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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
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 slope4
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 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 7
7 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
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"# $%
Equate the x-coordinates.
2 1 2
1
2
t
t
= +
=
Equate the y-coordinates.
1 3
2
t
t
= !
=
MHR • Calculus and Vectors 12 Solutions 841
Equate the z-coordinates.
3 7 3
4
3
t
t
! = ! +
=
Since the t values are not equal, the point P0(2, 1, –3) does not lie on the line.
c) The point P0(13, –3, 11) corresponds to the position vector [13, –3, 11]. Substitute the coordinates into the vector equation.
Since the t values are not equal, the point P0(–4, 0.5, –14.5) does not lie on the line.
Chapter 8 Section 1 Question 14 Page 438 a) A direction vector to the first line can be [6, 4] and a direction vector for the second line can be [3, 2].
[6, 4] = 2[3, 2]
Since one direction vector is a scalar multiple of the other, the two lines are parallel.
b) A direction vector to the first line can be [–9, 1], and a direction vector for the second line can be [1, 9].
[–9, 1]·[1, 9] = –9(1) + 1(9)
= 0
Since the dot product of these two direction vectors is zero, the two lines are perpendicular.
Chapter 8 Section 1 Question 15 Page 438 a) This represents a horizontal line in two-space with a y-intercept of 3.
b) This represents a line that lies along the z-axis in three-space.
c) This represents a vertical line in two-space with x-intercept 1.
d) This represents a line parallel to the y-axis and passing through the point (–1, 3, 2).
Chapter 8 Section 1 Question 16 Page 438 a) The line is parallel to the x-axis. Choose [ ]1, 0, 0i =
! as a direction vector.
Point (3, –8) is on the line and has position vector [ ]3, 8! .
A possible vector equation is
x, y!" #$ = 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 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
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.
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"# $%
Equate the x-coordinates. Equate the y-coordinates.
9 3 8
3
2
t
t
! = !
=
8 2 6
5
3
t
t
= ! +
=
Since the t values are not equal, the point (−9, 8) does not lie on the line.
Chapter 8 Section 1 Question 24 Page 439 a) The vectors will be perpendicular to the line if their dot product with the direction vector for the line is
Therefore, all vectors in part c) are parallel to the plane.
Chapter 8 Section 3 Question 21 Page 461
The required plane is perpendicular to the given planes. Therefore the normal vectors
[ ] [ ]1 22, 3, 0 and 1, 2, 2n n= ! = !
! ! are direction vectors for the required plane.
The vector equation of the required plane is
x, y, z!" #$ = 3, 1, %1!" #$ + s 2, % 3, 0!" #$ + t 1, 2, % 2!" #$ , s, t &! .
The yellow plane is the required plane in the image from 3D Grapher shown above.
MHR • Calculus and Vectors 12 Solutions 894
Chapter 8 Section 3 Question 22 Page 461
The required plane is perpendicular to the given planes. Therefore the normal vectors
[ ] [ ]1 23, 2, 1 and 4, 6, 5n n= ! = !
! ! are direction vectors for the required plane.
The normal to the required plane is necessary for the scalar equation.
n!
= n!
1 ! n!
2
= 3, " 2, 1#$ %& ! 4, 6, " 5#$ %&
= "2(–5)" 6(1), 1(4)" (–5)(3), 3(6)" 4(–2)#$ %&
= 4, 19, 26#$ %&
The scalar equation is of the form 4 19 26 0x y z D+ + + = .
A is a point on the plane. Determine the value of D.
4(2) +19(1) + 26(–5) + D = 0
D = 103
A scalar equation for the plane is 4 19 26 103 0x y z+ + + = .
Chapter 8 Section 3 Question 23 Page 461
a) The first two planes are parallel to the z-axis and perpendicular to the xy-plane. Call this direction
vertical.
The diagram below shows a top view of the situation.
The other two vertical planes will be of the form x y k! = . In particular choose, 0x y! = (blue) and
x y k! = (red)
The remaining plane is the top of the box. It will be of the form z q= where q is the length of one side
of the box.
MHR • Calculus and Vectors 12 Solutions 895
From the diagram:
2 2 2
2 2
2
2
2
2
2
q q k
q k
kq
kq
+ =
=
=
=
The required equations can be 0x y! = , x y k! = , and
2
kz = . Note that these planes are not
uniquely defined.
b) The sides of the cube are
2
k units long.
Chapter 8 Section 3 Question 24 Page 461
Answers may vary. For example:
The simplest answer is to choose three planes parallel to the coordinate axes: 3, 1, and 2x y z= = ! = ! .
Clearly these planes are not parallel (in fact, they are mutually perpendicular). It is also clear that the
coordinates of A(3, –1, –2) satisfy each of the equations.
Chapter 8 Section 3 Question 25 Page 461
From the diagram, three of the faces are coordinate planes: 0, 0, and 0x y z= = = .
The fourth plane has x-, y-, and z-intercepts of 5, –6, and 2 respectively and so has equation
15 6 2
x y z+ + =!
, which can also be written as 6 5 15 30 0x y z! + ! = .
MHR • Calculus and Vectors 12 Solutions 896
Chapter 8 Section 3 Question 26 Page 461 Answers may vary. For example:
a) If the plane is perpendicular to the x-axis, its equation is of the form x k= . In this case, B = C = 0 and
A, D !! .
b) The equation is 1 or 10 6 5 30 03 5 6
x y zx y z+ + = + + ! = . In this case, the previous equation can be
multiplied by any constant (on both sides of the equation).
Therefore A = 10k, B = 6k, C = 5k, and D = !30k for any k "! .
c) If the plane is parallel to the z-axis, it has no z-intercept and its equation is of the
form 0Ax BY D+ + = . In this case, C = 0 and A, B, D !! .
d) Perpendicular planes have perpendicular normals. The normals for these two planes are [A, B, C] and
[1, 4, –7].
A, B, C!" #$ % 1, 4, & 7!" #$ = 0
A + 4B & 7C = 0
Therefore A, B, and C must satisfy the equation 4 7 0A B C+ ! = and D can be any real number.
Chapter 8 Section 3 Question 27 Page 461 Answers may vary. For example:
Solution 1 Both equations have no y-term and so they have no y-intercept. They are parallel to the y-axis. Therefore
their normals are perpendicular to the y-axis. Any plane having these two vectors as direction vectors will
be perpendicular to the
y-axis.
Solution 2 The cross product of the normals to the given planes is [0, 31, 0] which is parallel to [0, 1, 0] which
would be the normal to the family of planes that would be perpendicular to the y-axis.
MHR • Calculus and Vectors 12 Solutions 897
Chapter 8 Section 3 Question 28 Page 461
Let x be the distance the bugs have travelled from A and C, arriving at P and Q respectively.
Triangles PBR and PQR are right angled.
From ΔPBR:
PR
2= x
2+ (12 ! x)
2
From ΔPQR:
PQ2
= x2
+ (12 ! x)2
+122
= 2x2! 24x + 288
To find the minimum distance, differentiate and set the derivative equal to zero.
4x ! 24 = 0
x = 6
The minimum distance occurs when 6x = . At this time:
PQ = 2(6)2! 24(6) + 288
= 216
= 6 6
! 14.7
The minimum distance between the bugs is approximately 14.7 cm.
MHR • Calculus and Vectors 12 Solutions 898
Chapter 8 Section 3 Question 29 Page 461
9 3log log 1x y= +
Convert all logarithms to the same base. Use base 9, so 9
1 log 9= .
3
1
2
9
9
2
3 9
log
3
9
1log
2
2log
log log
p
p
y p
y
y
y p
p y
y y
=
=
=
=
=
=
The equation then becomes:
2
9 9 9
2
9 9
2
2
log log log 9
log log 9
9
1
9
3
x y
x y
x y
y x
xy
= +
=
=
=
=
Note that logarithm functions have domain of the positive real numbers only.
MHR • Calculus and Vectors 12 Solutions 899
Chapter 8 Section 4 Intersections of Lines in Two-Space and Three-Space Chapter 8 Section 4 Question 1 Page 471 a) One, as the two lines have different direction vectors (i.e. different slopes).
b) One, as the two lines have different direction vectors (i.e. different slopes).
c) Zero solutions, as the two lines are parallel and distinct.
d) One, as the two lines have different direction vectors (i.e. different slopes).
Chapter 8 Section 4 Question 2 Page 471 a) Use elimination.
x + y = 8 !
x ! y = 13 "
2y = !5 !!"
y = !5
2
Substitute 5
2y = ! into equation .
x !5
2= 8
x =21
2
The point of intersection is 21 5
,2 2
! "#$ %& '
.
b) Use elimination.
3.5x ! 2.1y = 14 !
1.5x ! 0.3y = 8 "
! 7x = !42 !!7"
x = 6
Substitute 6x = into equation .
3.5(6)! 2.1y = 14
2.1y = !14 + 21
y =7
2.1
y =70
21 or
10
3
The point of intersection is
6, 10
3
!
"#$
%&.
MHR • Calculus and Vectors 12 Solutions 900
c) Use elimination.
!12 + 8s = 2 + 3t !
! 7 ! 5s = !1! 2t "
16s ! 6t = 28 2!
!15s + 6t = 18 3"
s = 46 2!+3"
!14 + 8(46) = 3t Substitute in !.
t = 118
Substitute 46 and 118s t= = in one of the original vector equations.
The point of intersection is (356, –237).
d) Use elimination.
16 + 5s = !7 ! 7t !
1+ s = 12 + 3t "
5s + 7t = !23 !
5s !15t = 55 5"
22t = !78 !!5"
t = !39
11
s = 11+ 3 !39
11
"
#$%
&'Substitute in ".
s =4
11
Substitute 4 39
and 11 11
s t= = ! in one of the original vector equations.
The point of intersection is
196
11,
15
11
!
"#$
%&.
Chapter 8 Section 4 Question 3 Page 471
a) The lines are not parallel since the slopes are 1
and 24
! ! .
Therefore the lines will intersect and there will be one solution.
b) The lines are parallel since the slopes are 12 8
and 21 14
, which are equal. Multiplying the second
equation by 3
2 gives the first equation except for the constant terms.
Therefore the lines are distinct and there will be no solutions.
MHR • Calculus and Vectors 12 Solutions 901
c) The lines are parallel since the direction vectors are scalar multiple ( [ ] [ ]2 3, 2 6, 4! ! = ! ). Letting
1s = in the first equation produces the point (4, 4).
Therefore the lines are coincident and there are an infinite number of solutions.
d) The lines are not parallel since the slopes are3 8
and 8 3
! .
Therefore the lines will intersect and there will be one solution.
e) Multiplying the first equation by –5 and the second equation by 3 produces the same result.
Therefore the lines are coincident and there are an infinite number of solutions.
f) The lines are not parallel since the direction vectors [2, 7] and [1, 4] are not scalar multiples of each
other.
Therefore the lines will intersect and there will be one solution.
Chapter 8 Section 4 Question 4 Page 471 a) Substitute the coordinates of [ ]5, 1, 3 into the second vector equation.
5, 1, 3!" #$ = 2, 3, 9!" #$ + t 2, 1, 7!" #$
Equate the x-coordinates. Equate the y-coordinates. Equate the z-coordinates.
5 2 2
3
2
t
t
= +
=
1 3
2
t
t
= +
= !
3 9 7
6
7
t
t
= +
= !
Since the t values are not equal, the point (5, 1, 3) does not lie on the second line and the lines are
distinct.
b) Substitute the coordinates of [ ]4, 1, 0 into the second vector equation.
Chapter 8 Section 4 Question 18 Page 473 a) Solve by elimination.
ax + by = c !
dx + ey = f "
aex + eby = ec e!
bdx + eby = bf b"
ae! bd( ) x = ec ! bf e!!b"
x =ec ! bf
ae! bd
Substitute for x in .
aec ! bf
ae! bd
"
#$%
&'+ by = c
by = cae! bd
ae! bd!
a ec ! bf( )
ae! bd
y =1
b
aec ! bcd ! aec + abf
ae! bd
y =
b af ! cd( )
b ae! bd( )
y =af ! cd
ae! bd
Note that the algebra assumes that no denominators are equal to zero.
Therefore, ( ), ,ec bf af cd
x yae bd ae bd
! !" #= $ %! !& '
.
MHR • Calculus and Vectors 12 Solutions 917
b) Equate the expressions for like coordinates for the two lines.
a + cs = e + gt
cs ! gt = e! a !
b + ds = f + ht
ds ! ht = f ! b !
Solve equations and for s and t.
chs ! ght = eh ! ah h!
dgs ! ght = fg ! bg g"
ch ! dg( )s = eh ! ah ! fg + bg h!!g"
s =eh ! ah ! fg + bg
ch ! dg
Substituting in .
ceh ! ah ! fg + bg
ch ! dg
"
#$%
&'! gt = e! a
gt = ceh ! ah ! fg + bg
ch ! dg
"
#$%
&'! e! a( )
ch ! dg
ch ! dg
gt =ceh ! ach ! cfg + bcg ! ceh + deg + ach ! adg
ch ! dg
t =!cf + bc + de! ad
ch ! dg
Use the value for s to find the intersection points.
x = a +eh ! ah ! fg + bg
ch ! dg
"
#$%
&'c
=
a ch ! dg( )
ch ! dg+
c eh ! ah ! fg + bg( )
ch ! dg
=ach ! adg + ceh ! ach ! cfg + bcg
ch ! dg
=!adg + ceh ! cfg + bcg
ch ! dg
y = b +eh ! ah ! fg + bg
ch ! dg
"
#$%
&'d
=
b ch ! dg( )
ch ! dg+
d eh ! ah ! fg + bg( )
ch ! dg
=bch ! bdg + deh ! adh ! dfg + bdg
ch ! dg
=bch + deh ! adh ! dfg
ch ! dg
Therefore, ( ), ,adg ceh cfg bcg bch deh adh dfg
x ych dg ch dg
! "# + # + + # #= $ %# #& '
.
MHR • Calculus and Vectors 12 Solutions 918
Chapter 8 Section 4 Question 19 Page 473 Examine the diagram carefully. There are 5 small right-angled triangles created by the overlap of the
rectangles. They are all similar.
Use the Pythagorean theorem in the lower left triangle to obtain the length 8.
The triangle in the upper left is similar and so must have sides 3 and 5.
Label the unknown part of the length of the second rectangle as x. The missing length of the part of the
top of the first rectangle is 11. The two triangles, upper and upper left, are similar.
Therefore,
x
11=
3
5
x =33
5
x = 6.6
5 + 6.6 = 11.6
The length of the second rectangle is 11.6 cm.
MHR • Calculus and Vectors 12 Solutions 919
Chapter 8 Section 4 Question 20 Page 473
If the angle between two planes is θ, then
cos ! =n!
1 "n!
2
n!
1 n!
2
.
Therefore:
cos 60o
=
1, 1, 2!" #$ % 2, &1, k!" #$
1, 1, 2!" #$ 2, &1, k!" #$
1
2=
2 &1+ 2k
6 5+ k2
2 + 4k = 30 + 6k2
4 +16k +16k2
= 30 + 6k2
10k2
+16k & 26 = 0
5k2
+ 8k &13 = 0
(5k +13)(k &1) = 0
So, k = 1 or k =
!13
5 or –2.6.
MHR • Calculus and Vectors 12 Solutions 920
Chapter 8 Section 5 Intersections of Lines and Planes Chapter 8 Section 5 Question 1 Page 479 Substitute the parametric equations into the scalar equation of the plane and solve for t.
(4 + t) + 5(2 ! 2t) + (6 + 3t)! 8 = 0
4 + t +10 !10t + 6 + 3t ! 8 = 0
!6t = !12
t = 2
Substitute this value of t in the parametric equations.
x = 4 + 2
= 6
y = 2 ! 2(2)
= !2
z = 6 + 3(2)
= 12
The point of intersection of the line and the plane is (6, –2, 12).
Chapter 8 Section 5 Question 2 Page 479 The line and the plane are parallel if 0n m! =
! "!.
a)
n!
!m"!
= 3, 5, 1"# $% ! 2, &1, &1"# $%
= 3(2) + 5(–1) +1(–1)
= 0
So, the line and the plane are parallel.
Check coincidence by checking if the point (1, 2, –8) on the line satisfies the plane equation.
L.S. = 3(1) + 5(2) + (–8)! 5
= 0
R.S. = 0
L.S. = R.S.
Therefore, the line and the plane are coincident.
b)
n!
!m"!
= 4, "1, 6#$ %& ! 7, "14, " 7#$ %&
= 4(7) + (–1)(–14) + 6(–7)
= 0
So, the line and the plane are parallel.
Check coincidence by checking if the point (4, 3, 10) on the line satisfies the plane equation.
L.S. = 4(4)! (3) + 6(10)!12
= 61
R.S. = 0
L.S. ≠ R.S.
Therefore, the line and the plane are parallel and distinct.
MHR • Calculus and Vectors 12 Solutions 921
c)
n!
!m"!
= 0, 3, 10"# $% ! 2, &10, 3"# $%
= 0(2) + 3(–10) +10(3)
= 0
So, the line and the plane are parallel.
Check coincidence by checking if the point (7, 1, –9) on the line satisfies the plane equation.
L.S. = 3(1) +10(–9) +1
= !86
R.S. = 0
L.S. ≠ R.S.
Therefore, the line and the plane are parallel and distinct.
d)
n!
!m"!
= 1, 2, " 5#$ %& ! 1, 2, 1#$ %&
= 1(1) + 2(2) + (–5)(1)
= 0
So, the line and the plane are parallel.
Check coincidence by checking if the point (10, 3, 4) on the line satisfies the plane equation.
L.S. = 10 + 2(3)! 5(4) + 4
= 0
R.S. = 0
L.S. = R.S.
Therefore the line and the plane are coincident.
Chapter 8 Section 5 Question 3 Page 479 a) Substitute the parametric equations into the scalar equation of the plane and solve for t.
3(3+ 7t)! (–11t) + 4(5 – 8t)! 8 = 0
9 + 21t +11t + 20 ! 32t ! 8 = 0
0t = !21
There are no values of t that make this equation true. The line and the plane do not intersect.
b) Substitute the parametric equations into the scalar equation of the plane and solve for t.
!2(5+ t) + 6(–1– 2t) + 4(4 + 3t)! 4 = 0
!10 ! 2t ! 6 !12t +16 +12t ! 4 = 0
!2t = 4
t = !2
Substitute this value of t in the parametric equations.
x = 5+ (–2)
= 3
y = !1! 2(–2)
= 3
z = 4 + 3(–2)
= !2
The point of intersection of the line and the plane is (3, 3, –2).
MHR • Calculus and Vectors 12 Solutions 922
c) Substitute the parametric equations into the scalar equation of the plane and solve for t.
5(4 + t) + 3(1+ 2t) + 4(5+ 3t)! 20 = 0
20 + 5t + 3+ 6t + 20 +12t ! 20 = 0
23t = !23
t = !1
Substitute this value of t in the parametric equations.
x = 4 + (–1)
= 3
y = 1+ 2(–1)
= !1
z = 5+ 3(–1)
= 2
The point of intersection of the line and the plane is (3, –1, 2).
d) Substitute the parametric equations into the scalar equation of the plane and solve for t.
5(10 + 2t)! 3(–5+ t) + 7(–2t) + 7 = 0
50 +10t +15! 3t !14t + 7 = 0
!7t = !72
t =72
7
Substitute this value of t in the parametric equations.
x = 10 + 272
7
!
"#$
%&
=214
7
y = !5+72
7
"
#$%
&'
=37
7
z = !272
7
"
#$%
&'
= !144
7
The point of intersection of the line and the plane is 214 37 144
, ,7 7 7
! "#$ %& '
.
e) Substitute the parametric equations into the scalar equation of the plane and solve for t.
9(4 + 2t)! 6(t) +12(–1– t)! 24 = 0
36 +18t ! 6t !12 !12t ! 24 = 0
0t = 0
This equation is true for all values of t. Any point on the line is a solution.
There are an infinite number of solutions.
MHR • Calculus and Vectors 12 Solutions 923
f) Substitute the parametric equations into the scalar equation of the plane and solve for t.
6(4 + 2t)! 2(12 ! 3t) + 3(–19 + 5t) + 6 = 0
24 +12t ! 24 + 6t ! 57 +15t + 6 = 0
33t = 51
t =17
11
Substitute this value of t in the parametric equations.
x = 4 + 217
11
!
"#$
%&
=78
11
y = 12 ! 317
11
"
#$%
&'
=81
11
z = !19 + 517
11
"
#$%
&'
= !124
11
The point of intersection of the line and the plane is 78 81 124
Since the triple scalar product is always zero, the normals for planes intersecting in a line will always be
coplanar.
Chapter 8 Section 6 Question 11 Page 492 a) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 3, 2, 1 2, 1, 2 2, 3, 4
3, 2, 1 2, 12, 8
22
n n n! " = # ! # " #
= # ! # # #
= #
! ! !
The normals are not coplanar; there is an intersection point.
Eliminate one of the variables from two pair of equations. Eliminate z.
6x + 4y ! 2z = !4 2!
2x + y ! 2z = 7 "
4x + 3y = !11 # 2!!"
12x + 8y ! 4z = !8 4!
2x ! 3y + 4z = !3 $
14x + 5y = !11 % 4!+$
Solve equations and for x and y.
20x +15y = !55 5!
42x +15y = !33 3"
!22x = !22 5!!3"
x = 1
Substitute x = 1 in ! .
14(1) + 5y = !11 !
y = !5
Substitute x = 1 and y = !5 in !.
3(1) + 2(–5)! z = !2
z = !5
The three planes intersect at the point (1, –5, –5).
MHR • Calculus and Vectors 12 Solutions 958
b) n!
1 = 3, ! 4, 2"# $% ; n!
2 = 6, ! 8, 4"# $% ; n!
3 = 15, ! 20, 10"# $%
10n!
1 = 5n!
2
= 2n!
3
So, the normals are parallel.
Examine the equations to determine if the planes are coincident or parallel and distinct.
30x ! 40y + 20z = 10 10!
30x ! 40y + 20z = 50 5"
30x ! 40y + 20z = !6 2#
The equations are not scalar multiples of each other.
The planes are parallel but distinct.
c)
n!
1 = 2, 1, 6!" #$ ; n!
2 = 5, 1, % 3!" #$ ; n!
3 = 3, 2, 15!" #$ By inspection, none of the normals are parallel.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 2, 1, 6 5, 1, 3 3, 2, 15
2, 1, 6 21, 84, 7
0
n n n! " = ! # "
= ! #
=
! ! !
Since the triple scalar product is zero, the normals are coplanar and the planes intersect either in a line
or not at all.
Solve the system algebraically to determine if there is a solution.
2x + y + 6z = 5 !
5x + y ! 3z = 1 "
3x + 2y +15z = 9 #
!3x + 9z = 4 $ !!"
7x ! 21z = !7 % 2"!#
!21x + 63z = 28 & 7$
21x ! 63z = !21 ' 3%
0 = 7 &+'
The system has no solution. The planes intersect in pairs only.
MHR • Calculus and Vectors 12 Solutions 959
d) n!
1 = 1, !1, 1"# $% ; n!
2 = 1, 1, 3"# $% ; n!
3 = 2, ! 5, !1"# $% By inspection, none of the normals are parallel.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 1, 1, 1 1, 1, 3 2, 5, 1
1, 1, 1 14, 7, 7
0
n n n! " = # ! " # #
= # ! #
=
! ! !
Since the triple scalar product is zero, the normals are coplanar and the planes intersect either in a line
or not at all.
Solve the system algebraically to determine if there is a solution.
x ! y + z = 20 !
x + y + 3z = !4 "
2x ! 5y ! z = !6 #
!2y ! 2z = 24 $ !!"
7 y + 7z = !2 % 2"!#
!14x !14z = 168 & 7$
14x +14z = !4 ' 2%
0 = 164 &+'
The system has no solution. The planes intersect in pairs only.
e)
n!
1 = 4, 8, 4!" #$ ; n!
2 = 5, 10, 5!" #$ ; n!
3 = 3, %1, % 4!" #$
Two of the normals are parallel since 1 25 4n n=
! !
However, 5! ! 4" . The equations are not scalar multiples of each other.
The first planes are parallel but distinct and the third plane intersects both of them.
There are no solutions to the system.
f) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 2, 2, 1 5, 4, 4 3, 5, 2
2, 2, 1 12, 2, 13
33
n n n! " = ! # " #
= ! #
=
! ! !
The normals are not coplanar; there is an intersection point.
MHR • Calculus and Vectors 12 Solutions 960
Eliminate one of the variables from two pair of equations. Eliminate z.
8x + 8y + 4z = 40 4!
5x + 4y ! 4z = 13 "
13x +12y = 53 # 4!+"
4x + 4y + 2z = 20 2!
3x + 5y ! 2z = 6 $
7x + 9y = 26 % 2!+$
Solve equations and for x and y.
39x + 36y = 159 3!
28x + 36y = 104 4"
11x = 55 3!!4"
x = 5
Substitute x = 5 in ! .
7(5) + 9y = 26 !
y = !1
Substitute x = 5 and y = !1 in !.
2(5) + 2(–1) + z = 10
z = 2
The three planes intersect at the point (5, –1, 2).
g) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 2, 5, 0 0, 4, 1 7, 0, 4
2, 5, 0 16, 7, 28
3
n n n! " = ! " #
= ! # #
=
! ! !
The normals are not coplanar; there is an intersection point.
Eliminate one of the variables from two pair of equations. Eliminate z.
16y + 4z = 4 4!
7x ! 4z = 1 "
7x +16y = 5 # 4!+"
MHR • Calculus and Vectors 12 Solutions 961
Solve equations and for x and y.
14x + 35y = 7 7!
14x + 32y = 10 2"
3y = !3 7!!2"
y = !1
Substitute y = –1 in ! .
2x + 5(!1) = 1 !
x = 3
Substitute y = !1 in !.
4(–1) + z = 1
z = 5
The three planes intersect at the point (3, –1, 5).
h) These planes have an intersection point at the origin since the constant term for each equation is zero.
Check to see if this point is unique or if the planes intersect in a line.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 3, 2, 1 1, 0, 1 3, 2, 5
3, 2, 1 2, 2, 2
0
n n n! " = # # ! # " #
= # # !
=
! ! !
The normals are coplanar and the planes intersect in a line.
Eliminate one of the variables from two pair of equations. Eliminate y.
3x ! 2y ! z = 0 !
3x + 2y ! 5z = 0 "
6x + ! 6z = 0 !+"
x ! z = 0 #
Equations and are identical.
Let z = t.
x ! t = 0!
x = t
Substitute and x t z t= = in
3t ! 2y ! t = 0
y = t
The planes intersect in the line
x, y, z!" #$ = 0, 0, 0!" #$ + t 1, 1, 1!" #$ , t %! .
MHR • Calculus and Vectors 12 Solutions 962
i) First examine the normals. 2 33 4n n=
! !. These planes are parallel.
However, 3! = 4" and therefore these planes are coincident.
Since the first plane is not parallel, the three planes will intersect in a line.
Since x does not appear in any equation, let x t= . Solve equations and for y and z.
4y + 8z = 44 4!
4y ! 4z = !16 "
12z = 60 4!!"
z = 5
Substitute z = 5 in ! .
y + 2(5) = 11 !
y = 1
The planes intersect in the line
x, y, z!" #$ = 0, 1, 5!" #$ + t 1, 0, 0!" #$ , t %! .
Chapter 8 Section 6 Question 12 Page 492 a) The planes intersect at the point (5, 10, –3).
b) The three planes intersect in a line. Let z = t. The equation of the line is
x, y, z!" #$ = %4, 8, 0!" #$ + t %1, 3, 1!" #$ , t &! .
c) There is no point on all three planes. The three planes intersect in pairs. Chapter 8 Section 6 Question 13 Page 493 a) The planes can intersect at a point or a line.
b) The planes can intersect at one line (if the two parallel planes are coincident) or in pairs (if the two
parallel planes are distinct). There are no intersection points for the system in the latter case.
c) The planes will either not intersect (if at least two of the planes are distinct) or have an infinite number
of solutions (if they are all coincident).
Chapter 8 Section 6 Question 14 Page 493 a) Check the dot product of pairs of normals.
[ ] [ ]
[ ] [ ]
[ ] [ ]
1 2
1 3
2 3
5, 1, 4 1, 3, 2 0
5, 1, 4 1, 1, 1 0
1, 3, 2 1, 1, 1 0
n n
n n
n n
! = ! " =
! = ! " " =
! = " ! " " =
! !
! !
! !
The planes are mutually perpendicular and must intersect in a single point.
Using algebra, the solution is (1, –3, 4).
MHR • Calculus and Vectors 12 Solutions 963
b) Check the dot product of pairs of normals.
[ ] [ ]
[ ] [ ]
[ ] [ ]
1 2
1 3
2 3
2, 6, 1 7, 3, 4 0
2, 6, 1 27, 1, 48 0
7, 3, 4 27, 1, 48 0
n n
n n
n n
! = " ! =
! = " ! " =
! = ! " =
! !
! !
! !
The planes are mutually perpendicular and must intersect in a single point.
Using algebra, the solution is (3, 4, 21).
Chapter 8 Section 6 Question 15 Page 493 Solutions for Achievement Checks are shown in the Teacher Resource.
Chapter 8 Section 6 Question 16 Page 493 a) All three equations have no y-coordinate. Any solution will have y = t and x and z values determined
by the three equations.
Solve and for x and z.
2x ! 6z = !6 2!
2x ! z = 4 "
! 5z = !10 2!!"
z = 2
Substitute z = 2 in ! .
2x ! 2 = 4 !
x = 3
For consistency, x = 3 and z = 2 must be a solution to .
L.S. = 3x + 5z
= 3(3) + 5(2)
= 19
For a consistent, dependent system, change the constant term in to 19.
b) Find the intersection of the first two planes. This will be a line since their normals are not scalar
multiples.
Solve using algebra.
Let z = t.
y + 5t = 20 !
y = 20 ! 5t
Substitute y = 20 – 5t in ! .
2x + 8(20 ! 5t) = !6 !
x = !83+ 20t
MHR • Calculus and Vectors 12 Solutions 964
The line of intersection is
x, y, z!" #$ = %83, 20, 0!" #$ + t 20, % 5, 1!" #$ , t &! .
If the system is to be consistent, the parametric equations for the line must satisfy the third equation.
L.S. = 3(!83+ 20t) +12(20 ! 5t) + 6t
= 6t ! 9
R.S. = –9
Clearly the z-term in the third equation must be 0z instead of 6z for the system to be consistent and
dependent.
Chapter 8 Section 6 Question 17 Page 493 Answers are not unique for any parts of this question.
Sketches are included to aid visualization. They are not drawn to correct scale or orientation.
a) Choose normals that are identical but D terms that are not.
A possible set of planes is x + y + z = 1, x + y + z = 4, and x + y + z = –11.
b) A simple set of three planes is 3, 1, and 9x y z= = = ! .
c) Choose three coplanar normals; 3 1 2n n n= +
! ! !will work. Then choose
3 1 2D D D= + for consistency of
the system. A possible set of planes is 1, 2 3 5, and 2 3 4 6x y z x y z x y z+ + = + + = + + = .
d) Choose three coplanar normals; 3 1 2n n n= +
! ! !will work. Then choose
3 1 2D D D! + for inconsistency of
the system. A possible set of planes is 1, 2 3 5, and 2 3 4 13x y z x y z x y z+ + = + + = + + = .
MHR • Calculus and Vectors 12 Solutions 965
e) The normal vectors for the planes must be perpendicular to the direction vector of the line (dot product
is zero).
Choose D terms for the planes so that the point (1, 3, –4) satisfies the equations.
A possible set of planes is 4 11, 9 31, and 9 4 25x y y z x z! = ! ! = ! = .
f) Choose three equations with no y-term. Choose planes with non-parallel normals to assure
intersections.
A possible set of planes is 3 4, 2, and 9 4 0x z x z x z+ = + = ! = .
g) The normal vectors for the planes must be perpendicular to each other (dot product is zero). Choose D
terms for the planes so that the point (–2, 4, –4) satisfies the equations.
A possible set of planes: 2, 4, and 4x y z= ! = = ! .
h) Choose planes with normals perpendicular to the z-axis, i.e., with no –z-term. Also points such as
(0, 0, 0) must satisfy the equations.
A possible set of planes is 0, 0, and 3 2 0x y x y= = ! = .
MHR • Calculus and Vectors 12 Solutions 966
Chapter 8 Section 6 Question 18 Page 493 The first three planes are parallel to the y-axis, forming a triangular prism.
The last two planes indicate that the prism is a right triangular prism and that the height is 14 units.
Find the intersection points of the planes with the base of the prism ( 4y = ! ).
x + z = !3 !
10x ! 3z = 22 "
13z = !52 10!!"
z = !4
x = 1
x + z = !3 !
4x ! 9z = !38 "
13z = 26 4!!#
z = 2
x = !5
10x ! 3z = 22 !
4x ! 9z = !38 "
26x = 104 3!!"
x = 4
z = 6
The points on the base are (1, –4, –4), (–5, –4, 2) and (4, –4, 6).
The lengths of the sides are:
((–5) –1)2
+ 02
+ (2 – (–4))2
= 72
! 8.49
(4 !1)2
+ 02
+ (6 – (–4))2
= 109
! 10.44
(4 – (–5))2
+ 02
+ (6 – 2)2
= 97
! 9.85
Find the area of the triangle using a convenient method (trigonometry, analytic geometry formula given
The third row indicates that there is no solution to the system.
The normals for the first and third planes are scalar multiples of each other; these planes are distinct.
b) 1 8 5 1
3 2 1 6
5 1 4 10
! "# $%# $# $%& '
1
2
3
R
R
R
1 8 5 1
0 26 14 3
0 39 21 15
! "# $
%# $# $& '
1
1 2
1 3
R
3R R
5R R
!
!
26 0 18 50
0 26 14 3
0 0 0 237
! "# $
%# $# $%& '
1 2
2
2 3
26R 8R
R
39R 8R
!
!
The third row indicates that there is no solution to the system.
The normals for the planes are not scalar multiples of each other; these planes only intersect in pairs as
in a triangular prism.
c) 1 6 2 3
1 3 1 1
2 12 4 6
!" #$ %!$ %$ %!& '
1
2
3
R
R
R
1 6 2 3
0 3 1 2
0 0 0 0
! "#$ %
#$ %$ %& '
1
1 2
1 3
R
R R
2R R
!
!
1 0 0 1
0 3 1 2
0 0 0 0
!" #$ %
!$ %$ %& '
1 2
2
3
R 2R
R
R
!
The third row is 0 = 0. This indicates the solution is consistent and dependent.
MHR • Calculus and Vectors 12 Solutions 974
There are ‘leading ones’ in columns 1 and 2. Let z = t.
From row 1, x = –1.
From row 2,
!3y + t = 2
y = !2
3+
1
3t
The three planes intersect in a line that can be defined as
x, y, z!" #$ = %1, %2
3, 0
!
"&
#
$' + t 0,
1
3, 1
!
"&
#
$' , t (! .
d) 4 5 2 7
3 2 1 6
1 0 1 4
!" #$ %!$ %$ %!& '
1
2
3
R
R
R
4 5 2 7
0 7 2 3
0 5 6 9
! "#$ %
# #$ %$ %# #& '
1
1 2
1 3
R
3R 4R
R 4R
!
!
28 0 4 64
0 7 2 3
0 0 32 48
! "# $
% %# $# $%& '
1 2
2
2 3
7R 5R
R
5R 7R
!
!
224 0 0 560
0 112 0 0
0 0 32 48
! "# $%# $# $%& '
1 3
2 3
3
8R +R
16R R
R
+
1 0 0 2.5
0 1 0 0
0 0 1 1.5
! "# $# $# $%& '
R1÷ 224
R2
÷ !112
R3÷ !32
The planes intersect at the point (2.5, 0, –1.5).
e) 2 1 3 10
10 5 15 3
6 3 9 4
!" #$ %! !$ %$ %!& '
1
2
3
R
R
R
2 1 3 10
0 0 0 53
! "#$ %$ %$ %& '
1
1 2
R
5R R!
Stop the row reduction process at this point. Row 2 shows there is no solution.
MHR • Calculus and Vectors 12 Solutions 975
Examination of the normals to the planes indicates that all three planes are parallel to each other.
(Normals are all scalar multiples of each other and the planes are distinct.)
f) 3 2 2 15
1 5 6 46
4 2 1 9
! "# $%# $# $& '
1
2
3
R
R
R
3 2 2 15
0 13 20 123
0 2 5 33
! "# $
% %# $# $& '
1
1 2
1 3
R
R 3R
4R 3R
!
!
39 0 66 51
0 13 20 123
0 0 105 183
!" #$ %
! !$ %$ %& '
1 2
2
2 3
13R +2R
R
2R +13R
4095 0 0 17433
0 1365 0 16575
0 0 105 183
!" #$ %! !$ %$ %& '
1 3
2 3
3
105R 66R
105R 20R
R
!
!
105 0 0 447
0 7 0 85
0 0 35 61
!" #$ %$ %$ %& '
R1÷ 39
R2
÷195
R3÷ 3
The planes intersect at the point 447 85 61
, ,105 7 35
! "#$ %& '
.
Chapter 8 Extension Question 6 Page 501 a) The three planes intersect at the point (–7, 4, –10).
b) The three planes do not intersect.
c) The planes intersect in a line that can be defined by
x, y, z!" #$ = 8, 1, 0!" #$ + t %2, % 7, 1!" #$ , t &!
d) The three planes do not intersect.
Chapter 8 Extension Question 7 Page 501 Answers may vary.
MHR • Calculus and Vectors 12 Solutions 976
Chapter 8 Extension Question 8 Page 501 This question is most efficiently solved using CAS.
The solution point is (63, 101, –146, –14).
Use matrices and row reduction,
1 3 2 3 10
4 2 3 1 2
2 1 1 5 11
2 4 1 10 8
! !" #$ %$ %$ %$ %! !& '
1
2
3
4
R
R
R
R
1 3 2 3 10
0 14 11 11 38
0 7 5 1 9
0 2 3 16 12
! !" #$ %! !$ %$ %! !$ %
! !& '
1
1 2
1 3
1 4
R
4R R
2R R
2R +R
!
!
14 0 5 9 26
0 14 11 11 38
0 0 1 9 20
0 0 10 101 46
! "# $% %# $# $%# $
% %& '
1 2
2
2 3
2 4
14R 3R
R
R 2R
R 7R
!
!
!
14 0 0 54 126
0 14 0 88 182
0 0 1 9 20
0 0 0 11 154
! "# $% % %# $# $%# $
%& '
1 3
2 3
3
3 4
R 5R
R 11R
R
10R R
+
!
+
154 0 0 0 9702
0 14 0 0 1414
0 0 11 0 1166
0 0 0 11 154
! "# $% %# $# $% %# $
%& '
1 4
2 4
3 4
4
11R 54R
R 8R
11R 9R
R
+
!
+
MHR • Calculus and Vectors 12 Solutions 977
1 0 0 0 63
0 1 0 0 101
0 0 1 0 !146
0 0 0 1 !14
"
#
$$$$
%
&
''''
R1÷154
R2
÷ !14
R3÷ !11
R4
÷ !11
The solution point is (63, 101, –146, –14) or w = 63, x = 101, y = !146, and z = !14 .
Chapter 8 Extension Question 9 Page 501 Take the 7 L jug and fill it up with water. Using this jug, fill the 3 L jug. Take the 4 L that now remain in
the 7 L jug and pour into a pail. Refill the 7 L jug and once again pour this into the pail. The 7 L being
added to the 4 L that was there before will make 11 L of water in the pail.
Chapter 8 Extension Question 10 Page 501 Fill the 3 oz glass using the 8 oz glass. Take the 3 oz glass and pour it into the 5 oz glass. Refill the 3 oz
glass using the 8 oz glass that now has 5 oz remaining. At this point the 8 oz glass will now only have 2
oz in it. Pour the 3 oz glass into the 5 oz glass that currently has 3 oz in it, until it is full. At this point, the
3 oz glass has 1 oz in it, so pour this one oz out. Take the 5 oz glass that is now full and fill up the 3 oz
glass. This will leave 2 oz in the 5 oz glass that can now be poured into the 8 oz glass that has 2 oz in it to
make a total of 4 oz.
MHR • Calculus and Vectors 12 Solutions 978
Chapter 8 Review Chapter 8 Review Question 1 Page 502 t !! for all equations.
a) Vector:
x, y!" #$ = %3, 2!" #$ + t 1, 2!" #$
Parametric equations:
x = !3+ t
y = 2 + 2t
b) Vector:
x, y, z!" #$ = %9, 0, 4!" #$ + t 6, 5, 1!" #$ Parametric equations:
x = !9 + 6t
y = 5t
z = 4 + t
c) Vector:
x, y, z!" #$ = 0, 0, 7!" #$ + t 1, 0, 0!" #$
Parametric equations:
x = 0
y = t
z = 7 + t
d) Lines perpendicular to the xy-plane are parallel to the z-axis and the vector [ ]0, 0, 1k =
!.
Vector:
x, y, z!" #$ = 3, 0, % 4!" #$ + t 0, 0, 1!" #$ Parametric equations:
x = 3
y = 0
z = !4 + t
Chapter 8 Review Question 2 Page 502 a) Find two points on the line. If x = 3, y = 3; if x = 1, y = –2.
The vector equation is
x, y!" #$ = 3, 3!" #$ + t 2, 5!" #$ , t %! .
b) Find two points on the line. If x = 3, y = 1; if x = 10, y = 0.
The vector equation is
x, y!" #$ = 3, 1!" #$ + t %7, 1!" #$ , t &! .
c) Find two points on the line. x = 8, y = 2; x = 8, y = 3.
The vector equation is
x, y!" #$ = 8, 2!" #$ + t 0, 1!" #$ , t %! .
d) Find two points on the line. If x = 4, y = 1; if x = 0, y = 0.
The vector equation is
x, y!" #$ = 4, 1!" #$ + t 4, 1!" #$ , t %! .
MHR • Calculus and Vectors 12 Solutions 979
Chapter 8 Review Question 3 Page 502 a) The normal and the direction vector are perpendicular. Therefore, [ ]7, 2n = !
!.
The equation is of the form 7 2 0x y C! + = . Substituting (1, 4) into the equation gives C = 1. The scalar equation is 7x – 2y + 1 = 0.
b) The normal and the direction vector are perpendicular. Therefore, [ ]7, 5n =
!.
The equation is of the form 7 5 0x y C+ + = . Substituting (10, –3) into the equation gives C = –55. The scalar equation is 7x + 5y – 55 = 0.
Chapter 8 Review Question 4 Page 502 a) The parametric equations are:
x = 1+ 3t
y = !1+ 4t
z = 5+ 7t, t "!
b) Substitute the coordinates into the parametric equations.
13 1 3
4
t
t
= +
=
15 1 4
4
t
t
= ! +
=
23 5 7
18
7
t
t
= +
=
Since the t-values are not identical, the point does not lie on the line.
Chapter 8 Review Question 5 Page 502 The direction vectors for the sides are:
AB
! "!!
= 4, 2!" #$
= 2, 1!" #$
BC
! "!!
= 4, ! 4"# $%
= 1, !1"# $%
The vector equations are:
AB:
x, y!" #$ = %1, 4!" #$ + t 2, 1!" #$ , t &!
BC:
x, y!" #$ = 3, 6!" #$ + t 1, %1!" #$ , t &!
CD:
x, y!" #$ = 7, 2!" #$ + t 2, 1!" #$ , t %!
DA:
x, y!" #$ = %1, 4!" #$ + t 1, %1!" #$ , t &!
MHR • Calculus and Vectors 12 Solutions 980
Chapter 8 Review Question 6 Page 502 For the x-intercept of the first line, let y = 0 and z = 0.
0 8 4
2
t
t
= +
= !
0 14 7
2
t
t
= +
= !
When t = –2:
x = !21!12(!2)
= 3
The x-intercept of the first line is 3.
For the y-intercept of the second line, let x = 0 and z = 0.
0 = 6 + 2s
s = !3
0 = 12 + 4s
s = !3
When s = –3:
y = !8! 5(!3)
= 7
The y-intercept of the second line is 7.
Since (3, 0, 0) and (0, 7, 0) are points on the required line, a direction vector is [3, –7, 0].
Parametric equations for the line are:
x = 3+ 3t
y = !7t
z = 0, t "!
Chapter 8 Review Question 7 Page 502 a) When s = 0 and t = 0, (3, 4, –1) is a point on the plane.
When s = 1 and t = 0, (4, 5, –5) is a point on the plane.
When s = 1 and t = 0, (2, 3, 3) is a point on the plane.
b) When x = 0 and y = 0, (0, 0, 12) is a point on the plane.
When y = 0 and z = 0, (–12, 0, 0) is a point on the plane.
When x = 0 and z = 0, (0, –6, 0) is a point on the plane.
c) When k = 1 and p = 1, (7, –6, 3) is a point on the plane.
When k = 0 and p = 0, (0, –5, 2) is a point on the plane.
When k = –1 and p = –1, (–7, –4, 1) is a point on the plane.
MHR • Calculus and Vectors 12 Solutions 981
Chapter 8 Review Question 8 Page 502 A direction vector for the plane is [ ] [ ] [ ]5, 1, 3 2, 9, 10 3, 10, 7! ! = ! .
A vector equation for the plane is
x, y, z!" #$ = 2, % 9, 10!" #$ + s 3, % 8, 7!" #$ + t 3, 10, % 7!" #$ , s, t &! .
Parametric equations for the plane are:
x = 2 + 3s + 3t
y = !9 ! 8s +10t
z = 10 + 7s ! 7t, s, t "!
Chapter 8 Review Question 9 Page 502 a) If P(–3, 4, –5) lies on the plane, there exists a single set of s- and t-values that satisfy the equation.
!3 = 1+ 2s + t !
4 = !5+ s + 7t "
!5 = 6 + 3s + t #
Solve and for s and t.
! 4 = 2s + t !
18 = 2s +14t 2"
!22 = !13s !!2"
s =22
13
Substitute s =
22
13 in ! .
!4 = 222
13
"
#$%
&'+ t !
t = !96
13
Now check if these values satisfy equation .
L.S. = 5
R.S. = 6 + 322
13
!
"#$
%&+ '
96
13
!
"#$
%&
= '48
13
L.S. ≠ R.S.
Therefore, P(–3, 4, –5) does not lie on the plane.
b) Substitute the coordinates in the scalar equation.
L.S. = 4(–3) + 4 ! 2(–5)! 2
= 0
R.S. = 0
L.S. = R.S.
Therefore, the point P(–3, 4, –5) lies on the plane.
MHR • Calculus and Vectors 12 Solutions 982
Chapter 8 Review Question 10 Page 502 Calculate direction vectors.
AB
! "!!
= OB
! "!!
!OA
! "!!
= !1, !1, 10"# $% ! 2, 1, 5"# $%
= !3,! 2, 5"# $%
AC
! "!!
= OC
! "!!
!OA
! "!!
= 8, 5, ! 5"# $% ! 2, 1, 5"# $%
= 6, 4, !10"# $%
AC
! "!!
= 2AB
! "!!
Since the direction vectors are scalar multiples of each other, the three points are collinear and do not
define a unique plane.
Chapter 8 Review Question 11 Page 502 a) The x-, y-, and z- intercepts are 16, –4, and 8. (Points: (16, 0, 0), (0, –4, 0), and (0, 0, 8))
Possible direction vectors are [–16, –4, 0] or [4, 1, 0] and [0, 4, 8] or [0, 1, 2].
b) (16, 0, 0), (0, –4, 0), and (0, 0, 8), so x-intercept = 16, y = intercept = –4, and z-intercept = 8.
c) Vector equation is
x, y, z!" #$ = 16, 0, 0!" #$ + s 4, 1, 0!" #$ + t 0, 1, 2!" #$ , s, t %! .
Parametric equations are:
x = 16 + 4s
y = s + t
z = 2t, s, t !!
Chapter 8 Review Question 12 Page 502 The scalar equation is of the form 2 9 0x y z D+ ! + = ,
Substitute the coordinates of P to determine D.
3+ 2(–4)! 9(0) + D = 0
D = 5
The equation of the plane is x + 2y – 9z + 5 = 0.
Chapter 8 Review Question 13 Page 502 Since and j k
! !are direction vectors of the plane:
n!
= j!
! k!
= i!
= 1, 0, 0"# $%
.
The equation is of the form x + D = 0.
Since (5, 4, –7) is on the plane, D = –5.
The equation of the plane is x – 5 = 0 or x = 5.
MHR • Calculus and Vectors 12 Solutions 983
Chapter 8 Review Question 14 Page 502 a) By observation, equation is x = 4.
b) [3, –7, 1] and [0, 1, 0] are direction vectors for the plane.
n!
= 3, ! 7, 1"# $% & 0, 1, 0"# $%
= !1, 0, 3"# $%
The scalar equation is of the form 3 0x z D! + + = .
The point (1, 2, 4) is on the plane.
!1+ 3(4) + D = 0
D = !11
The equation is 3 11 0x z! + = .
Chapter 8 Review Question 15 Page 502 a) The direction vectors are [5, 2] and [7, 3] which are not parallel.
Substitute the parametric equations in the scalar equation.
( ) ( )2 9 7 5 4 3 6
18 14 20 15 6
4
t t
t t
t
! + ! ! + =
! + + ! =
= !
Substitute 4t = ! in the parametric equations.
The one solution is (–37, –16).
MHR • Calculus and Vectors 12 Solutions 984
b) Equate the corresponding coordinates and solve for s and t.
9 + s = 3t !
4 + s = 9 ! 4t "
5 = !9 + 7t !!"
t = 2
Substitute t = 2 in ! .
9 + s = 6 !
s = !3
Substitute either 3 in given equation with parameter or 2 in given equation with parameter .s s t t= ! =
Then one solution is (6, 1).
Chapter 8 Review Question 16 Page 503 Scalar multiples and sums or differences of these equations will pass through the intersection point.
Therefore, add the equations to get 2 4x y! = . Subtract the equations to get 4 7 32x y! = ! .
These two equations will pass through the intersection point as shown in the graph.
Chapter 8 Review Question 17 Page 503 a) The two direction vectors are equal. The lines are parallel and either coincident or distinct.
Check if (1, 5, –2) is on the second line.
1= !3+ t
t = 4
5 = !23+ 7t
t = 4
!2 = 10 ! 3t
t = 4
Since the t-values are equal, the point lies on the second line and the two lines are coincident.
There are an infinite number of points of intersection.
MHR • Calculus and Vectors 12 Solutions 985
b) The direction vectors are not scalar multiples of each other. The lines must intersect.
To find the intersection point, equate like coordinates.
15+ 4s = 13! 5t !
2 + s = !5+ 2t "
!1! s = !4 + 3t #
Solve and for s and t.
4s + 5t = !2 !
4s ! 8t = !28 4"
13t = 26 !!4"
t = 2
Substitute t = 2 in .
s ! 2(2) = !7 !
s = !3
Check if these values satisfy .
L.S. = !1! (!3)
= 2
R.S. = !4 + 3(2)
= 2
L.S. = R.S.
Substitute 3 or 2 s t= ! = to find the point of intersection.
Chapter 8 Review Question 19 Page 503 a) Substitute the parametric equations into the scalar equation of the plane and solve for t.
5(!17 + 4t)! 2(7 + t) + 4(!6 ! 3t) = 23
!85+ 20t !14 ! 2t ! 24 !12t = 23
6t = 146
t =73
3
Substitute this value of t in the parametric equations.
x = !17 + 473
3
"
#$%
&'
=241
3
y = 7 +73
3
=94
3
z = !6 ! 373
3
"
#$%
&'
= !79
The point of intersection of the line and the plane is 241 94
, , 793 3
! "#$ %& '
.
b) Substitute the parametric equations into the scalar equation of the plane and solve for t.
(!1+ 3t) + 4(!9 + 3t) + 3(16 ! 5t) = 11
!1+ 3t ! 36 +12t + 48!15t = 11
0t = 0
This equation is true for every value of t.
The two lines intersect at every point on the line; the line is on the plane.
Chapter 8 Review Question 20 Page 503
The distance between a point P and a plane is given by
d =
n!
!PQ
" !""
n! where Q is any point on the plane
with normal n!
.
PQ! "!!
= 0, ! 2, 0"# $% ! 3, ! 2, 0"# $%
= !3, 0, 0"# $%
d =
4, !1, 8"# $% & !3, 0, 0"# $%
42
+ (–1)2
+ 82
=12
9
=4
3
! 1.33
The distance is
4
3 or approximately 1.3 units.
MHR • Calculus and Vectors 12 Solutions 987
Chapter 8 Review Question 21 Page 503 Use elimination.
12x + 4y + 4z = 10 4!
5x + 4y ! 2z = 31 "
7x + 6z = !21 # 4!!"
Let z = t.
becomes
x =
!21! 6t
7
becomes:
5!21! 6t
7
"
#$%
&'+ 4y ! 2t = !21
y =22t ! 21
14
The parametric equations for the line are:
x = !3!6
7t
y = !3
2+
11
7t
z = t, t "!
A vector equation for the line is
x, y, z!" #$ = %3, %1.5, 0!" #$ + t %6, 11, 7!" #$ , t &! .
Chapter 8 Review Question 22 Page 503 a) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 2, 5, 2 1, 2, 3 2, 1, 5
2, 5, 2 13, 11, 3
35
n n n! " = ! # "
= ! # #
= #
! ! !
The normals are not coplanar; there is an intersection point.
Eliminate one of the variables from two pairs of equations. Eliminate x.
2x + 5y + 2z = 3 !
2x + 4y ! 6z = !22 2"
y + 8z = 25 # !!2"
2x + 5y + 2z = 3 !
2x + y + 5z = 8 $
4y ! 3z = !5 % !!$
MHR • Calculus and Vectors 12 Solutions 988
Solve equations and for y and z.
4y + 32z = 100 4!
4y ! 3z = !5 "
35z = 105 4!!"
z = 3
Substitute z = 3 in ! .
y + 8(3) = 25 !
y = 1
Substitute y = 1 and z = 3 in !.
x + 2(1)! 3(3) = !11
x = !4
The three planes intersect at the point (–4, 1, 3).
b) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 1, 3, 2 3, 5, 1 6, 4, 7
1, 3, 2 39, 15, 42
0
n n n! " = ! # "
= ! # #
=
! ! !
Since the triple scalar product is zero, the normals are coplanar and the planes intersect either in a line
or not at all.
Solve the system algebraically to determine if there is a solution.
x + 3y + 2z = 10 !
3x ! 5y + z = 1 "
6x + 4y + 7z = !5 #
14y + 5z = 29 $ 3!!"
14y + 5z = 65 % 6!!#
0 = !36 $!%
The system has no solution. The planes intersect in pairs only.
c) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 1, 3, 1 3, 1, 1 5, 7, 1
1, 3, 1 6, 2, 16
16
n n n! " = # ! "
= # ! #
= #
! ! !
The normals are not coplanar; there is an intersection point.
MHR • Calculus and Vectors 12 Solutions 989
Eliminate one of the variables from two pairs of equations. Eliminate z.
x + 3y ! z = !2 !
3x + y + z = 14 "
4x + 4y = 12 # !+"
x + 3y ! z = !2 !
5x + 7 y + z = 10 $
6x +10y = 8 % !+$
Solve equations and for x and y.
12x +12y = 36 3!
12x + 20y = 16 2"
!8y = 20 3!!2"
y = !2.5
Substitute y = –2.5 in ! .
4x + 4(!2.5) = 12 !
x = 5.5
Substitute x = 5.5 and y = !2.5 in !.
3(5.5) + (!2.5) + z = 14
z = 0
The three planes intersect at the point (5.5, –2.5, 0).
Chapter 8 Review Question 23 Page 503 a) First examine the normals. None are scalar multiples of each other. The planes are not parallel.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 2, 5, 3 1, 3, 6 3, 2, 9
2, 5, 3 39, 9, 11
0
n n n! " = ! # "
= ! #
=
! ! !
Since the triple scalar product is zero, the normals are coplanar and the planes intersect either in a line
or not at all.
The usual way to check this is to solve the system algebraically. However, in this case, observe that
1 2 3n n n+ =
! ! !
If the planes intersect in a line, then + = , but 0 19 7+ ! " .
Therefore, the three planes do not have a common intersection.
MHR • Calculus and Vectors 12 Solutions 990
b) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 8, 20, 16 3, 15, 12 2, 5, 4
8, 20, 16 120, 12, 15
1440
n n n! " = ! " #
= !
=
! ! !
The normals are not coplanar. These planes will intersect at a point.
MHR • Calculus and Vectors 12 Solutions 991
Chapter 8 Practice Test Chapter 8 Practice Test Question 1 Page 504 D is the best answer.
[ ] [ ]A, B 3, 5= is the normal to the line.
Check which vector has dot product of zero with this normal. [ ] [ ]3, 5 5, 3 0! " =
Chapter 8 Practice Test Question 2 Page 504 B is the best answer.
All equations have parallel direction vectors. None can be eliminated on this criterion.
Check if ( )1, 8! is a point on each line.
A: 1
4t = gives 1, but 8x y= ! " .
B: 3
2t = ! gives 1 and 8x y= = ! .
C: 11
12t = gives 1, but 8x y= ! " .
D: 3t = gives 1, but 8x y= ! " .
Chapter 8 Practice Test Question 3 Page 504 B is the best answer.
Substitute the point in each equation.
A: 3(10) + 6(–3)! 2(5)! 2 = 0
B: 1(10) + (–3)!1(5)!12 = !10
C: 2(10)! 2(–3)! 3(5)!11= 0
D: 4(10) + 5(–3) + (5)! 30 = 0
Chapter 8 Practice Test Question 4 Page 504 D is the best answer.
Lines in three-space have vector, parametric, and symmetric equations.
The slope-intercept equation only exists for lines in two-space.
Chapter 8 Practice Test Question 5 Page 504 D is the best answer.
Scalar equations represent lines in two-space and planes in three-space.
MHR • Calculus and Vectors 12 Solutions 992
Chapter 8 Practice Test Question 6 Page 504 A is the best answer.
Check which vector is a scalar multiple of [ ]1, 2, 3n = !
!.
A: [ ] [ ]2, 3, 4 1, 2, 3k! " !
B: [ ] [ ]1, 2, 3 1, 2, 3! ! = ! !
C: [ ] [ ]2, 4, 6 2 1, 2, 3! = !
D: [ ] [ ]3, 6, 9 3 1, 2, 3! = !
Chapter 8 Practice Test Question 7 Page 504 D is the best answer.
Substitute each point in the equation.
A: 3(!8)! 4(–9) + (0)!12 = 0
B: 3(4)! 4(1) + (4)!12 = 0
C: 3(16)! 4(10) + (4)!12 = 0
D: 3(18)! 4(12) + (2)!12 = !4
Chapter 8 Practice Test Question 8 Page 504 B is the best answer.
The normals [ ] [ ]10, 7 and 4, 5! are not scalar multiples of each other. Therefore, the lines must not be
parallel and they intersect in a unique point.
Chapter 8 Practice Test Question 9 Page 504 C is the best answer.
The lines have the same direction vector. They are either parallel or coincident.
Check if (4, –2, 3) is a point on the second line.
4 = 1+ 5t
t =3
5
!2 = !3! 3t
t = !1
3
3 = 1+ 2t
t = 1
Since the t-values are different, the point is not on the line and the lines are parallel and distinct.
A: means there is an intersection point.
B: means the lines are identical.
C: means the lines do not intersect
D: means the lines do not intersect but they are also not parallel
MHR • Calculus and Vectors 12 Solutions 993
Chapter 8 Practice Test Question 10 Page 504 A is the best answer.
Planes can intersect in a line, be parallel ands distinct, or be coincident.
Two planes cannot intersect in just one point.
Chapter 8 Practice Test Question 11 Page 504 a)
AB
! "!!
= OB
! "!!
!OC
! "!!
= 2, ! 9, 0"# $% ! 1, 5, ! 4"# $%
= 1, !14, 4"# $%
A possible vector equation is
x, y, z!" #$ = 1, 5, % 4!" #$ + t 1, %14, 4!" #$ , t &! .
Possible parametric equations are:
x = 1+ t
y = 5!14t
z = !4 + 4t, t "!
b) Let 2 and 1t t= = ! . The resulting points are: (3, –23, 4) and (0, 19, –8).
Chapter 8 Practice Test Question 12 Page 504 The normal to the first line is [4, 8] or [1, 2]. This vector will be a direction vector for the required line.
To find the x-intercept, let 0y = .
0 = 7 + 3t
t = !7
3
When
t = !
7
3:
x = 2 + !7
3
"
#$%
&'!10( )
=76
3
The x-intercept is 76
3.
The parametric equations of the required line are:
x =76
3+ t
y = 2t, t !!
MHR • Calculus and Vectors 12 Solutions 994
Chapter 8 Practice Test Question 13 Page 504 For a perpendicular direction, find the cross product of the direction vectors of the two given lines.
Chapter 8 Practice Test Question 14 Page 504 A plane parallel to the xz-plane has equation of the form y = k.
Since the plane must contain the point (3, –1, 5), the required equation is 1 or 1 0y y= ! + = .
(Note that the line is actually parallel to the xz-plane since every point has y = –1.)
Chapter 8 Practice Test Question 15 Page 505 a) Substitute the parametric equations in the scalar equation of the first line.
6(4 + t) + 2(–7 – 3t) = 5
24 + 6t !14 ! 6t = 5
10 = !5
Since there are no solutions for t, the lines do not intersect
b) Use elimination.
2x + 3y = 21 !
4x ! y = 7 "
14x = 42 !+3"
x = 3
Substitute x = 3 in ! .
4(3)! y = 7 !
y = 5
The intersection point is (3, 5).
c) The direction vectors are not parallel since the direction vectors are not scalar multiples of each other.
Equate the expressions for like coordinates
!2 + 3s = 7 + t
3s ! t = 9 !
4 ! 3s = 10 + 4t
3s + 4t = !6 !
!1+ s = 4 + 3t
s ! 3t = 5 !
MHR • Calculus and Vectors 12 Solutions 995
Solve equations and for s and t.
3s ! t = 9 !
3s + 4t = !6 "
! 5t = 15 !!"
t = !3
Substitute t = –3 in
3s + 4(!3) = !6
3s = 6
s = 2
Check that s and t satisfy .
L.S. = 2 ! 3(!3)
= 11
R.S. = 5
L.S. ≠ R.S.
Therefore, the lines do not intersect.
d) The direction vectors are not parallel since the direction vectors are not scalar multiples of each other.
Equate the expressions for like coordinates
3+ 5s = 1+10t
5s !10t = !2
s ! 2t = !1 !
4 + 2s = !4 + 4t
2s ! 4t = !8
s ! 2t = !4 !
!6 ! 2s = 4 ! 4t
!2s + 4t = 10
s ! 2t = !5 !
Since the same expression has three different values, the lines do not intersect.
Chapter 8 Practice Test Question 16 Page 505 a) The direction vectors are not parallel since the direction vectors are not scalar multiples of each other.
Now show that the lines do not intersect.
Equate the expressions for like coordinates
2 + 5s = 1! 2t
5s + 2t = !1 !
6 + s = 5+ 4t
s ! 4t = !1 !
1+ 3s = !3+ t
3s ! t = !4 !
Solve equations and for s and t.
10s + 4t = !2 2!
s ! 4t = !1 "
11s = !3 2!+"
s = !3
11
MHR • Calculus and Vectors 12 Solutions 996
Substitute in s =
!3
11 in .
34 1
11
2
11
t
t
! ! = !
=
Check that s and t satisfy.
L.S. = 3 !3
11
"
#$%
&'!
2
11
"
#$%
&'
= !1
R.S. = –4
L.S. ≠ R.S.
Therefore the lines do not intersect.
Since the two lines do not intersect and are not parallel, they are skew lines.
Chapter 6 to 8 Review Question 33 Page 507 a) Substitute the parametric equations for the line in the equation for the plane and solve for s.
2(4 + 3s) + 3(4 + 3s)! 5(!1! 2s) = 3
8 + 6s +18 +12s + 5+10s = 3
28s = !28
s = !1
Substitute s = 1 in the parametric equations.
x = 4 + 3(!1) y = 6 + 4(!1) z = !1! 2(!1)
= 1 = 2 = 1
The point of intersection of the line and the plane is (1, 2, 1) .
b) Substitute the parametric equations into the scalar equation of the plane and solve for s.
(8 + 4s) + 3(!2 ! 2s)! 2(!2 ! s) = 6
8 + 4s ! 6 ! 6s + 4 + 2s = 6
0s = 0
This equation is true for every value of s.
The two lines intersect at every point on the line; the line is on the plane.
Chapter 6 to 8 Review Question 34 Page 507
The distance between a point P and a plane is given by n PQ
dn
!=
! """!
! where Q is any point on the plane with
normal n!
. Choose Q(1, 0, 0).
PQ
! "!!
= 1, 0, 0!" #$ % 3, % 2, 5!" #$
= %2, 2, % 5!" #$
d =
2, 4, %1!" #$ & %2, 2, % 5!" #$
22
+ 42
+ (–1)2
=9
21
# 1.96
The distance is about 1.96 units.
MHR • Calculus and Vectors 12 Solutions 1017
Chapter 6 to 8 Review Question 35 Page 507 a) The normals are identical but the equations are different.
The planes are parallel and distinct.
b) The normals are scalar multiples of each other. 2n!
1 = n!
2 and 2! = " .
The planes are parallel and coincident. Chapter 6 to 8 Review Question 36 Page 507 a) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 2, 3, 5 5, 1, 2 1,7, 12
2, 3, 5 2, 58, 34
0
n n n! " = # ! # " # #
= # ! #
=
! ! !
The normals are coplanar; there may be a line of intersection.
Use elimination to eliminate one of the variables, say y, from two pairs of equations.
2x + 3y ! 5z = 9 !
15x ! 3y + 6z = !9 3"
17x + z = 0 # !+3"
35x ! 7 y +14z = !21 7"
!x + 7 y !12z = 21 $
34x + 2z = 0 % 7"+$
Solve equations and for x and z.
34x + 2z = 0 2!
34x + 2z = 0 "
0 = 0 2!!"
Let z = t,
17x + t = 0 !
x = !1
17t
Substitute
x = !
1
17t and z = t in ! .
5 !1
17t
"
#$%
&'! y + 2t = !3 !
y = 3+29
17t
MHR • Calculus and Vectors 12 Solutions 1018
The three planes intersect at a line that can be defined as
x = !1
17t
y = 3+29
17t
z = t, t "!
These equations can be simplified by multiplying the direction vector by 17. The simplified equations
are (t ∈ R):
x = –t y = 3 + 29t z = 17t
b) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 2, 3, 1 1, 4, 2 7, 6, 5
2, 3, 1 32, 19, 22
15
n n n! " = ! # "
= ! # #
= #
! ! !
The normals are not coplanar; there is an intersection point.
Use elimination to eliminate one of the variables, say x, from two pairs of equations.
2x + 3y + z = 5 !
2x + 8y ! 4z = 20 2"
! 5y + 5z = !15 !!2"
! y + z = !3 #
7x + 28y !14z = 70 7"
7x + 6y + 5z = 7 $
22y !19z = 63 % 7"!$
Solve equations and for y and z.
!22y + 22z = !66 22!
22y !19z = 63 "
3z = !3 22!+"
z = !1
Substitute z = –1 in ! .
! y + (!1) = !3 !
y = 2
Substitute y = 2 and z = !1 in !.
x + 4(2)! 2(!1) = 10
x = 0
The three planes intersect at the point (0, 2, –1).
MHR • Calculus and Vectors 12 Solutions 1019
c) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
[ ] [ ] [ ]
[ ] [ ]
1 2 3 3, 1, 1 4, 2, 3 8, 6, 1
3, 1, 1 20, 20, 40
0
n n n! " = # ! # # " #
= # ! #
=
! ! !
The normals are coplanar; there may be a line of intersection.
Use elimination to eliminate one of the variables, say y, from two pairs of equations.
6x + 2y ! 2z = 8 2!
4x ! 2y ! 3z = 5 "
10x ! 5z = 13 # 2!+"
18x + 6y ! 6z = 24 6!
8x + 6y ! z = 7$
10x ! 5z = 17 % 6!!$
Solve equations and for x and z.
10x ! 5z = 13 !
10x ! 5z = 17 "
0 = !4 !!"
This equation is never true.
The three planes do not have any common points. The planes intersect in pairs.