The slope and a point on the line are known. Use the point-slope form of the equation of a line.

y ! 0

x ! 4=

2

3

3y = 2x ! 8

y =2

3x !

8

3

The equation of the line is 2 8

3 3y x= ! .

d) Perpendicular lines have negative reciprocal slopes. The line 77 5 20 4

5x y y x+ = ! = " + has

slope 7

5! .

The required line will have slope 57

.

The y-intercept of 66 5 15 3

5x y y x! = " = ! is 3! . Therefore the point

(0,–3) is on the required line.

The slope and a point on the line are known. Use the point-slope form of the equation of a line.

y ! (–3)

x ! 0=

5

7

7 y + 21= 5x

y =5

7x ! 3

The equation of the line is 53

7y x= ! .

e) Lines parallel to the y-axis have the form x = a. Since the required line passes through (–3, 0), the required equation is x = –3.


Chapter 8 Prerequisite Skills Question 7 Page 427 a)

a

!

!b!

= 3, 1"# $% ! 5, 7"# $%

= 3(5)+1(7)

= 22

Since 0a b! "

! !, and a b

! ! are not perpendicular.

b)

a

!

!b!

= "4, 5#$ %& ! "9, 1#$ %&

= (–4)(–9)+ 5(1)

= 41

Since 0a b! "

! !, and a b


c)

a

!

!b!

= 6, 1"# $% ! &2, 12"# $%

= 6(–2)+1(12)

= 0

Since 0a b! =

! !, and a b

! ! are perpendicular.

d)

a

!

!b!

= 1, 9, " 4#$ %& ! 3, " 6, " 2#$ %&

= 1(3)+ 9(–6)+ (–4)(–2)

= "43

Since 0a b! "

! !, and a b


e)

a

!

!b!

= 3, 4, 1"# $% ! 1, &1, 1"# $%

= 3(1)+ 4(–1)+1(1)

= 0

Since 0a b! =

! !, and a b

! ! are perpendicular.

f)

a

!

!b!

= 7, " 3, 2#$ %& ! 1, 8, 10#$ %&

= 7(1)+ (–3)(8)+ 2(10)

= 3

Since 0a b! "

! !, and a b



Chapter 8 Prerequisite Skills Question 8 Page 427 a)

a

!

! b

!

= 2, " 7, 3#$ %& ! 1, 9, 6#$ %&

= (–7)(6)" 9(3), 3(1)" 6(2), 2(9)"1(–7)#$ %&

= "69, " 9, 25#$ %&

b)

a

!

! b

!

= 8, 2, " 4#$ %& ! 3, 7, "1#$ %&

= 2(–1)" 7(–4), " 4(3)" (–1)(8), 8(7)" 3(2)#$ %&

= 26, " 4, 50#$ %&

c)

a

!

! b

!

= 3, 3, 5"# $% ! 5, 1, &1"# $%

= 3(–1)&1(5), 5(5)& (–1)(3), 3(1)& 5(3)"# $%

= &8, 28, &12"# $%

d)

a

!

! b

!

= 2, 0, 0"# $% ! 0, 7, 0"# $%

= 0(0)& 7(0), 0(0)& 0(2), 2(7)& 0(0)"# $%

= 0, 0, 14"# $%

Chapter 8 Prerequisite Skills Question 9 Page 427 The vectors are not unique, as any vector that is a scalar multiple of the given vector will be parallel. a) [2, 10] b) [–30, 40] c) [4, 2, 14] d) [1, 4, –5] Chapter 8 Prerequisite Skills Question 10 Page 427 These vectors are not unique, as any vector that produces zero in a dot product with the given direction vector will be perpendicular. a) [–5, 1] since [–5, 1]·[1, 5] = 0 b) [4, 3] since [4, 3]·[–3, 4] = 0 c) [3, 1, –1] since [3, 1, –1]·[2, 1, 7] = 0 d) [1, 1, 1] since [1, 1, 1]·[–1, –4, 5] = 0


Chapter 8 Prerequisite Skills Question 11 Page 427 Use the formula for dot product.

a)

cos ! =a

!

"b!

a

!

b

!

cos ! =

1, 3#$ %& " 2, 5#$ %&

12+ 3

22

2+ 5

2

cos ! "17

17.0294

! " cos'1 17

17.0294

(

)*+

,-

! " 3.4o

b)

cos ! =a

!

"b!

a

!

b

!

cos ! =

#4, 1$% &' " 7, 2$% &'

(–4)2+12 72

+ 22

cos ! "#26

30.0167

! " cos#1 #26

30.0167

(

)*+

,-

! " 150.0o

c)

cos ! =a

!

"b!

a

!

b

!

cos ! =

1, 0, 2#$ %& " 5, 3, 4#$ %&

12+ 0

2+ 2

25

2+ 3

2+ 4

2

cos ! "13

15.8114

! " cos'1 13

15.8114

(

)*+

,-

! " 34.7o


d)

cos ! =a

!

"b!

a

!

b

!

cos ! =

#3, 2, # 8$% &' " 1, # 2, 6$% &'

(–3)2+ 22

+ (–8)2 12+ (–2)2

+ 62

cos ! "#55

56.1872

! " cos#1 #55

56.1872

(

)*+

,-

! " 168.2o


Chapter 8 Section 1 Equations of Lines in Two-Space and Three-Space Chapter 8 Section 1 Question 1 Page 437 a)

x, y!" #$ = 2, 7!" #$ + t 3, 1!" #$ , t !!

b)

x, y!" #$ = 10, % 4!" #$ + t %2, 5!" #$ , t !!

c)

x, y, z!" #$ = 9, % 8, 1!" #$ + t 10, % 3, 2!" #$ , t !!

d)

x, y, z!" #$ = %7, 1, 5!" #$ + t 0, 6, %1!" #$ , t !!

Chapter 8 Section 1 Question 2 Page 437 a) First determine the direction vector.

m

!"

= OB! "!!

!OA! "!!

= 4, 10"# $% ! 1, 7"# $%

= 3, 3"# $%

A possible vector equation of the line is [x, y] = [4, 10] + t[3, 3], t !! . b) First determine the direction vector.

m

!"

= OB! "!!

!OA! "!!

= !2, ! 8"# $% ! !3, 5"# $%

= 1, !13"# $%

A possible vector equation of the line is [x, y] = [–2, –8] + t[1, –13], t !! . c) First determine the direction vector.

m

!"

= OB! "!!

!OA! "!!

= 9, 2, 8"# $% ! 6, 2, 5"# $%

= 3, 0, 3"# $%

A possible vector equation of the line is [x, y, z] = [9, 2, 8] + t[3, 0, 3], t !! . d) First determine the direction vector.

m

!"

= OB! "!!

!OA! "!!

= 1, !1, ! 5"# $% ! 1, 1, ! 3"# $%

= 0, ! 2, ! 2"# $%

A possible vector equation of the line is [x, y, z] = [1, –1, –5] + t[0, –2, –2], t !! .


Chapter 8 Section 1 Question 3 Page 437 a)

b)

c)

d)


Chapter 8 Section 1 Question 4 Page 437 Use The Geometer’s Sketchpad® file 8.1 VectorEquation.gsp. Some possible screens are shown below. a)

6

4

2

-2

-4

-10 -5 5 10

Double click on the above values tochange them to match any vector

equation. Alternately you may clickon one (or more) of the values to

select it and use the + and - keys tochange them incrementally.

(x, y) = ( 1.0, -3.0 ) + t( 2.0 , 5.0)Py = -3.0Px = 1.0

my = 5.0mx = 2.0

B

A

b)

6

4

2

-2

-4

-10 -5 5 10

(x, y) = ( -5.0 , 2.0) + t( 4.0 , -1.0 )Py = 2.0Px = -5.0

my = -1.0mx = 4.0

B

A

c)

6

4

2

-2

-4

5 10 15

(x, y) = ( 2.0, 5.0) + t( 4.0, -3.0 )Py = 5.0Px = 2.0

my = -3.0mx = 4.0

B

A


d)

4

2

-2

-4

-6

-10 -5 5 10

(x, y) = ( -2.0 , -1.0 ) + t( -5.0 , 2.0)Py = -1.0Px = -2.0

my = 2.0mx = -5.0

B

A

Chapter 8 Section 1 Question 5 Page 437 a) The point P(–1, 11) corresponds to the position vector [–1, 11]. Substitute the coordinates into the vector equation.

!1, 11"# $% = 3, 1"# $% + t !2, 5"# $%

Equate the x-coordinates.

1 3 2

2

t

t

! = !

=

Equate the y-coordinates.

11 1 5

2

t

t

= +

=

Since the t values are equal, the point P(–1, 11) does lie on the line. b) The point P(9, –15) corresponds to the position vector [9, –15]. Substitute the coordinates into the vector equation.

9, !15"# $% = 3, 1"# $% + t !2, 5"# $%


9 3 2

3

t

t

= !

= !


15 1 5

16

5

t

t

! = +

= !

Since the t values are not equal, the point P(9, –15) does not lie on the line.


c) The point P(–9, 21) corresponds to the position vector [–9, 21]. Substitute the coordinates into the vector equation.

!9, 21"# $% = 3, 1"# $% + t !2, 5"# $%


9 3 2

6

t

t

! = !

=


21 1 5

4

t

t

= +

=

Since the t values are not equal, the point P(–9, 21) does not lie on the line. d) The point P(−2, 13.5) corresponds to the position vector [−2, 13.5]. Substitute the coordinates into the vector equation.

!2, 13.5"# $% = 3, 1"# $% + t !2, 5"# $%


2 3 2

5

2

t

t

! = !

=


13.5 1 5

5

2

t

t

= +

=

Since the t values are equal, the point P(−2, 13.5) does lie on the line. Chapter 8 Section 1 Question 6 Page 437 t !! for all equations. a)

x = 10 +13t

y = 6 + t

b)

x = 12t

y = 5! 7t

c)

x = 3+ 6t

y = !9t

z = !1+ t


d)

x = 11+ 3t

y = 2

z = 0


x, y!" #$ = 3, 9!" #$ + t 5, 7!" #$ , t !! .

b)

x, y!" #$ = %5, 0!" #$ + t %6, 11!" #$ , t !! .

c)

x, y, z!" #$ = 1, % 6, 2!" #$ + t 4, 1, % 2!" #$ , t !! .

d)

x, y, z!" #$ = 7, 0, 0!" #$ + t 0, %1, 0!" #$ , t !! .

Chapter 8 Section 1 Question 8 Page 437 a) Isolate t in each of the parametric equations.

1 2

1

2

x t

xt

= +

!=

1 3

1

3

y t

yt

= !

!=

!

1 1

2 3

3 3 2 2

3 2 5 0

x y

x y

x y

! !=

!

! + = !

+ ! =

b) Isolate t in each of the parametric equations.

x = !2 + t

t = x + 2

y = 4 + 5t

t =y ! 4

5

42

5

5 10 4

5 14 0

yx

x y

x y

!+ =

+ = !

! + =


c) Isolate t in each of the parametric equations.

5 7

5

7

x t

xt

= +

!=

2 4

2

4

y t

yt

= ! !

+=

!

5 2

7 4

4 20 7 14

4 7 6 0

x y

x y

x y

! +=

!

! + = +

+ ! =

d) Isolate t in each of the parametric equations.

0.5 0.3

0.5

0.3

x t

xt

= +

!=

1.5 0.2

1.5

0.2

y t

yt

= !

!=

!

0.5 1.5

0.3 0.2

0.2 0.1 0.3 0.45

0.2 0.3 0.55 0

2 3 5.5 0

4 6 11 0

x y

x y

x y

x y

x y

! !=

!

! + = !

+ ! =

+ ! =

+ ! =

Chapter 8 Section 1 Question 9 Page 437 a) Choose two points on the graph. The x-intercept (–12) and the y-intercept (–6) are easy to calculate.

b) Find two points on the graph by letting t = 0 and 1, say. This gives the points (–1, 7) and (1, 2).


c) Find two points on the graph by letting t = 0 and 1, say. This gives the points (4, –2) and (5, 1).

d) Choose two points on the graph. The x-intercept (2.5) and the y-intercept

!2

3

"

#$%

&' are easy to calculate.

Chapter 8 Section 1 Question 10 Page 437 The scalar equation of a line in two-space is of the form [ ]0 where ,Ax By C n A B+ + = =

! is a normal

vector for the line. a) The scalar equation is of the form 3 0x y C+ + = . Substitute the point (2, 4) to determine the value of C.

3(2)+ 4 +C = 0

C = !10

A scalar equation for the line is 3 2 10 0x y+ ! = . b) The scalar equation is of the form x – y + C = 0. Substitute the point (–5, 1) to determine the value of C.

!5!1+C = 0

C = 6

A scalar equation for the line is 6 0x y! + = . c) The scalar equation is of the form y + C = 0. Substitute the point (–3, –7) to determine the value of C.

!7 +C = 0

C = 7

A scalar equation for the line is 7 0y + = .


d) The scalar equation is of the form 1.5 3.5 0x y C! + = . Substitute the point (0.5, –2.5) to determine the value of C.

1.5(0.5)! 3.5(!2.5)+C = 0

C = !9.5

A scalar equation for the line is 1.5 3.5 9.5 0x y! ! = or 3 7 19 0x y! ! = Chapter 8 Section 1 Question 11 Page 437 t !! for all equations. a) Choose two points on the line, say (0, 3) and (6, 0). (Hint: consider the intercepts) The vector joining these two points is a possible direction vector.

m

!"

= 6, 0!" #$ % 0, 3!" #$

= 6, % 3!" #$

A possible vector equation is

x, y!" #$ = 0, 3!" #$ + t 6, % 3!" #$ . Possible parametric equations are

x = 6t, y = 3! 3t.

b) Choose two points on the line, say (0, –12) and (3, 0). (Hint: consider the intercepts) The vector joining these two points is a possible direction vector.

m

!"

= 3, 0!" #$ % 0, %12!" #$

= 3, 12!" #$


x, y!" #$ = 3, 0!" #$ + t 3, 12!" #$ . Possible parametric equations are

x = 3+ 3t, y = 12t.

c) Choose two points on the line, say (1, –4) and (3, 1). (Hint: consider convenient values for x and solve for y.) The vector joining these two points is a possible direction vector.

m

!"

= 3, 1!" #$ % 1, % 4!" #$

= 2, 5!" #$


x, y!" #$ = 1, % 4!" #$ + t 2, 5!" #$ . Possible parametric equations are

x = 1+ 2t, y = !4 + 5t.

d) Choose two points on the line, say (–9, 3) and (0, –5). (Hint: consider convenient values for x and solve for y.) The vector joining these two points is a possible direction vector.

m

!"

= !9, 3"# $% ! 0, ! 5"# $%

= !9, 8"# $%


x, y!" #$ = 0, % 5!" #$ + t %9, 8!" #$ . Possible parametric equations are

x = !9t, y = !5+ 8t.


Chapter 8 Section 1 Question 12 Page 437 The vector joining the two given two points is a possible direction vector.

m

!"

= 3, 4, – 5!" #$ % 9, % 2, 7!" #$

= %6, 6, %12!" #$


x, y, z!" #$ = 3, 4, % 5!" #$ + t %6, 6, %12!" #$ Possible parametric equations are

x = 3! 6t, y = 4 + 6t, and z = !5!12t.

t !! for all equations. Chapter 8 Section 1 Question 13 Page 437 a) The point P0(7, 0, 2) corresponds to the position vector [7, 0, 2]. Substitute the coordinates into the vector equation.

7, 0, 2!" #$ = 1, 3, % 7!" #$ + t 2, %1, 3!" #$


7 1 2

3

t

t

= +

=


0 3

3

t

t

= !

=

Equate the z-coordinates.

2 7 3

3

t

t

= ! +

=

Since the t values are equal, the point P0(7, 0, 2) does lie on the line. b) The point P0(2, 1, –3) corresponds to the position vector [2, 1, –3]. Substitute the coordinates into the vector equation.

2, 1,! 3"# $% = 1, 3, ! 7"# $% + t 2, !1, 3"# $%


2 1 2

1

2

t

t

= +

=


1 3

2

t

t

= !

=



3 7 3

4

3

t

t

! = ! +

=

Since the t values are not equal, the point P0(2, 1, –3) does not lie on the line. c) The point P0(13, –3, 11) corresponds to the position vector [13, –3, 11]. Substitute the coordinates into the vector equation.

13, ! 3, 11"# $% = 1, 3, ! 7"# $% + t 2, !1, 3"# $%


13 1 2

6

t

t

= +

=


3 3

6

t

t

! = !

=


11 7 3

6

t

t

= ! +

=

Since the t values are equal, the point P0(13, –3, 11) does lie on the line. d) The point P0(–4, 0.5, –14.5) corresponds to the position vector [–4, 0.5, –14.5]. Substitute the coordinates into the vector equation.

!4, 0.5,!14.5"# $% = 1, 3, ! 7"# $% + t 2, !1, 3"# $%


4 1 2

5

2

t

t

! = +

= !


0.5 = 3! t

t = 2.5

t =5

2



!14.5 = !7 + 3t

t = !7.5

3

t = !5

2

Since the t values are not equal, the point P0(–4, 0.5, –14.5) does not lie on the line. Chapter 8 Section 1 Question 14 Page 438 a) A direction vector to the first line can be [6, 4] and a direction vector for the second line can be [3, 2]. [6, 4] = 2[3, 2] Since one direction vector is a scalar multiple of the other, the two lines are parallel. b) A direction vector to the first line can be [–9, 1], and a direction vector for the second line can be [1, 9]. [–9, 1]·[1, 9] = –9(1) + 1(9) = 0 Since the dot product of these two direction vectors is zero, the two lines are perpendicular. Chapter 8 Section 1 Question 15 Page 438 a) This represents a horizontal line in two-space with a y-intercept of 3. b) This represents a line that lies along the z-axis in three-space. c) This represents a vertical line in two-space with x-intercept 1. d) This represents a line parallel to the y-axis and passing through the point (–1, 3, 2). Chapter 8 Section 1 Question 16 Page 438 a) The line is parallel to the x-axis. Choose [ ]1, 0, 0i =

! as a direction vector.

Point (3, –8) is on the line and has position vector [ ]3, 8! .


x, y!" #$ = 3, % 8!" #$ + t 1, 0!" #$ , t !! . b) A normal to the given line is [ ]4, 3n = !

!. This is a direction vector for the new line.

Point (–2, 4) is on the line and has position vector [ ]2, 4!


x, y!" #$ = %2, 4!" #$ + t 4, % 3!" #$ , t !! . c) The line is parallel to the z-axis. Choose [ ]0, 0, 1k =

! as a direction vector.

Point (1, 5, 10) is on the line and has position vector [ ]1, 5, 10


x, y, z!" #$ = 1, 5, 10!" #$ + t 0, 0, 1!" #$ , t !! .


d) The given line has direction vector [ ]3, 5, 9m = ! !

!".

The position vector for the x-intercept of 10 is [ ]10, 0, 0! .


x, y, z!" #$ = %10, 0, 0!" #$ + t 3, % 5, % 9!" #$ , t !! . e) The position vector for the x-intercept of the first line is [ ]3, 0, 0 . Let x = 0 and solve for t.

0 = 6 + 3t

t = !2

Let y = 0 and solve for t.

0 = !2 ! t

t = !2

Substitute t = –2 and solve for z.

z = !3+ (–2)(–2)

z = 1

Thus, the position vector for the z-intercept is [0, 0, 1]. A direction vector for the line is [ ] [ ] [ ]3, 0, 0 0, 0, 1 3, 0, 1m = ! = !

!", t !! .


x, y, z!" #$ = 0, 0, 1!" #$ + t 3, 0, %1!" #$ . Chapter 8 Section 1 Question 17 Page 438 Answers may vary. For example: A scalar equation in three-space would be of the form 0Ax By Cz D+ + + = . For such an equation, you could let y and z equal 0 and solve for x to find a unique x-intercept. Letting x and z equal 0 would lead to a unique y-intercept and letting x and y equal 0 would lead to a unique z-intercept. But it is easy to imagine lines in three-space that do not intersect even one axis to form an intercept. Therefore the original assumption must be wrong. A scalar equation in three-space must not represent a line. Chapter 8 Section 1 Question 18 Page 438 a) No.


b) Yes.

c) No.

Chapter 8 Section 1 Question 19 Page 438 a) The following are the answers to the questions in 8.1ChapterProblem.gsp. 2. The variable x is always twice the value of t and the variable y is always equal to t. A possible equation is 2 2 ,x y t t= = ! ! . 3. The rectangle seems to rotate 60º clockwise for every 1 increase in the value of t. The centre of the rectangle seems to follow the line 0.5y x= . 4. Yes. It agrees with the observations from part 3. If the formula is changed for x or y, the rotation

continues in the same way but the centre of rotation follows a different line. (If 2y t= , the centre

follows the line y x= ; this suggests that the centre follows the line

y =x parameter

y parameterx.!

"#

5. If you double the rotation angle (120º), the motion is twice as fast. If you halve the rotation angle

(60º), the motion is half as fast. 6. If you change x to 2

t , the rectangle (centre) will follow a parabolic path.


b) Answers may vary. A sample solution is shown. Changing the value of a parameter can change both the location and the orientation of a figure

(rectangle) defined by parametric equations. Chapter 8 Section 1 Question 20 Page 438 a) The point P(7, –21, 7) corresponds to the position vector [7, –21, 7]. Substitute the coordinates into the vector equation.

7, ! 21,7"# $% = 4, ! 3, 2"# $% + t 1, 8, ! 3"# $%


7 4

3

t

t

= +

=


!21= !3+ 8t

t = !18

8

t = !9

4


7 2 3

5

3

t

t

= !

= !

Since the t values are not equal, the point P(7, –21, 7) does not lie on the line. b) If the lines intersect, there is one t-value and one s-value that will produce the same point. Equate the x-coordinates. Equate the y-coordinates. Equate the z-coordinates.

4 + t = 2 + 4s

4s ! t = 2 !

!3+ 8t = !19 ! 5s

5s + 8t = !16 !

2 ! 3t = 8! 9s

9s ! 3t = 6 !

Solve and for s and t.

4s ! t = 2 !

5s + 8t = !16 "

37s = 0 8!+"

s = 0

Substitute 0s = into equation .

4(0)! t = 2

t = !2


When t = –2, the first line gives the point:

(4 + (–2)(1), ! 3+ (–2)(8), 2 + (–2)(–3)) = (2, !19, 8)

When s = 0, the second line gives the point:

(2 + 0(4), !19 + (0)(–5), 8+ 0(–9)) = (2,–19,8)

Therefore, the two lines intersect at the point (2, –19, 8). c) If the lines intersect, there is one t-value and one v-value that will produce the same point. Equate the x-coordinates. Equate the y-coordinates. Equate the z-coordinates.

4 + t = 1+ 4v

4v ! t = 3 !

!3+ 8t = 0 ! 5v

5v + 8t = 3 !

2 ! 3t = 3! 9v

9v ! 3t = 1 !

Solve and for s and v.

4v ! t = 3 !

5v + 8t = 3 "

37v = 27 8!+"

v =27

37

Substitute 27

37v = into equation .

427

37

!

"#$

%&' t = 3

t =108

37' 3

t = '3

37

When 3

37t = ! , the first line gives the point:

4 + !3

37

"

#$%

&'(1), ! 3+ !

3

37

"

#$%

&'(8), 2 + !

3

37

"

#$%

&'(–3)

"

#$%

&'=

145

37,

134

37,

83

37

"

#$%

&'.

When 27

37v = , the second line gives the point:

2 +27

37

!

"#$

%&(4), '19 +

27

37

!

"#$

%&(–5), 8+

27

37

!

"#$

%&(–9)

!

"#$

%&=

182

37,'838

37,

53

37

!

"#$

%&.

Therefore, the two lines do not intersect.


Chapter 8 Section 1 Question 21 Page 438 a) The lines are parallel if their direction vectors are scalar multiples of each other. [ ] [ ]4, 5 7,p k! = Equate the x-coordinates.

4 7

4

7

p

p

=

=


5

457

35

4

pk

k

k

! =

! =

= !

The lines are parallel if 35

4k = ! .

b) The lines are perpendicular if the dot product of the direction vectors is zero.

[ ] [ ]

( ) ( )

4, 5 7, 0

4 7 5 0

28

5

k

k

k

! " =

+ ! =

=

The lines are perpendicular when 28

5k = .

Chapter 8 Section 1 Question 22 Page 438 The direction vectors are scalar multiples of each other:

!2 3, !1, 4"# $% = 2 !3, 1, ! 4"# $%

= !6, 2, ! 8"# $%.

The lines are at least parallel. Does (–13, 6, –10) lie on

1! ?

Substitute the coordinates into the vector equation.

!13, 6, !10"# $% = 11, ! 2, 17"# $% + t 3, !1, 4"# $%

Equate the x-coordinates. Equate the y-coordinates. Equate the z-coordinates.

13 11 3

8

t

t

! = +

= !

6 2

8

t

t

= ! !

= !

10 17 4

27

4

t

t

! = +

= !


Since the t values are not equal, the point (–13, 6, –10) does not lie on the line1! and so

1 2 and ! ! are

distinct lines. Does (–4, 3, –3) lie on

1! ?

Substitute the coordinates into the vector equation.

!4, 3, ! 3"# $% = 11, ! 2, 17"# $% + t 3, !1, 4"# $%

Equate the x-coordinates. Equate the y-coordinates. Equate the z-coordinates. 4 11 3

5

t

t

! = +

= !

3 2

5

t

t

= ! !

= !

3 17 4

5

t

t

! = +

= !

Since the t values are equal, the point (–4, 3, –3) does lie on the line

1! and so

1 3 and ! ! are different

representations of the same line. Chapter 8 Section 1 Question 23 Page 439 a)

AB! "!!

= OB! "!!

!OA! "!!

= !5, 4"# $% ! 3, ! 2"# $%

= !8, 6"# $%

This is a direction vector for the line. b) A possible vector equation is

x, y!" #$ = 3, % 2!" #$ + t %8, 6!" #$ , t &! .

Possible parametric equations are x = 3! 8t and y = !2 + 6t, t "!.

c) Choose a vector that makes a zero dot product with [ ]8, 6! . For example, [ ]3, 4 . The scalar equation is of the form 3 4 0x y C+ + = . Point A is on the line. Substitute its coordinates and solve for C.

3(3)+ 4(–2)+C = 0

C = !1

The scalar equation is 3 4 1 0x y+ ! = d) Choose convenient values for t and substitute in the parametric equations. If t = –1, (11, –8) is a point on the line. If t = –2, (19, –14) is a point on the line. If t = 2, (–13, 10) is a point on the line.

If t = 1

2! , (7, –5) is a point on the line.


e) Substitute the coordinates for (35, –26) into the vector equation.

35, ! 26"# $% = 3, ! 2"# $% + t !8, 6"# $%

Equate the x-coordinates. Equate the y-coordinates.

35 3 8

4

t

t

= !

= !

26 2 6

4

t

t

! = ! +

= !

Since the t values are equal, the point (35, –26) does lie on the line. Substitute the coordinates for (−9, 8) into the vector equation.

!9, 8"# $% = 3,! 2"# $% + t !8, 6"# $%

Equate the x-coordinates. Equate the y-coordinates.

9 3 8

3

2

t

t

! = !

=

8 2 6

5

3

t

t

= ! +

=

Since the t values are not equal, the point (−9, 8) does not lie on the line. Chapter 8 Section 1 Question 24 Page 439 a) The vectors will be perpendicular to the line if their dot product with the direction vector for the line is zero. i)

2, ! 3, !1"# $% & 1, !1, 5"# $% = 2(1)+ (–3)(–1)+ (–1)(5)

= 0

This vector is perpendicular to the line. ii)

2, ! 3, !1"# $% & 2, 2, 2"# $% = 2(2)+ (–3)(2)+ (–1)(2)

= !4

This vector is not perpendicular to the line. iii)

2, ! 3, !1"# $% & !4, ! 7, 13"# $% = 2(–4)+ (–3)(–7)+ (–1)(13)

= 0

This vector is perpendicular to the line. b) Choose vectors that have a dot product of zero with the direction vector for the line. Choose arbitrary

values for x and y, then calculate z so that the dot product is zero. Be careful not to choose scalar multiples of the vectors in part i) or iii) above.

If 1 and 1x y= = , then [1, 1, –1] is a vector perpendicular to the line. If 5 and 2x y= = , then [5, 2, 4], is a vector perpendicular to the line. If 2 and 0x y= = , then [2, 0, 4], is a vector perpendicular to the line.


Chapter 8 Section 1 Question 25 Page 439 a) First find the direction vectors for the sides.

AB! "!!

= OB! "!!

!OA! "!!

= 4, 3"# $% ! 7, 4"# $%

= !3, !1"# $%

AC! "!!

= OC! "!!

!OA! "!!

= 6, ! 3"# $% ! 7, 4"# $%

= !1, ! 7"# $%

BC! "!!

= OC! "!!

!OB! "!!

= 6, ! 3"# $% ! 4, 3"# $%

= 2, ! 6"# $%

Vector equations for the sides are:

AB : x, y!" #$ = 7, 4!" #$ + t %3, %1!" #$ , 0 & t &1, t '!

AC : x, y!" #$ = 7, 4!" #$ + s %1, % 7!" #$ , 0 & s &1, s '!

BC : x, y!" #$ = 4, 3!" #$ + v 2, % 6!" #$ , 0 & v &1, v '!

Note that the restrictions on the parameters limit the points on the lines to just those within the triangle. b) First find the direction vectors for the sides.

DE! "!!

= OE! "!!

!OD! "!!

= 1, !1, 8"# $% ! 1, ! 3, 2"# $%

= 0, 2, 6"# $%

DF! "!!

= OF! "!!

!OD! "!!

= 5, !17, 0"# $% ! 1, ! 3, 2"# $%

= 4, !14, ! 2"# $%

EF! "!

= OF! "!!

!OE! "!!

= 5, !17, 0"# $% ! 1, !1, 8"# $%

= 4, !16, ! 8"# $%

It is possible to choose simpler (scalar multiples) for the direction vectors.


Vector equations for the sides are:

DE : x, y, z!" #$ = 1, % 3, 2!" #$ + t 0, 1, 3!" #$ , 0 & t & 2, t '!

DF : x, y, z!" #$ = 1, % 3, 2!" #$ + s 2, % 7, %1!" #$ , 0 & s & 2, s '!

EF : x, y, z!" #$ = 1, %1, 8!" #$ + v 1, % 4, % 2!" #$ , 0 & v & 4, v '!

Chapter 8 Section 1 Question 26 Page 439 a) Let x = 0. Let y = 0. Let z = 0.

0 = 4 + 3t

t = !4

3

0 = 1+ t

t = !1

0 = !2 ! 5t

t = !2

5

For x = 0 and y = 0, the t-values are different. So, there is no z-intercept. For x = 0 and z = 0, the t-values are different. So, there is no y-intercept. For y = 0 and z = 0, the t-values are different. So, there is no x-intercept. This line has no x-, y-, or z-intercepts. b) A line parallel to the given line would be of the form [ ] [ ] [ ], , , , 3, 1, 5x y z a b c t= + ! . i) To have an x-intercept, setting y = 0 and z = 0 must lead to consistent t-values. Let y = 0. Let z = 0.

0 = b+ t

t = !b

0 = c ! 5t

t =1

5c

∴ c = –5b is the necessary condition. ii) To have a y-intercept, setting x = 0 and z = 0 must lead to consistent t-values. Let x = 0. Let z = 0.

0 = a + 3t

t = !1

3a

0 = c ! 5t

t =1

5c

∴ 3c = –5a is the necessary condition. iii) To have a z-intercept, setting x = 0 and y = 0 must lead to consistent t-values. Let x = 0. Let y = 0.

0 = a + 3t

t = !1

3a

0 = b+ t

t = !b

∴ a = 3b is the necessary condition. c) The line needs to contain the origin since a line intersecting two axes can only intersect the third axis if

the intersections occur at the origin. The equation will be

x, y, z!" #$ = 0, 0, 0!" #$ + t 3, 1, % 5!" #$ , t &!.


Chapter 8 Section 1 Question 27 Page 439

6

4

2

-2

-5 5

y = 2.00

x = -0.03

t = 6.27 Reset

t

a) The graph is a circle with radius 2 units. b) The value 2 is the radius of the circle. c) If the coefficients are changed the curve becomes an ellipse, with the coefficient on x becoming the

semi-x-axis for the ellipse and the coefficient of y becoming the semi-y-axis for the ellipse. The example shows the coefficients as 3 and 2 for x and y respectively.

6

4

2

-2

-5 5

y = 2.00

x = -0.04

t = 6.27 Reset

t

d) The resulting curve is the line segment defined by y x= , restricted so that 2 , 2x y! " " .

6

4

2

-2

-5 5

y = -0.03

x = -0.03

t = 6.27 Reset

t


Chapter 8 Section 1 Question 28 Page 439 The curve is a spiral (an Archimedean spiral).

6

4

2

-2

-5 5

y = 3.18

x = 0.64

t = 6.48 Reset

t


6

4

2

-2

-5 5 10

y = 1.72

x = 7.95

t = 8.65 Reset

t

a) The curve is periodic, repeating itself every 6.28 (2π) along the x-axis. b) The maximum y-value is 2, as determined by the coefficients of y in this example. (cos (t) varies

between –1 and +1).


Chapter 8 Section 1 Question 30 Page 439 a) Parametric equations for a tricuspoid are of the form:

x = a(2cos t + cos 2t); y = a(2sin t ! sin 2t)

The curve below is with a = 1. This curve was first discovered by Euler in 1745. The curve is sometimes called a deltoid. It looks like an equilateral triangle with sides that curve inwards instead of being straight.

6

4

2

-2

-5 5 10

y = 0.00

x = 3.00

t = 6.30 Reset

t

b) Parametric equations for a lissajous curve are of the form:

x = A cos !

xt "#

x( ); y = Bcos !yt "#

y( )

The curve below has A = 2, B = 3, 4x=! , 3

y=! ,

2x=!

" , and 2

y=!

" .

6

4

2

-2

-5 5

y = -0.11

x = 0.10

t = 6.27 Reset

t

This family of curves models simple harmonic motion. They were investigated first by Nathaniel

Bowditch in 1815 and more extensively by J.A. Lissajous in 1857. The curves are described as like a knot in three-space.


c) Parametric equations for a epicycloid are of the form:

x = a + b( )cos ! " bcos a + b

b!

#

$%&

'(; y = a + b( )sin ! " bsin

a + b

b!

#

$%&

'(.

The curve below has 2 and 1a b= = .

8

6

4

2

-2

-4

-6

-5 5

y = 0.00

x = 2.00

t = 6.27 Reset

t

Some of the early mathematicians exploring this family of curves were Galileo and Mersenne (1599).

The shape of the curve is somewhat gear like. Chapter 8 Section 1 Question 31 Page 439 a) Neither. Since [ ] [ ]4, 6, 15 8, 12, 20k! ! " ! for any k !! , the direction vectors are not scalar multiples

of each other. So the lines are not parallel. Also:

4, ! 6, !15"# $% & !8, 12, 20"# $% = 4(–8)+ (–6)(12)+ (–15)(20)

= !404

Since this dot product is not zero, the direction vectors and the lines are not perpendicular. b) Since [ ] [ ]5, 1, 5 1, 5, 2k! " for any k !! , the direction vectors are not scalar multiples of each other.

So the lines are not parallel. However:

5, 1, ! 5"# $% & 1, 5, 2"# $% = 5(1)+1(5)+ (–5)(2)

= 0

Since the dot product is zero, the direction vectors and the lines are perpendicular.



x = 3+ 2t

y = 4 + 5t, t !!

b) 3

2

4

5

xt

yt

!=

!=

c) 3 4

2 5

x y! !=

d) The components of the direction vector become the denominators of each fraction, with the x-value of

the vector under the fraction involving x and the y-value of the vector under the fraction involving y. As well, the x-value of the position vector exists with an opposite sign beside the x in the numerator of corresponding fraction and the

y-value of the position vector exists with an opposite sign beside the y in the numerator of the other fraction.

e) i) 1 7

3 8

x y! !=

ii) 24

9

yx

+! =

iii) ( )5 2

3 4

x y! ! !=

! !

This can also be written as 5 2

3 4

x y+ !=

!.


m

!"

= 2, 7!" #$ , r"

0 = 6, 9!" #$

The vector equation is

x, y!" #$ = 6, 9!" #$ + t 2, 7!" #$ , t %!. For the scalar equation, start with the symmetric equation and then simplify.

x ! 6

2=

y ! 9

7

7x ! 42 = 2y !18

7x ! 2y ! 24 = 0


b) m

!"

= 4, ! 5"# $% , r"

0 = !3, ! 9"# $%


x, y!" #$ = %3, % 9!" #$ + t 4, % 5!" #$ , t &!. For the scalar equation, start with the symmetric equation and then simplify.

x – (–3)

4=

y ! (–9)

!5

!5x !15 = 4y + 36

5x + 4y + 51= 0

c)

m

!"

= !7, 1"# $% , r"

0 = 4, !10"# $%


x, y!" #$ = 4, %10!" #$ + t %7, 1!" #$ , t &! . For the scalar equation, start with the symmetric equation and then simplify.

x ! 4

!7=

y ! (!10)

1

x ! 4 = !7 y ! 70

x + 7 y + 66 = 0


a) i)

x !1

5=

y ! 3

4=

z ! 9

2

ii) 4 17

2 8

x yz

+ += = !

!

iii) 1 95

3 11

y zx

! +! = =

b) i)

x, y, z!" #$ = 4, 12 ,15!" #$ + t 8, 5, 2!" #$ , t %!

ii)

x, y, z!" #$ = 6, %1, % 5!" #$ + t 1, 7, % 3!" #$ , t &!

iii)

x, y, z!" #$ = 5, % 3, 0!" #$ + t %6, %10 ,11!" #$ , t &!



a)

cos ! =m

!"

1 "m!"

2

m

!"

1 m

!"

2

cos ! =

1, 5#$ %& " '2, 7#$ %&1, 5#$ %& '2, 7#$ %&

cos ! =1(–2)+ 5(7)

12+ 52 (–2)2

+ 72

cos ! #33

37.1214

! # cos'1 33

37.1214

(

)*+

,-

! # 27.3o

b)

cos! =m

!"

1 "m!"

2

m

!"

1 m

!"

2

cos ! =

2,# 2, 3$% &' " 1, # 3, 5$% &'2,# 2, 3$% &' 1, # 3, 5$% &'

cos ! =2(1)+ (–2)(–3)+ 3(5)

22+ (–2)2

+ 32 12+ (–3)2

+ 52

cos ! #23

24.3926

! # cos#1 23

24.3926

(

)*+

,-

! # 19.5o

Chapter 8 Section 1 Question 36 Page 440 To find a direction vector perpendicular to both lines, calculate 1 2m m!

!" !".

3, !1, 1"# $% & 1, ! 3, 7"# $% = !1(7)! (–3)(1), 1(1)! 7(3), 3(–3)!1(–1)"# $%

= !4, ! 20, ! 8"# $%

Using a simpler vector for this direction,[ ]1, 5, 2 , the parametric equations are:

x = 6 + t

y = !2 + 5t

z = 1+ 2t, t "!


Chapter 8 Section 1 Question 37 Page 440 Answers may vary. For example:

l1: x, y, z!" #$ = 3, 1, %1!" #$ + t 1, 0, 0!" #$ , t &!

l2

: x, y, z!" #$ = 3, 1, %1!" #$ + s 0, 1, 0!" #$ , s &!


Consider one triangle (area 6 cm2) in the hexagon. Drawing an altitude creates a right angled triangle. The altitude is 3x . Then,

A =1

22x( ) 3x( )

6 = 3x2

x2=

6

3

x ! 1.86

The value of a is found by examining a regular pentagon, finding the distance from the centre to one of its sides. The pentagon can be divided into 10 triangles meeting at its centre. Each triangle will have a 36º at the centre of the pentagon.


x

a= tan36

o

a !1.86

tan36o

a ! 2.56

Finally, to find h, use the Pythagorean theorem with the red triangle above.

h2+ (2.56)2

= 3 1.86( )( )2

h2+ 6.5536 = 10.3788

h = 3.8252

! 1.96

The height of the pyramid will be about 1.96 cm. Chapter 8 Section 1 Question 39 Page 440 Thee area of the quadrilateral can be found by considering an upper triangle and a lower triangle, separated by the x-axis.

30 =1

2(4)(4k)+

1

2(4)(k)

30 = 10k

k = 3


Chapter 8 Section 2 Equations of Planes Chapter 8 Section 2 Question 1 Page 451 Answers may vary. For example: a) (8, 1, 2), (8, –2, 5), (8, 7, 6) All points must have the first coordinate 8. b) (1, 0, 3), (5, 9, 0), (2, 6, 1) Choose any value for z and then solve for y; x can be any value. c) (0, 1, –1), (1, 0, –11), (1, 2, 3) Choose any values for x and y, then solve for z. d) (0, 0, –2), (4, 0, 0), (2, 0, –1) Choose any values for x and y, then solve for z. Chapter 8 Section 2 Question 2 Page 451 Answers may vary. For example: Find vectors between the points found in question 1. a)

8, ! 2, 5"# $% ! 8, 1, 2"# $% = 0, ! 3, 3"# $%

8, 7, 6"# $% ! 8, 1, 2"# $% = 0, 6, 4"# $%

b)

5, 9, 0!" #$ % 1, 0, 3!" #$ = 4, 9, % 3!" #$

2, 6, 1!" #$ % 1, 0, 3!" #$ = 1, 6, % 2!" #$

c)

1, 0, !11"# $% ! 0, 1, !1"# $% = 1, !1, !10"# $%

1, 2, 3"# $% ! 1, 0, !11"# $% = 0, 2, 14"# $%

d)

2, 0, !1"# $% ! 0, 0, ! 2"# $% = 2, 0, 1"# $%

4, 0, 0"# $% ! 0, 0, ! 2"# $% = 4, 0, 2"# $%

Chapter 8 Section 2 Question 3 Page 451 Substitute the coordinates for each point into the equation 4 3 5 10x y z+ ! = . a)

L.S.=4(1)+ 3(2)! 5(0)

= 10

R.S. = 10

L.S. = R.S. Point A(1, 2, 0) lies on the plane.


b)

L.S.=4(–7)+ 3(6)! 5(4)

= !30

R.S. = 10

L.S. ≠ R.S. Point B(–7, 6, 4) does not lie on the plane. c)

L.S.=4(–2)+ 3(1)! 5(–3)

= 10

R.S. = 10

LS = RS Point C(–2, 1, –3) lies on the plane. d)

L.S.=4(1.2)+ 3(–2.4)! 5(6.2)

= !33.4

R.S. = 10

L.S. ≠ R.S. Point D(1.2, –2.4, 6.2) does not lie on the plane. Chapter 8 Section 2 Question 4 Page 451 a) Let y = 0 and z = 0. Let x = 0 and z = 0. Let x = 0 and y = 0.

3x ! 2(0)+ 4(0) = 12

x = 4

3(0)! 2y + 4(0) = 12

y = !6

3(0)! 2(0)+ 4z = 12

z = 3

The x-intercept is 4. The y-intercept is –6, and the z-intercept is 3. b) Let y = 0 and z = 0. Let x = 0 and z = 0. Let x = 0 and y = 0.

x + 5(0)! 6(0) = 30

x = 30

0 + 5y ! 6(0) = 30

y = 6

0 + 5(0)! 6z = 30

z = !5

The x-intercept is 30. The y-intercept is 6, and the z-intercept is –5. c) Let y = 0 and z = 0. Let x = 0 and z = 0. Let x = 0 and y = 0.

4x + 2(0)! 7(0)+14 = 0

x = !7

2

4(0)+ 2y ! 7(0)+14 = 0

y = !7

4(0)+ 2(0)! 7z +14 = 0

z = 2

The x-intercept is

!7

2. The y-intercept is –7, and the z-intercept is 2.

d) Let y = 0 and z = 0. Let x = 0 and z = 0. Let x = 0 and y = 0.

3x + 6(0)+18 = 0

x = !6

3(0)+ 6(0)+18 = 0

No values possible for y.

3(0)+ 6z +18 = 0

z = !3

The x-intercept is –6 and the z-intercept is –3. There is no y-intercept.


Chapter 8 Section 2 Question 5 Page 451 s, t !! for all equations. a)

x = 1! 3s + 9t

y = 3+ 4s + 2t

z = !2 ! 5s ! t

b)

x = s

y = !4 +10s + 3t

z = 1! s + 4t

c)

x = t

y = 3s

z = 5+ 5t


x, y, z!" #$ = 9, 4, %1!" #$ + s 3, % 7, % 5!" #$ + t %2, 1, % 4!" #$ , s, t &!

b)

x, y, z!" #$ = 2, 0, 11!" #$ + s 1, 12, 6!" #$ + t 7, % 8, 0!" #$ , s, t &!

c)

x, y, z!" #$ = %6, 0, 5!" #$ + s 0, 8, 0!" #$ + t 0, 0, %13!" #$ , s, t &!

Chapter 8 Section 2 Question 7 Page 451 a) If P(10, –19, 15) lies on the plane there exists a single set of t- and s-values that satisfy the equation.

10 = 6 + s + 2t !

!19 = !7 + 3s ! 2t "

15 = 10 ! s + t #


4 = s + 2t !

!12 = 3s ! 2t "

! 8 = 4s !+"

s = !2

4 = !2 + 2t !

t = 3

Now check if these values satisfy equation . L.S. = 15

R.S.= 10 ! (–2)+ 3

= 15

L.S. = R.S. Thus, P(10, –19, 15) lies on the plane.


b) If P(–4, –13, 10) lies on the plane there exists a single set of t- and s-values that satisfy the equation.

! 4 = 6 + s + 2t !

!13= !7 + 3s ! 2t "

10 = 10 ! s + t #


!10 = s + 2t !

! 6 = 3s ! 2t "

!16 = 4s !+"

s = !4

!10 = !4 + 2t !

t = !3


R.S.= 10 ! (–4)+ (–3)

= 11

L.S. ≠ R.S. Thus, P(–4, –13, 10) does not lie on the plane c) If P(8.5, –3.5, 9) lies on the plane there exists a single set of t- and s-values that satisfy the equation.

8.5 = 6 + s + 2t !

!3.5 = !7 + 3s ! 2t "

9 = 10 ! s + t #


2.5 = s + 2t !

3.5 = 3s ! 2t "

6 = 4s !+"

s = 1.5

2.5 = 1.5+ 2t !

t = 0.5


R.S.= 10 !1.5+ 0.5

= 9

L.S. = R.S. Thus, P(8.5, –3.5, 9) lies on the plane.


Chapter 8 Section 2 Question 8 Page 451 Answers may vary. For example: Let 1 and 1s t= = ! . P(5, –2, 8) is a point on the plane. Let 1 and 0s t= = . P(7, –4, 9) is a point on the plane. Let 0 and 0s t= = . P(6, –7, 10) is a point on the plane. Chapter 8 Section 2 Question 9 Page 451 a) To find the x-intercept, let y = 0 and z = 0. Solve for s and t.

x = 1+ s + 2t !

0 = 8!12s + 4t "

0 = 6 !12s ! 3t #


! 8 = !12s + 4t !

! 6 = !12s ! 3t "

! 2 = 7t !!"

t = !2

7

! 6 = !12s +6

7"

s =4

7

Now substitute in equation .

x = 1+4

7+ 2 !

2

7

"

#$%

&'

= 1

The x-intercept is 1. To find the y-intercept, let x = 0 and z = 0. Solve for s and t.

0 = 1+ s + 2t !

y = 8!12s + 4t "

0 = 6 !12s ! 3t #



!1= s + 2t !

! 6 = !12s ! 3t "

!18 = 21t 12!+"

t = !6

7

! 6 = !12s +18

7"

s =5

7


y = 8!125

7

"

#$%

&'+ 4 !

6

7

"

#$%

&'

= !4

The y-intercept is –4. To find the z-intercept, let x = 0 and y = 0. Solve for s and t.

0 = 1+ s + 2t !

0 = 8!12s + 4t "

z = 6 !12s ! 3t #


!1= s + 2t !

!8 = !12s + 4t "

6 = 14s 2!!"

s =3

7

!1=3

7+ 2t !

t = !5

7


z = 6 !123

7

"

#$%

&'! 3 !

5

7

"

#$%

&'

= 3

The z-intercept is 3.


b) To find the x-intercept, let y = 0 and z = 0. Solve for s and t.

x = 6 + s + 3t !

0 = !9 ! 4s + 3t "

0 = !8! 4s + 8t #


9 = !4s + 3t !

8 = !4s + 8t "

1 = !5t !!"

t = !1

5

8 = !4s + 8 !1

5

"

#$%

&'"

s = !12

5


x = 6 + !12

5

"

#$%

&'+ 3 !

1

5

"

#$%

&'

= 3

The x-intercept is 3. To find the y-intercept, let x = 0 and z = 0. Solve for s and t.

0 = 6 + s + 3t !

y = !9 ! 4s + 3t "

0 = !8! 4s + 8t #


! 6 = s + 3t !

8 = !4s + 8t "

!16 = 20t 4!+"

t = !4

5

8 = !4s + 8 !4

5

"

#$%

&'"

s = !18

5



y = !9 ! 4 !18

5

"

#$%

&'+ 3 !

4

5

"

#$%

&'

= 3

The y-intercept is 3. To find the z-intercept, let x = 0 and y = 0. Solve for s and t.

0 = 6 + s + 3t !

0 = !9 ! 4s + 3t "

z = !8! 4s + 8t #


! 6 = s + 3t !

9 = !4s + 3t "

!15 = 5s !!"

s = !3

! 6 = !3+ 3t !

t = !1


z = !8! 4(–3)+ 8(–1)

= !4

The z-intercept is –4. Chapter 8 Section 2 Question 10 Page 451

s, t !! for all equations. a) Vector:

x, y, z!" #$ = 6, %1, 0!" #$ + s 2, 0, % 5!" #$ + t 1, % 3, 1!" #$

Parametric:

x = 6 + 2s + t

y = !1! 3t

z = !5s + t

b) If the plane is parallel to a line, then it can have the same direction vector as the line. In this case, [1, –1, 1] and [–6, 2, 5] are direction vectors for the plane. Vector:

x, y, z!" #$ = 9, 1, % 2!" #$ + s 1, %1, 1!" #$ + t %6, 2, 5!" #$

Parametric:

x = 9 + s ! 6t

y = 1! s + 2t

z = !2 + s + 5t


c) Direction vectors for the plane can be:

For AB! "!!

= OB! "!!

!OA! "!!

: 3, ! 9, 7"# $% ! 1, 3, ! 2"# $% = 2, !12, 9"# $% .

For AC! "!!

= OC! "!!

!OA! "!!

: 4, ! 4, 5"# $% ! 1, 3, ! 2"# $% = 3, ! 7, 7"# $% .

Vector:

x, y, z!" #$ = 1, 3, % 2!" #$ + s 2, %12 ,9!" #$ + t 3, % 7 ,7!" #$ Parametric:

x = 1+ 2s + 3t

y = 3!12s ! 7t

z = !2 + 9s + 7t

d) Points on the plane include A(8, 0, 0), B(0, –3, 0), and C(0, 0, 2). Direction vectors for the plane can be:

For BA! "!!

= OA! "!!

!OB! "!!

: 8, 0, 0"# $% ! 0, ! 3, 0"# $% = 8, 3, 0"# $% .

For CA! "!!

= OA! "!!

!OC! "!!

: 8, 0, 0"# $% ! 0, 0, 2"# $% = 8, 0, ! 2"# $% .

Vector:

x, y, z!" #$ = 8, 0 ,0!" #$ + s 8, 3, 0!" #$ + t 8, 0, % 2!" #$ Parametric:

x = 8+ 8s + 8t

y = 3s

z = !2t

Chapter 8 Section 2 Question 11 Page 451 a) This plane is parallel to the yz-plane with all points having an x-coordinate of 5. b) This plane is parallel to the xz-plane with all points having a y-coordinate of –7. c) This plane is parallel to the xy-plane with all points having a z-coordinate of 10. d) This plane is parallel to the z-axis and contains points where the x and y values add to 8. The z-

coordinate can be any real number. e) This plane is parallel to the y-axis and contains points where the sum of the x value and two times the z

value is 4. The y-coordinate can be any real number. f) This plane is parallel to the x-axis and contains points where the result of two times the z-value

subtracted from three times the y-value is 12. The x-coordinate can be any real number. g) This plane contains points where the x-, y-, and z-values add to 0. The plane passes through the origin. h) This plane has an x-intercept of 2, a y-intercept of 3, and a z-intercept of –6. i) This plane has an x-intercept of –5, a y-intercept of 4, and a z-intercept of –10.


Chapter 8 Section 2 Question 12 Page 452 Answers for part e) to part g) may vary. a) y = 1 b)

z = k, k !!

c) First find two direction vectors for the plane.

For AB! "!!

= OB! "!!

!OA! "!!

: 5, ! 3, 2"# $% ! 2, 1, 1"# $% = 3, ! 4, 1"# $% .

For AC! "!!

= OC! "!!

!OA! "!!

: 0, !1, 4"# $% ! 2, 1, 1"# $% = !2, ! 2, 3"# $% .

A vector equation is

x, y, z!" #$ = 2, 1, 1!" #$ + s 3, % 4, 1!" #$ + t %2, % 2, 3!" #$ , s, t &! . d) Direction vectors for the plane are perpendicular to a

!

and so have a dot product of zero with a!

. Two possibilities are [5, –4, 0] and [1, 0, 2]. A possible vector equation is

x, y, z!" #$ = 1, 0, 0!" #$ + s 5, % 4, 0!" #$ + t 1, 0, 2!" #$ , s, t &! .

Verify that P0(1, 0, 0) does not create a plane passing through the origin.

0 = 1+ 5s + t !

0 = !4s "

0 = 2t #


s = 0 !

t = 0 "


R.S.= 1+ .5(0)+ (0)

= 1

L.S. ≠ R..S. The origin, (0, 0, 0) is not on the plane. e) Direction vectors for this plane include [4, 1, –3] and [1, 0, 0]. Any point on the x-axis is on the plane. Use P(3, 0, 0). A possible vector equation is

x, y, z!" #$ = 3, 0, 0!" #$ + s 4, 1, % 3!" #$ + t 1, 0, 0!" #$ , s, t &! .

f) 3 10x z+ = is a possible scalar equation. (In general Ax + Bz = C where none of A, B, C ! 0 .) g) 2 1x y+ = is a possible scalar equation. (In general

Ax + By = C where none of A, B, C ! 0 .)


Chapter 8 Section 2 Question 13 Page 452 a) The points (0, –3, 0), R(2, –3, 4), and S(–2, –3, –4) are collinear. There are many planes passing through any one line. b) [ ] [ ] [ ]

[ ] [ ] [ ]

5, 15, 7 2, 0, 1 3, 15, 6

0, 10, 3 2, 0, 1 2, 10, 4

3

2

AB

AC

AB AC

= ! ! = !

= ! ! = ! !

= !

!!!"

!!!"

!!!" !!!"

Therefore A, B, and C are collinear and many planes pass through the points.

c) Since[ ] [ ]4

8, 12, 4 6, 9, 33

! = ! ! ! , the direction vectors are parallel.

The two direction vectors cannot be parallel if the plane is to be unique. d) Check to see if the point is on the line. Substitute the coordinates into the vector equation. [ ] [ ] [ ]3, 1, 4 5, 3, 10 4, 2, 3t! = ! + ! ! Equate the x-coordinates. Equate the y-coordinates. Equate the y-coordinates.

3 5 4

2

t

t

= ! +

=

1 3 2

2

t

t

! = !

=

4 10 3

2

t

t

= !

=

Since the t values are equal, the point P(3, –1, 4) does lie on the line. There are many planes passing through any one line. e) Since [ ] [ ]2 2, 3, 1 4, 6, 2! ! = ! ! , the direction vectors are parallel which does not define a unique plane. f) Check to see if the lines are identical. Since[ ] [ ]1, 5, 2 1, 5, 2! ! = ! ! , the lines are at least parallel. Check if the point P0(0, 1, 1) is on the second line. Substitute the coordinates into the vector equation.

0, 1, 1!" #$ = %1, 6, %1!" #$ + t 1, % 5, 2!" #$

Equate the x-coordinates. Equate the y-coordinates. Equate the y-coordinates.

0 1

1

t

t

= ! +

=

1 6 5

1

t

t

= !

=

1 1 2

1

t

t

= ! +

=

Since the t values are equal, the point P0(0, 1, 1) lies on the second line. There are many planes passing through any one line.


Chapter 8 Section 2 Question 14 Page 452 The equation for the front wall is x = 8 and for the back wall is x = 0. The equation for the left wall is y = 0 and for the right wall is y = 6. The equation for the floor is z = 0 and for the ceiling is z = 4. Chapter 8 Section 2 Question 15 Page 452

If the plane is perpendicular to the lines, then the lines are perpendicular to the plane. If A is on the plane, then AP lies in the plane and AP

! "!!

will be perpendicular to the lines.

AP! "!!

= OP! "!!

!OA! "!!

= !1, 4, ! 2"# $% ! 7, 10, 16"# $%

= !8, ! 6, !18"# $%

Check perpendicularity.

!8, ! 6, !18"# $% & 1, 2, !1"# $% = !8 1( ) + !6( ) 2( ) + !18( ) !1( )= !2

Since the dot product is not zero, AP

! "!!

is not perpendicular to the first line and therefore A does not lie on the plane. Chapter 8 Section 2 Question 16 Page 452 First find direction vectors.

AB! "!!

= OB! "!!

!OA! "!!

= 2, ! 3, 1"# $% ! 3, 0, 4"# $%

= !1, ! 3, ! 3"# $%

AC! "!!

= OC! "!!

!OA! "!!

= !5, 8, ! 4"# $% ! 3, 0, 4"# $%

= !8, 8, ! 8"# $%


The plane containing A, B, and C has vector equation:

x, y, z!" #$ = 3, 0, 4!" #$ + s 1, 3, 3!" #$ + t 1, %1, 1!" #$ , s, t &!

Check if D(1, 4, 3) lies on the plane. Substitute the coordinates in the vector equation

1= 3+ s + t !

4 = 3s ! t "

3= 4 + 3s + t #


!2 = s + t !

4 = 3s ! t "

2 = 4s !+"

s =1

2

!2 =1

2+ t !

t = !5

2

Now check if these values satisfy equation .

L.S. = 3

R.S.= 4 + 31

2

!

"#$

%&+ '

5

2

!

"#$

%&

= 3

L.S. = R.S. The point D lies on the same plane as A, B, and C. Chapter 8 Section 2 Question 17 Page 452 First find direction vectors.

AB! "!!

= OB! "!!

!OA! "!!

= 0, 1, 0"# $% ! 4, ! 2, 6"# $%

= !4, 3, ! 6"# $%

AC! "!!

= OC! "!!

!OA! "!!

= 1, 0, ! 5"# $% ! 4, ! 2, 6"# $%

= !3, 2, !11"# $%

Chapter 8 Lines and Planes Chapter 8 Prerequisite …8...MHR • Calculus and Vectors 12 Solutions 819 Chapter 8 Lines and Planes Chapter 8 Prerequisite Skills Chapter 8 Prerequisite

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