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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
MHR Calculus and Vectors 12 Solutions 820
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 77
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
MHR Calculus and Vectors 12 Solutions 821
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).
MHR Calculus and Vectors 12 Solutions 822
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.
MHR Calculus and Vectors 12 Solutions 823
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).
MHR Calculus and Vectors 12 Solutions 824
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).
MHR Calculus and Vectors 12 Solutions 825
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 45
! .
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 is4 22
5 5y x= ! + .
MHR Calculus and Vectors 12 Solutions 826
c) Perpendicular lines have negative reciprocal slopes. The line 3 62
y x= ! + has slope3
2! .
The required line will have slope 2
3.
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 is2 8
3 3y x= ! .
d) Perpendicular lines have negative reciprocal slopes. The line 77 5 20 45
x y y x+ = ! = " + has
slope7
5! .
The required line will have slope 5
7.
The y-intercept of 6
6 5 15 35
x 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 is5
37
y 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.
MHR Calculus and Vectors 12 Solutions 827
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! !
are not perpendicular.
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! !
are not perpendicular.
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! !
are not perpendicular.
MHR Calculus and Vectors 12 Solutions 828
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"# $%
Chapter 8 Prerequisite Skills Question 9 Page 427 The vectors are not unique, as any vector that is a scalar multiple of the given vector will be parallel.
a) [2, 10] b) [30, 40] c) [4, 2, 14] d) [1, 4, 5]
Chapter 8 Prerequisite Skills Question 10 Page 427 These vectors are not unique, as any vector that produces zero in a dot product with the given direction
vector will be perpendicular.
a) [5, 1] since [5, 1][1, 5] = 0
b) [4, 3] since [4, 3][3, 4] = 0 c) [3, 1, 1] since [3, 1, 1][2, 1, 7] = 0 d) [1, 1, 1] since [1, 1, 1][1, 4, 5] = 0
MHR Calculus and Vectors 12 Solutions 829
Chapter 8 Prerequisite Skills Question 11 Page 427
Use the formula for dot product.
a)
cos ! =a!
"b!
a!
b!
cos ! =1, 3#$ %& " 2, 5#$ %&
12
+ 32
22
+ 52
cos ! "17
17.0294
! " cos'117
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
+ 02
+ 22
52
+ 32
+ 42
cos ! "13
15.8114
! " cos'113
15.8114
(
)*+
,-
! " 34.7o
MHR Calculus and Vectors 12 Solutions 830
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
MHR Calculus and Vectors 12 Solutions 831
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 !! .
MHR Calculus and Vectors 12 Solutions 832
Chapter 8 Section 1 Question 3 Page 437 a)
b)
c)
d)
MHR Calculus and Vectors 12 Solutions 833
Chapter 8 Section 1 Question 4 Page 437 Use The Geometers 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
MHR Calculus and Vectors 12 Solutions 834
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"# $%
Equate the x-coordinates.
9 3 2
3
t
t
= !
= !
Equate the y-coordinates.
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.
MHR Calculus and Vectors 12 Solutions 835
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"# $%
Equate the x-coordinates.
9 3 2
6
t
t
! = !
=
Equate the y-coordinates.
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"# $%
Equate the x-coordinates.
2 3 2
5
2
t
t
! = !
=
Equate the y-coordinates.
13.5 1 5
5
2
t
t
= +
=
Since the t values are equal, the point P(2, 13.5) does lie on the line.
Chapter 8 Section 1 Question 6 Page 437 t !! for all equations.
a)
x = 10 +13t
y = 6 + t
b)
x = 12t
y = 5! 7t
c)
x = 3+ 6t
y = !9t
z = !1+ t
MHR Calculus and Vectors 12 Solutions 836
d)
x = 11+ 3t
y = 2
z = 0
Chapter 8 Section 1 Question 7 Page 437 a)
x, y!" #$ = 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
4
25
5 10 4
5 14 0
yx
x y
x y
!+ =
+ = !
! + =
MHR Calculus and Vectors 12 Solutions 837
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).
MHR Calculus and Vectors 12 Solutions 838
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
"
#$%
&' are easy to calculate.
Chapter 8 Section 1 Question 10 Page 437 The scalar equation of a line in two-space is of the form [ ]0 where ,Ax By C n A B+ + = =
! is a normal
vector for the line.
a) The scalar equation is of the form 3 0x y C+ + = . Substitute the point (2, 4) to determine the value of C.
3(2) + 4 + C = 0
C = !10
A scalar equation for the line is 3 2 10 0x y+ ! = .
b) The scalar equation is of the form x y + C = 0. Substitute the point (5, 1) to determine the value of C.
!5!1+ C = 0
C = 6
A scalar equation for the line is 6 0x y! + = .
c) The scalar equation is of the form y + C = 0. Substitute the point (3, 7) to determine the value of C.
!7 + C = 0
C = 7
A scalar equation for the line is 7 0y + = .
MHR Calculus and Vectors 12 Solutions 839
d) The scalar equation is of the form 1.5 3.5 0x y C! + = . Substitute the point (0.5, 2.5) to determine the value of C.
1.5(0.5)! 3.5(!2.5) + C = 0
C = !9.5
A scalar equation for the line is 1.5 3.5 9.5 0x y! ! = or 3 7 19 0x y! ! =
Chapter 8 Section 1 Question 11 Page 437 t !! for all equations. a) Choose two points on the line, say (0, 3) and (6, 0). (Hint: consider the intercepts) The vector joining these two points is a possible direction vector.
m!"
= 6, 0!" #$ % 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!" #$
A possible vector equation is
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!" #$
A possible vector equation is
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"# $%
A possible vector equation is
x, y!" #$ = 0, % 5!" #$ + t %9, 8!" #$ . Possible parametric equations are
x = !9t, y = !5+ 8t.
MHR Calculus and Vectors 12 Solutions 840
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!" #$
A possible vector equation is
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!" #$
Equate the x-coordinates.
7 1 2
3
t
t
= +
=
Equate the y-coordinates.
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"# $%
Equate the x-coordinates.
2 1 2
1
2
t
t
= +
=
Equate the y-coordinates.
1 3
2
t
t
= !
=
MHR Calculus and Vectors 12 Solutions 841
Equate the z-coordinates.
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"# $%
Equate the x-coordinates.
13 1 2
6
t
t
= +
=
Equate the y-coordinates.
3 3
6
t
t
! = !
=
Equate the z-coordinates.
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"# $%
Equate the x-coordinates.
4 1 2
5
2
t
t
! = +
= !
Equate the y-coordinates.
0.5 = 3! t
t = 2.5
t =5
2
MHR Calculus and Vectors 12 Solutions 842
Equate the z-coordinates.
!14.5 = !7 + 3t
t = !7.5
3
t = !5
2
Since the t values are not equal, the point P0(4, 0.5, 14.5) does not lie on the line.
Chapter 8 Section 1 Question 14 Page 438 a) A direction vector to the first line can be [6, 4] and a direction vector for the second line can be [3, 2]. [6, 4] = 2[3, 2]
Since one direction vector is a scalar multiple of the other, the two lines are parallel.
b) A direction vector to the first line can be [9, 1], and a direction vector for the second line can be [1, 9]. [9, 1][1, 9] = 9(1) + 1(9)
= 0
Since the dot product of these two direction vectors is zero, the two lines are perpendicular.
Chapter 8 Section 1 Question 15 Page 438 a) This represents a horizontal line in two-space with a y-intercept of 3.
b) This represents a line that lies along the z-axis in three-space. c) This represents a vertical line in two-space with x-intercept 1. d) This represents a line parallel to the y-axis and passing through the point (1, 3, 2).
Chapter 8 Section 1 Question 16 Page 438 a) The line is parallel to the x-axis. Choose [ ]1, 0, 0i =
! as a direction vector.
Point (3, 8) is on the line and has position vector [ ]3, 8! .
A possible vector equation is
x, y!" #$ = 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!
A possible vector equation is
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
A possible vector equation is
x, y, z!" #$ = 1, 5, 10!" #$ + t 0, 0, 1!" #$ , t !! .
MHR Calculus and Vectors 12 Solutions 843
d) The given line has direction vector [ ]3, 5, 9m = ! !!"
.
The position vector for the x-intercept of 10 is [ ]10, 0, 0! .
A possible vector equation is
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 !! .
A possible vector equation is
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.
MHR Calculus and Vectors 12 Solutions 844
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 60 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 (120), the motion is twice as fast. If you halve the rotation angle
(60), the motion is half as fast.
6. If you change x to 2t , the rectangle (centre) will follow a parabolic path.
MHR Calculus and Vectors 12 Solutions 845
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"# $%
Equate the x-coordinates.
7 4
3
t
t
= +
=
Equate the y-coordinates.
!21= !3+ 8t
t = !18
8
t = !9
4
Equate the z-coordinates.
7 2 3
5
3
t
t
= !
= !
Since the t values are not equal, the point P(7, 21, 7) does not lie on the line.
b) If the lines intersect, there is one t-value and one s-value that will produce the same point.
Equate the x-coordinates. Equate the y-coordinates. Equate the z-coordinates.
4 + t = 2 + 4s
4s ! t = 2 !
!3+ 8t = !19 ! 5s
5s + 8t = !16 !
2 ! 3t = 8! 9s
9s ! 3t = 6 !
Solve and for s and t.
4s ! t = 2 !
5s + 8t = !16 "
37s = 0 8!+"
s = 0
Substitute 0s = into equation .
4(0)! t = 2
t = !2
MHR Calculus and Vectors 12 Solutions 846
When t = 2, the first line gives the point:
(4 + (2)(1), ! 3+ (2)(8), 2 + (2)(3)) = (2, !19, 8)
When s = 0, the second line gives the point:
(2 + 0(4), !19 + (0)(5), 8 + 0(9)) = (2,19,8)
Therefore, the two lines intersect at the point (2, 19, 8).
c) If the lines intersect, there is one t-value and one v-value that will produce the same point.
Equate the x-coordinates. Equate the y-coordinates. Equate the z-coordinates.
4 + t = 1+ 4v
4v ! t = 3 !
!3+ 8t = 0 ! 5v
5v + 8t = 3 !
2 ! 3t = 3! 9v
9v ! 3t = 1 !
Solve and for s and v.
4v ! t = 3 !
5v + 8t = 3 "
37v = 27 8!+"
v =27
37
Substitute 27
37v = into equation .
427
37
!
"#$
%&' t = 3
t =108
37' 3
t = '3
37
When3
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
"
#$%
&'.
When27
37v = , the second line gives the point:
2 +27
37
!
"#$
%&(4), '19 +
27
37
!
"#$
%&(5), 8 +
27
37
!
"#$
%&(9)
!
"#$
%&=
182
37,'83837
,53
37
!
"#$
%&.
Therefore, the two lines do not intersect.
MHR Calculus and Vectors 12 Solutions 847
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
=
=
Equate the y-coordinates.
5
45
7
35
4
pk
k
k
! =
! =
= !
The lines are parallel if35
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 when28
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 on1! ?
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
! = +
= !
MHR Calculus and Vectors 12 Solutions 848
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 on1! ?
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 line1! 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.
MHR Calculus and Vectors 12 Solutions 849
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.
MHR Calculus and Vectors 12 Solutions 850
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"# $%
It is possible to choose simpler (scalar multiples) for the direction vectors.
MHR Calculus and Vectors 12 Solutions 851
Vector equations for the sides are:
DE : x, y, z!" #$ = 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 &!.
MHR Calculus and Vectors 12 Solutions 852
Chapter 8 Section 1 Question 27 Page 439
6
4
2
-2
-5 5
y = 2.00
x = -0.03
t = 6.27 Resett
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 Resett
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 Resett
MHR Calculus and Vectors 12 Solutions 853
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 Resett
Chapter 8 Section 1 Question 29 Page 439
6
4
2
-2
-5 5 10
y = 1.72
x = 7.95
t = 8.65 Resett
a) The curve is periodic, repeating itself every 6.28 (2) 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).
MHR Calculus and Vectors 12 Solutions 854
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 Resett
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
=! , 3y
=! , 2
x=!
" , and 2
y=!
" .
6
4
2
-2
-5 5
y = -0.11
x = 0.10
t = 6.27 Resett
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.
MHR Calculus and Vectors 12 Solutions 855
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"# $% = 5(1) +1(5) + (5)(2)
= 0
Since the dot product is zero, the direction vectors and the lines are perpendicular.
MHR Calculus and Vectors 12 Solutions 856
Chapter 8 Section 1 Question 32 Page 440
a)
x = 3+ 2t
y = 4 + 5t, t !!
b) 32
4
5
xt
yt
!=
!=
c) 3 42 5
x y! !=
d) The components of the direction vector become the denominators of each fraction, with the x-value of
the vector under the fraction involving x and the y-value of the vector under the fraction involving y.
As well, the x-value of the position vector exists with an opposite sign beside the x in the numerator of
corresponding fraction and the
y-value of the position vector exists with an opposite sign beside the y in the numerator of the other
fraction.
e) i) 1 73 8
x y! !=
ii) 249
yx
+! =
iii) ( )5 2
3 4
x y! ! !=
! !
This can also be written as 5 23 4
x y+ !=
!.
Chapter 8 Section 1 Question 33 Page 440
a) m!"
= 2, 7!" #$ , 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
MHR Calculus and Vectors 12 Solutions 857
b) m!"
= 4, ! 5"# $% , r"
0 = !3, ! 9"# $%
The vector equation is
x, y!" #$ = %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"# $%
The vector equation is
x, y!" #$ = 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
Chapter 8 Section 1 Question 34 Page 440
a) i)
x !1
5=
y ! 3
4=
z ! 9
2
ii) 4 1 72 8
x yz
+ += = !
!
iii) 1 953 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 &!
MHR Calculus and Vectors 12 Solutions 858
Chapter 8 Section 1 Question 35 Page 440
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'133
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#123
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"# $%
Using a simpler vector for this direction,[ ]1, 5, 2 , the parametric equations are:
x = 6 + t
y = !2 + 5t
z = 1+ 2t, t "!
MHR Calculus and Vectors 12 Solutions 859
Chapter 8 Section 1 Question 37 Page 440
Answers may vary. For example:
l1: x, y, z!" #$ = 3, 1, %1!" #$ + t 1, 0, 0!" #$ , t &!
l2
: x, y, z!" #$ = 3, 1, %1!" #$ + s 0, 1, 0!" #$ , s &!
Chapter 8 Section 1 Question 38 Page 440
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 36 at
the centre of the pentagon.
MHR Calculus and Vectors 12 Solutions 860
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
MHR Calculus and Vectors 12 Solutions 861
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"# $%
Chapter 8 Section 2 Question 3 Page 451
Substitute the coordinates for each point into the equation 4 3 5 10x y z+ ! = .
a)
L.S.=4(1) + 3(2)! 5(0)
= 10
R.S. = 10
L.S. = R.S.
Point A(1, 2, 0) lies on the plane.
MHR Calculus and Vectors 12 Solutions 862
b)
L.S.=4(7) + 3(6)! 5(4)
= !30
R.S. = 10
L.S. R.S.
Point B(7, 6, 4) does not lie on the plane.
c)
L.S.=4(2) + 3(1)! 5(3)
= 10
R.S. = 10
LS = RS
Point C(2, 1, 3) lies on the plane.
d)
L.S.=4(1.2) + 3(2.4)! 5(6.2)
= !33.4
R.S. = 10
L.S. R.S.
Point D(1.2, 2.4, 6.2) does not lie on the plane.
Chapter 8 Section 2 Question 4 Page 451
a) Let y = 0 and z = 0. Let x = 0 and z = 0. Let x = 0 and y = 0.
3x ! 2(0) + 4(0) = 12
x = 4
3(0)! 2y + 4(0) = 12
y = !6
3(0)! 2(0) + 4z = 12
z = 3
The x-intercept is 4. The y-intercept is 6, and the z-intercept is 3.
b) Let y = 0 and z = 0. Let x = 0 and z = 0. Let x = 0 and y = 0.
x + 5(0)! 6(0) = 30
x = 30
0 + 5y ! 6(0) = 30
y = 6
0 + 5(0)! 6z = 30
z = !5
The x-intercept is 30. The y-intercept is 6, and the z-intercept is 5.
c) Let y = 0 and z = 0. Let x = 0 and z = 0. Let x = 0 and y = 0.
4x + 2(0)! 7(0) +14 = 0
x = !7
2
4(0) + 2y ! 7(0) +14 = 0
y = !7
4(0) + 2(0)! 7z +14 = 0
z = 2
The x-intercept is
!7
2. The y-intercept is 7, and the z-intercept is 2.
d) Let y = 0 and z = 0. Let x = 0 and z = 0. Let x = 0 and y = 0.
3x + 6(0) +18 = 0
x = !6
3(0) + 6(0) +18 = 0
No values possible for y.
3(0) + 6z +18 = 0
z = !3
The x-intercept is 6 and the z-intercept is 3. There is no y-intercept.
MHR Calculus and Vectors 12 Solutions 863
Chapter 8 Section 2 Question 5 Page 451
s, t !! for all equations.
a)
x = 1! 3s + 9t
y = 3+ 4s + 2t
z = !2 ! 5s ! t
b)
x = s
y = !4 +10s + 3t
z = 1! s + 4t
c)
x = t
y = 3s
z = 5+ 5t
Chapter 8 Section 2 Question 6 Page 451
a)
x, y, z!" #$ = 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 #
Solve and for s and 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.
MHR Calculus and Vectors 12 Solutions 864
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 #
Solve and for s and t.
!10 = s + 2t !
! 6 = 3s ! 2t "
!16 = 4s !+"
s = !4
!10 = !4 + 2t !
t = !3
Now check if these values satisfy equation . L.S. = 10
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 #
Solve and for s and t.
2.5 = s + 2t !
3.5 = 3s ! 2t "
6 = 4s !+"
s = 1.5
2.5 = 1.5+ 2t !
t = 0.5
Now check if these values satisfy equation . L.S. = 9
R.S. = 10 !1.5+ 0.5
= 9
L.S. = R.S.
Thus, P(8.5, 3.5, 9) lies on the plane.
MHR Calculus and Vectors 12 Solutions 865
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 #
Solve and for s and t.
! 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
"
#$%
&'
= 1
The x-intercept is 1.
To find the y-intercept, let x = 0 and z = 0. Solve for s and t.
0 = 1+ s + 2t !
y = 8!12s + 4t "
0 = 6 !12s ! 3t #
MHR Calculus and Vectors 12 Solutions 866
Solve and for s and t.
!1= s + 2t !
! 6 = !12s ! 3t "
!18 = 21t 12!+"
t = !6
7
! 6 = !12s +18
7"
s =5
7
Now substitute in equation .
y = 8!125
7
"
#$%
&'+ 4 !
6
7
"
#$%
&'
= !4
The y-intercept is 4.
To find the z-intercept, let x = 0 and y = 0. Solve for s and t.
0 = 1+ s + 2t !
0 = 8!12s + 4t "
z = 6 !12s ! 3t #
Solve and for s and t.
!1= s + 2t !
!8 = !12s + 4t "
6 = 14s 2!!"
s =3
7
!1=3
7+ 2t !
t = !5
7
Now substitute in equation .
z = 6 !123
7
"
#$%
&'! 3 !
5
7
"
#$%
&'
= 3
The z-intercept is 3.
MHR Calculus and Vectors 12 Solutions 867
b) To find the x-intercept, let y = 0 and z = 0. Solve for s and t.
x = 6 + s + 3t !
0 = !9 ! 4s + 3t "
0 = !8! 4s + 8t #
Solve and for s and t.
9 = !4s + 3t !8 = !4s + 8t "1 = !5t !!"
t = !1
5
8 = !4s + 8 !1
5
"
#$%
&'"
s = !12
5
Now substitute in equation .
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 #
Solve and for s and t.
! 6 = s + 3t !8 = !4s + 8t "
!16 = 20t 4!+"
t = !4
5
8 = !4s + 8 !4
5
"
#$%
&'"
s = !18
5
MHR Calculus and Vectors 12 Solutions 868
Now substitute in equation .
y = !9 ! 4 !18
5
"
#$%
&'+ 3 !
4
5
"
#$%
&'
= 3
The y-intercept is 3.
To find the z-intercept, let x = 0 and y = 0. Solve for s and t.
0 = 6 + s + 3t !
0 = !9 ! 4s + 3t "
z = !8! 4s + 8t #
Solve and for s and t.
! 6 = s + 3t !
9 = !4s + 3t "
!15 = 5s !!"
s = !3
! 6 = !3+ 3t !
t = !1
Now substitute in equation .
z = !8! 4(3) + 8(1)
= !4
The z-intercept is 4.
Chapter 8 Section 2 Question 10 Page 451
s, t !! for all equations.
a) Vector:
x, y, z!" #$ = 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
MHR Calculus and Vectors 12 Solutions 869
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.
MHR Calculus and Vectors 12 Solutions 870
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 #
Solve and for s and t.
s = 0 !
t = 0 "
Now check if these values satisfy equation . L.S. = 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 .)
MHR Calculus and Vectors 12 Solutions 871
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.
MHR Calculus and Vectors 12 Solutions 872
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"# $%
MHR Calculus and Vectors 12 Solutions 873
The plane containing A, B, and C has vector equation:
x, y, z!" #$ = 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 #
Solve and for s and 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"# $%
MHR Calculus and Vectors 12 Solutions 874
AB
! "!!
! AC
! "!!
gives a vector perpendicular to the plane.
AB
! "!!
! AC! "!!
= "4, 3, " 6#$ %& ! "3, 2, "11#$ %&
= 3(11)" 2(6), (6)(3)" (11)(4), (4) 2( )" (3)(3)#$%&
= "21, " 26, 1#$ %&
If D lies on the plane:
BD
! "!!
= OD
! "!!
!OB! "!!
= 1, k, ! 2"# $% ! 0, 1, 0"# $%
= 1, k !1, ! 2"# $%
Then BD must be perpendicular to AB
! "!!
! AC
! "!!
. The dot product of these two vectors must be zero.
!21, ! 26, 1"# $% & 1, k !1, ! 2"# $% = 0
!21! 26(k !1)! 2 = 0
!26k = !3
k =3
26
Chapter 8 Section 2 Question 18 Page 452
Answers may vary. For example:
a) Choose arbitrary values for two of the variables and solve for the third. Let x = 0 and y = 0. The resulting point is J(0, 0, D),
Let y = 0 and z = 0. The resulting point is K(D, 0, 0),
Let y = 1 and z = 0. The resulting point is L(D 2, 1, 0),
Let y = 0 and z = 1. The resulting point is M(D + 1, 0, 1)
b)
JK
! "!
= OK
! "!!
!OJ! "!
= D, 0, 0"# $% ! 0, 0, ! D"# $%
= D, 0, D"# $%
JL
!"!
= OL
! "!!
!OJ! "!
= D ! 2, 1, 0"# $% ! 0, 0, ! D"# $%
= D ! 2, 1, D"# $%
JM
! "!!
= OM
! "!!!
!OJ! "!
= D +1, 0, 1"# $% ! 0, 0, ! D"# $%
= D +1, 0, D +1"# $%
MHR Calculus and Vectors 12 Solutions 875
c)
JK
! "!
! JL!"!
" JM! "!!
= D,0, D#$ %& ! D ' 2, 1, D#$ %& " D +1, 0, D +1#$ %&
= D,0, D#$ %& ! 1(D +1)' 0(D), D(D +1)' (D +1)(D 2), (D 2)(0)' (D +1)(1)#$ %&
= D,0, D#$ %& ! D +1, D2
+ D ' D2 + D + 2, ' D '1#$%&
= D,0, D#$ %& ! D +1, 2D + 2, ' D '1#$ %&
= D D +1( ) + 0(2D + 2) + D('D '1)
= D2
+ D ' D2 ' D
= 0
Since the triple scalar product is zero, the volume of the parallelepiped defined by the three vectors is
zero, which implies that the three vectors are coplanar.
d) Find the cross product of two of the vectors.
JK
! "!
! JL!"!
= D, 0, D"# $% ! D & 2, 1, D"# $%
= 0(D)&1(D), D(D & 2)& D(D), D(1)& (D & 2)(0)"# $%
= &D, & 2D, D"# $%
Choose any vector parallel to this vector. A simple vector is [1, 2, 1]. Note the similarity to the
coefficients in the original scalar equation.
Chapter 8 Section 2 Question 19 Page 453 Answers may vary. For example:
Yes, as an example, a plane defined by 3x + 2y + 4z = 12 can be written as 14 6 3
x y z+ + = . The plane has
an x-intercept of 4, a y-intercept of 6 and a z-intercept of 3.
MHR Calculus and Vectors 12 Solutions 876
Chapter 8 Section 2 Question 20 Page 453 To find the number, create placeholders for the nine digits (ddddddddd) and replace them when you can.
The fifth digit must be 5 or 0 to get divisibility by 5. Not using 0, so: dddd5dddd.
The second, fourth, sixth, and eighth digits must be even. Thus, the other digits are odd: OEOE5EOEO.
Numbers are divisible by 4 if the last two digits are divisible by 4. Since the third and seventh digits are
odd, the fourth and eighth digits must then be 2 and 6 in some order since O4 and O8 cannot produce
multiples of 4. Therefore, the second and sixth digits are 4 and 8 in some order. This gives dAdB5AdBd0
where A = 4 or 8 and B = 2 or 6.
Since the sixth number is even, the seventh and eighth digits form a number divisible by 8.
The first three digits must add to a multiple of three to get divisibility by 3, and likewise so must the next
three digits and the last three digits.
Consider the second set of three, where the first digit is 2 or 6, the next is 5, and the last is 4 or 8. From
these combinations, only 654 and 258 work. Each of these forces the second and eighth digits. This gives
either d8d654d2d0 or d4d258d6d.
Now, in the 654 case, the seventh digit must be 3 or 7 to give divisibility by 8, and in the 258 case, the
seventh digit must by 1 or 9 (or 5, but that is taken).
What numbers can go in the first and third positions? Some pair of 1, 3, 7, 9 must go there such that the
first three add up to a number divisible by three, and a valid number must be left for the seventh position.
For 654, these begining numbers are divisible by three: 183, 189, 381, 387, 783, 789, 981, 987. Must
discard 387 and 783 because the seventh digit must be 3 or 7 in this case.
For 258, there are only two possible begining numbers: 147, 741.
Now for the remaining eight cases, fill in the one of the two possible seventh digits and test the numbers
formed by the first seven digits for divisibility by 7.
1836547, 1896543, and 1896547 are not divisible by 7.
3816547 works, so the answer might be 381654729.
7896543, 9816543, 9816547, and 9876543 are not divisible by 7.
1472589 and 7412589 are not divisible by 7.
Divisibility by 9 is guaranteed since the digits 1 to 9 have a sum of 45 which is divisible by 9.
Therefore, the number is 381 654 729.
MHR Calculus and Vectors 12 Solutions 877
Chapter 8 Section 2 Question 21 Page 453
Let the dimensions of the flag be x and y as shown.
From the 60 and 30 right-angled triangle:
tan 60o
=10
8! y
8! y =10
tan 60o
y ! 2.23
From the 76 and 14 right-angled triangle:
tan 14o
=y
10 ! x
10 ! x =2.23
tan 14o
x ! 1.06
2.23 1.06 ! 2.36
The area of the flag is approximately 2.36 m2.
MHR Calculus and Vectors 12 Solutions 878
Chapter 8 Section 3 Properties of Planes Chapter 8 Section 3 Question 1 Page 459
Substitute the coordinates in the equation 2 3 5 0x y z+ ! ! = .
a)
L.S. = 5+ 2(3)! 3(2)! 5
= 0
R.S. = 0
L.S. = R.S.
Therefore, M lies on the plane. b)
L.S. = 3+ 2(2)! 3(1)! 5
= 5
R.S. = 0
L.S. R.S. Therefore, N does not lie on the plane. c)
L.S. = !7 + 2(0)! 3(4)! 5
= 0
R.S. = 0
L.S. = R.S.
Therefore, P lies on the plane. d)
L.S. = 6 + 2(1)! 3(1)! 5
= 0
R.S. = 0
L.S. = R.S.
Therefore, Q lies on the plane. e)
L.S. = 0 + 2(0)! 3(5)! 5
= !20
R.S. = 0
L.S. R.S. Therefore, R does not lie on the plane. f)
L.S. = 1+ 2(2)! 3(3)! 5
= 9
R.S. = 0
L.S. R.S. Therefore, S does not lie on the plane.
MHR Calculus and Vectors 12 Solutions 879
Chapter 8 Section 3 Question 2 Page 459
Answers may vary. For example:
a) [ ] [ ]1 21, 2, 2 ; 2, 4, 4n n= = ! ! !! !
b) [ ] [ ]1 26, 1, 4 ; 12, 2, 8n n= ! = !! !
c) [ ] [ ]1 25, 0, 2 ; 15, 0, 6n n= =! !
d) [ ] [ ]1 20, 5, 0 ; 0, 1, 0n n= =! !
e) [ ] [ ]1 23, 4, 0 ; 6, 8, 0n n= = ! !! !
f) [ ] [ ]1 21, 3, 1 ; 1, 3, 1n n= ! ! = !! !
Chapter 8 Section 3 Question 3 Page 459
Answers may vary. Choose any vector that has a dot product of zero with n!
. For example:
a) [2, 0, 1] b) [1, 2, 1] c) [2, 0, 5] d) [2, 0, 1] e) [4, 3, 10] f) [0, 1, 3]
Chapter 8 Section 3 Question 4 Page 459
a) The scalar equation has the form 0x y z D! + + = .
P lies on the plane. Substitute its coordinates and solve for D.
2 ! (1) + 8 + D = 0
D = !11
The scalar equation is 11 0x y z! + ! = .
MHR Calculus and Vectors 12 Solutions 880
b) The scalar equation has the form 3 7 0x y z D+ + + = .
P lies on the plane. Substitute its coordinates and solve for D.
3(3) + 7(6) + 4 + D = 0
D = 29
The scalar equation is 3 7 29 0x y z+ + + = .
c) The scalar equation has the form 2 5 0x z D! + = .
P lies on the plane. Substitute its coordinates and solve for D.
2(1)! 5(3) + D = 0
D = !17
The scalar equation is 2 5 17 0x z! ! = .
d) The scalar equation has the form 9 0x D! + = .
P lies on the plane. Substitute its coordinates and solve for D.
!9(2) + D = 0
D = !18
The scalar equation is 9 18 0x! ! = or 2 0x + = .
e) The scalar equation has the form 4 3 2 0x y z D! + + = .
P lies on the plane. Substitute its coordinates and solve for D.
4(6)! 3(3) + 2(4) + D = 0
D = !7
The scalar equation is 4 3 2 7 0x y z! + ! = .
f) The scalar equation has the form 4 3 4 0x y z D! + + = .
P lies on the plane. Substitute its coordinates and solve for D.
4(2)! 3(5) + 4(3) + D = 0
D = 11
The scalar equation is 4 3 4 11 0x y z! + + = .
MHR Calculus and Vectors 12 Solutions 881
Chapter 8 Section 3 Question 5 Page 459
Answers for part b) and part c) may vary.
a) A possible normal vector is [ ]1, 4, 2n = !!
.
b) Let x = 0 and y = 0.
!(0) + 4(0) + 2z + 6 = 0
z = !3
S(0, 0, 3) is one point.
Let y = 0 and z = 0.
!x + 4(0) + 2(0) + 6 = 0
x = 6
T(6, 0, 0) is another point.
c)
ST
! "!
= OT
! "!!
!OS! "!!
= 6, 0, 0"# $% ! 0, 0, ! 3"# $%
= 6, 0, 3"# $%
d)
ST
! "!
!n"
= 6, 0, 3"# $% ! &1, 4, 2"# $%
= 6(1) + 0(4) + 3(2)
= 0
Therefore, ST! "!
! n"
.
Chapter 8 Section 3 Question 6 Page 459
a) Find a normal vector to the plane.
n!
= 1, 2, !1"# $% & 1, ! 2, 3"# $%
= 2(3)! (2)(1), (1)(1)! 3(1), 1(2)!1(2)"# $%
= 4, ! 4, ! 4"# $%
Use [1, 1, 1]. The scalar equation has the form 0x y z D! ! + = .
Use the point (3, 7, 5) to determine D.
3! 7 ! (5) + D = 0
D = !1
A scalar equation of the plane is 1 0x y z! ! ! = .
MHR Calculus and Vectors 12 Solutions 882
b) Find a normal vector to the plane.
n!
= 3, ! 2, 4"# $% & 5, ! 2, 6"# $%
= !2(6)! (2)(4), 4(5)! 6(3), 3(2)! 5(2)"# $%
= !4, 2, 4"# $%
Use [2, 1, 2]. The scalar equation has the form 2 2 0x y z D! ! + = .
Use the point (5, 2, 3) to determine D.
2(5)! (2)! 2(3) + D = 0
D = !6
A scalar equation of the plane is 2 2 6 0x y z! ! ! = . c) Find a normal vector to the plane.
n!
= 2, !1, !1"# $% & 1, 3, 3"# $%
= !1(3)! 3(1), (1)(1)! 3(2), 2(3) 3( )!1(1)"#$%
= 0, ! 7, 7"# $%
Use [0, 1, 1]. The scalar equation has the form 0y z D! + = .
Use the point (6, 8, 2) to determine D.
8! 2 + D = 0
D = !6
A scalar equation of the plane is 6 0y z! ! = . d) Find a normal vector to the plane.
n!
= 6, 5, 2!" #$ % 3, & 3, 1!" #$
= 5(1)& (3)(2), 2(3)&1(6), 6(3)& 3(5)!" #$
= 11, 0, & 33!" #$
Use [1, 0, 3]. The scalar equation has the form 3 0x z D! + = .
Use the point (9, 1, 8) to determine D.
9 ! 3(8) + D = 0
D = !33
A scalar equation of the plane is 3 33 0x z! ! = .
MHR Calculus and Vectors 12 Solutions 883
e) Find a normal vector to the plane.
n!
= 0, 1, 0!" #$ % 0, 0, &1!" #$
= 1(1)& 0(0), 0(0)& (1)(0), 0(0)& 0(1)!" #$
= &1, 0, 0!" #$
Use [1, 0, 0]. The scalar equation has the form 0x D+ = .
Use the point (0, 0, 1) to determine D.
0 + D = 0
D = 0
A scalar equation of the plane is 0x = . f) Find a normal vector to the plane.
n!
= 2, 0, 3!" #$ % 3, 0, 2!" #$
= 0(2)& 0(3), 3(3)& 2(2), 2(0)& 3(0)!" #$
= 0, 5, 0!" #$
Use [0, 1, 0]. The scalar equation has the form 0y D+ = .
Use the point (3, 2, 1) to determine D.
2 + D = 0
D = !2
A scalar equation of the plane is 2 0y ! = . Chapter 8 Section 3 Question 7 Page 459
a) Find a normal vector to the plane.
n!
= !2, 3, !1"# $% & 2, 1, ! 2"# $%
= 3(2)!1(1), !1(2)! (2)(2), ! 2(1)! 2(3)"# $%
= !5, ! 6, ! 8"# $%
Use [5, 6, 8]. The scalar equation has the form 5 6 8 0x y z D+ + + = .
Use the point (3, 1, 5) to determine D.
5(3) + 6(1) + 8(5) + D = 0
D = !61
A scalar equation of the plane is 5 6 8 61 0x y z+ + ! = .
MHR Calculus and Vectors 12 Solutions 884
b) Find a normal vector to the plane.
n!
= 0, !1, 1"# $% & 5, ! 2, 0"# $%
= !1(0)! (2)(1), 1(5)! 0(0), 0(2)! 5(1)"# $%
= 2, 5, 5"# $%
Use [2, 5, 5]. The scalar equation has the form 2 5 5 0x y z D+ + + = .
Use the point (1, 3, 2) to determine D.
2(1) + 5(3) + 5(2) + D = 0
D = !3
A scalar equation of the plane is 2 5 5 3 0x y z+ + ! = . c) Find a normal vector to the plane.
n!
= !1, !1, 3"# $% & 2, 4, 1"# $%
= !1(1)! 4(3), 3(2)!1(1), !1(4)! 2(1)"# $%
= !13, 7, ! 2"# $%
Use [13, 7, 2]. The scalar equation has the form 13 7 2 0x y z D! + + = .
Use the point (1, 1, 2) to determine D.
13(1)! 7(1) + 2(2) + D = 0
D = 16
A scalar equation of the plane is 13 7 2 16 0x y z! + + = . d) Find a normal vector to the plane.
n!
= 1, 2, 4!" #$ % 1, 0, 2!" #$
= 2(2)& 0(4), 4(1)& 2(1), 1(0)&1(2)!" #$
= 4, 2, & 2!" #$
Use [2, 1, 1]. The scalar equation has the form 2 0x y z D+ ! + = .
Use the point (2, 3, 5) to determine D.
2(2) + 3! 5+ D = 0
D = !2
A scalar equation of the plane is 2 2 0x y z+ ! ! = .
MHR Calculus and Vectors 12 Solutions 885
e) Find a normal vector to the plane.
n!
= 5, 3, !1"# $% & 2, 0, 2"# $%
= 3(2)! 0(1), !1(2)! 2(5), 5(0)! 2(3)"# $%
= 6, !12, ! 6"# $%
Use [1, 2, 1]. The scalar equation has the form 2 0x y z D! ! + = .
Use the point (3, 2, 1) to determine D.
!3! 2(2)! (1) + D = 0
D = 6
A scalar equation of the plane is 2 6 0x y z! ! + = . f) Find a normal vector to the plane.
n!
= 3, !1, 2"# $% & 1, ! 5, 3"# $%
= !1(3)! (5)(2), 2(1)! 3(3), 3(5)!1(1)"# $%
= 7, ! 7, !14"# $%
Use [1, 1, 2]. The scalar equation has the form 2 0x y z D! ! + = .
Use the point (0, 2, 1) to determine D.
!2 ! 2(1) + D = 0
D = 4
A scalar equation of the plane is 2 4 0x y z! ! + = . Chapter 8 Section 3 Question 8 Page 460
Answers may vary. For example:
a) The direction vector[ ]3, 5, 3! for the line can be the normal vector for the plane. The scalar equation has the form 3 5 3 0x y z D+ ! + =
Use the point (4, 2, 7) to determine D.
3(4) + 5(2)! 3(7) + D = 0
D = 19
A scalar equation of the plane is 3 5 3 19 0x y z+ ! + = .
For direction vectors for the plane, choose any two non-collinear vectors whose dot product with
[ ]3, 5, 3! is zero.
Two such vectors are [ ] [ ]1, 0, 1 and 5, 3, 0! .
A vector equation for the plane is
x, y, z!" #$ = 4, % 2, 7!" #$ + s 1, 0, 1!" #$ + t 5, % 3, 0!" #$ , s, t &! .
MHR Calculus and Vectors 12 Solutions 886
b) Since the plane is parallel to the yz-plane, its equation is of the form 0x D+ = . Since the point (1, 2, 5) is on the plane, a scalar equation of the plane is 1 0x + = .
Since the plane is parallel to the yz-plane, the vectors and j k! !
are direction vectors.
A vector equation for the plane is
x, y, z!" #$ = %1, % 2, 5!" #$ + s 0, 0, 1!" #$ + t 0, 1, 0!" #$ , s, t &! .
c) Parallel planes have the same normals. Here [ ]3, 9, 1n = !!
.
The scalar equation of the plane has the form 3 9 0x y z D! + + =
Use the point (3, 7, 1) to determine D.
3(3)! 9(7) +1+ D = 0
D = 71
A scalar equation of the plane is 3 9 71 0x y z! + + = .
For direction vectors for the plane, choose any two non-collinear vectors whose dot product with
[ ]3, 9, 1n = !!
is zero.
Two such vectors are [ ] [ ]3, 1, 0 and 1, 0, 3! .
A vector equation for the plane is
x, y, z!" #$ = %3, 7, 1!" #$ + s 3, 1, 0!" #$ + t 1, 0, % 3!" #$ , s, t &! .
d) Since the lines are in the plane, their points and direction vectors apply to the plane. However, the direction vectors are parallel. Find the vector between the two position vectors for the
lines.
m!"
= !2, 3, 12"# $% ! 1, ! 4, 4"# $%
= !3, 7, 8"# $%
A vector equation for the plane is
x, y, z!" #$ = 1, % 4, 4!" #$ + s %2, 1, 5!" #$ + t %3, 7, 8!" #$ , s, t &! .
Need a normal vector for a scalar equation. Find the cross product of the vectors in the plane.
!2, 1, 5"# $% & !3, 7, 8"# $% = 1(8)! 7(5), 5(3)! 8(2), (2)(7)! (3)(1)"# $%
= !27, 1, !11"# $%
Use [ ]27, 1, 11! . The scalar equation of the plane has the form 27 11 0x y z D! + + =
Use the point (1, 4, 4) to determine D.
27(1)! (4) +11(4) + D = 0
D = !75
A scalar equation of the plane is 27 11 75 0x y z! + ! = .
MHR Calculus and Vectors 12 Solutions 887
Chapter 8 Section 3 Question 9 Page 460
Answer may vary. For example:
In general, it was easier to determine the scalar equation of each line. However, in part d) it was easier to
write the vector equation.
Chapter 8 Section 3 Question 10 Page 460
a) n!
1 = 4, ! 5, 1"# $% ; n!
2 = 2, ! 9, 1"# $%
n!
1 !n!
2 = 4(2) + (5)(9) +1(1)
= 54
" 0
n!
1 # n!
2 = $5(1)$ (9)(1), 1(2)$1(4), 4(9)$ 2(5)%& '(
" 0!
Since neither the dot product not the cross product is zero (vector), the planes are neither parallel nor
perpendicular.
b)
n!
1 = 5, ! 6, 2"# $% ; n!
2 = 2, ! 5, ! 20"# $%
n!
1 &n!
2 = 5(2) + (6)(5) + 2(20)
= 0
Since the dot product is zero, the normals are perpendicular and, therefore, the planes are
perpendicular.
c)
n!
1 = 4, ! 2, 1"# $% ; n!
2 = !2, 1, ! 3"# $%
n!
1 !n!
2 = 4(2) + (2)(1) +1(3)
= "13
# 0
n!
1 $ n!
2 = "2(3)"1(1), 1(2)" (3)(4), 4(1)" (2)(2)%& '(
# 0!
Since neither the dot product nor the cross product is zero (vector), the planes are neither parallel nor
perpendicular.
Chapter 8 Section 3 Question 11 Page 460
Since the line and plane are perpendicular, the direction vector of the line will be the same as the normal
for the plane.
The vector equation is
x, y, z!" #$ = 3, 9, % 2!" #$ + t 3, % 7, 3!" #$ , s, t &! .
MHR Calculus and Vectors 12 Solutions 888
Chapter 8 Section 3 Question 12 Page 460
a) Find a point on the plane. Let 5 and 2x z= = . P(5, 3, 2) is a point on the plane. Choose two direction vectors by choosing 2 non-collinear vectors whose dot product with [ ]0, 1, 0n =
!
is zero.
Two possible vectors are [ ] [ ]4, 0, 1 and 1, 0, 5! .
The vector equation of the plane is
x, y, z!" #$ = 5, 3, 2!" #$ + s 4, 0, 1!" #$ + t 1, 0, % 5!" #$ , s, t &! .
b) Find a point on the plane. Let 0 and 0x z= = . P(0, 8, 0) is a point on the plane. Choose two direction vectors by choosing 2 non-collinear vectors whose dot product with [ ]1, 1, 0n =
!
is zero.
Two possible vectors are [ ] [ ]1, 1, 1 and 1, 1, 2! ! .
The vector equation of the plane is
x, y, z!" #$ = 0, 8, 0!" #$ + s 1, %1, 1!" #$ + t 1, %1 ,2!" #$ , s, t &! .
c) Find a point on the plane. Let 0 and 0x y= = . P(0, 0, 10) is a point on the plane.
Choose two direction vectors by choosing 2 non-collinear vectors whose dot product with [ ]1, 1, 1n =!
is zero.
Two possible vectors are[ ] [ ]2, 1, 3 and 2, 1, 1! ! ! .
The vector equation of the plane is
x, y, z!" #$ = 0, 0, 10!" #$ + s 2, 1, % 3!" #$ + t 2, %1, %1!" #$ , s, t &! .
d) Find a point on the plane. Let 0 and 0y z= = . P(1, 0, 0) is a point on the plane. Choose two direction vectors by choosing 2 non-collinear vectors whose dot product with
[ ]4, 1, 8n = !!
is zero.
Two possible vectors are [ ] [ ]1, 4, 1 and 2, 0, 1! ! ! .
The vector equation of the plane is
x, y, z!" #$ = 1, 0, 0!" #$ + s 1, % 4, %1!" #$ + t 2, 0, %1!" #$ , s, t &! .
e) Find a point on the plane. Let 0 and 0y z= = . P(4, 0, 0) is a point on the plane. Choose two direction vectors by choosing 2 non-collinear vectors whose dot product with
[ ]3, 2, 1n = !!
is zero.
Two possible vectors are [ ] [ ]1, 2, 1 and 4, 1, 10! ! ! .
The vector equation of the plane is
x, y, z!" #$ = 4, 0, 0!" #$ + s 1, % 2, %1!" #$ + t 4, %1, 10!" #$ , s, t &! .
f) Find a point on the plane. Let 0 and 0y z= = . P(15, 0, 0) is a point on the plane. Choose two direction vectors by choosing 2 non-collinear vectors whose dot product with
[ ]2, 5, 3n = ! !!
is zero.
Two possible vectors are [ ] [ ]5, 2, 0 and 3, 0, 2 .
The vector equation of the plane is
x, y, z!" #$ = 15, 0, 0!" #$ + s 5, 2, 0!" #$ + t 3, 0, 2!" #$ , s, t %! .
MHR Calculus and Vectors 12 Solutions 889
Chapter 8 Section 3 Question 13 Page 460
Use the 3D Grapher or other software to verify your answers.
Chapter 8 Section 3 Question 14 Page 460
a) The bottom plane of the ramp has scalar equation 0z = . The right side plane has equation 0y = .
The left side plane has equation 2y = .
The back plane of the ramp has equation 5x = .
The slanted top plane of the ramp has equation 3 5 0x z! = .
b) The line passes t
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