b) equate the corresponding coordinates and...

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.


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. The two lines intersect at the point (3, –1, 2). Chapter 8 Review Question 18 Page 503

For the distance between skew lines,1 2PP n

d

n

!

=

!!!!" "

" where 1 21 1 2 2, , and P P n m m! ! = "

! "! "!# # .

Let P1(1, 0, –1), P2(8, 1, 3), [ ] [ ]1 22, 3, 4 , and 4, 5, 1m m= ! = !

!" !".

P

1P

2

! "!!!

= 7, 1, 4!" #$

n!

= m

"!

1 ! m

"!

2

= "17, "18, " 22#$ %&

d =

7, 1, 4!" #$ % &17, &18, & 22!" #$

&17, &18, & 22!" #$

d =225

1097

d ! 6.8


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.


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 % !!$


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.


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.


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.


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.


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


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 763

.

The parametric equations of the required line are:

x =76

3+ t

y = 2t, t !!


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.

[ ] [ ] [ ]4, 6, 3 3, 2, 4 30, 25, 10m = ! " = ! !

!". Use [ ]6, 5, 2! ! .

The parametric equations are

x = !6 + 6t

y = 4 ! 5t

z = 3! 2t, t "!

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 !


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


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.

b) The distance between skew lines is 1 2PP n

d

n

!

=

!!!!" "

" where 1 21 1 2 2, , and P P n m m! ! = "

! "! "!# #

Let P1(2, 6, 1), P2(1, 5, –3), [ ] [ ]1 25, 1, 3 , and 2, 4, 1m m= = !

!" !".

P

1P

2

! "!!!

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

n

!

= m

"!

1 ! m

"!

2

= "11, "11, 22#$ %&

d =

!1, !1, ! 4"# $% & !11, !11, 22"# $%

!11, !11, 22"# $%

d =66

726

d ! 2.45

The distance between the lines is approximately 2.45 units.


Chapter 8 Practice Test Question 17 Page 505 Find the scalar equation of the plane.

AB! "!!

= !9, ! 4, ! 2"# $%

BC! "!!

= 13, ! 3, !13"# $%

Find the normal vector.

AB! "!!

! BC! "!!

= "4("13)" ("3)("2)," 2(13)" ("13)("9), " 9("3)"13("4)#$ %&

= 46, "143, 79#$ %&

The equation is of the form 46 143 79 0x y z D! + + = . Use the coordinates of A to find D.

46(2)!143(5)+ 79(6)+ D = 0

D = 149

The scalar equation of the plane is 46 143 79 149 0x y z! + + = . Now check if the line is on the plane.

46(!3+ 22k)!143(!6 + k)+ 79(!11!11k)+149 = 0

!138+1012k + 858!143k ! 869 ! 869k +149 = 0

0k = 0

Since this equation is true for all values of t, the line does lie in the plane. Chapter 8 Practice Test Question 18 Page 505 a) The normal vectors are not scalar multiples of each other and so the planes are not parallel. The two planes intersect at a line. b) The normal vectors are not scalar multiples of each other and so the planes are not parallel. Check if the normals are coplanar.

[ ] [ ] [ ]

[ ] [ ]

1 2 3 1, 1, 1 2, 3, 4 2, 2, 1

1, 1, 1 11, 6, 10

27

n n n! " = # ! " #

= # ! #

=

! ! !

The normals are not coplanar. These planes will intersect at a point. c) Check the normals. 1 32n n=

! !, but 2 times equation does not equal equation .

Therefore two planes are parallel and distinct and the system has no solutions. The third plane is not parallel to the other two planes and intersects both of them. d) The normal vectors are not scalar multiples of each other and so the planes are not parallel. The planes intersect at a line.


Chapter 8 Practice Test Question 19 Page 505 a)

BA! "!!

= 3, 19, ! 3"# $% and CA! "!!

= 2, 14, 5"# $%

A vector equation is

x, y, z!" #$ = 1, 13 ,2!" #$ + s 3, 19, % 3!" #$ + t 2, 14 ,5!" #$ , s, t &! . Parametric equations are:

x = 1+ 3s + 2t

y = 13+19s +14t

z = 2 ! 3s + 5t, s, t "!

b) First, find the normal to the plane.

n

!

= BA" !""

!CA" !""

= 137, " 21, 4#$ %&

The scalar equation is of the form 137 21 4 0x y z D! + + = Use one of the points to find D. Use C.

137(–1)! 21(–1)+ 4(–3)+ D = 0

D = 128

The scalar equation is 137 21 4 128 0x y z! + + = . c) When s = 1 and t = 1, a point is (6, 46, 4). When s = 0 and t = 1, a point is (3, 27, 7). Chapter 8 Practice Test Question 20 Page 505 a) Substitute the coordinates for the origin in the parametric equations.

0 = 2 + s + 2t

s + 2t = !2 !

0 = 1+ s

s = !1 !

0 = 3+ s + 2t

s + 2t = !3 !

From and , 11 and

2s t= ! = ! .


L.S.= (!1)+ 2 !1

2

"

#$%

&'

= !2

R.S. = –3

L.S. ≠ R.S. Therefore the origin is not a point on the line.


b) 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!

.

n

!

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

= 2, 0, & 2!" #$

Choose Q(2, 1, 3).

PQ! "!!

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

= 2, 1, 3!" #$

d =

2, 0, ! 2"# $% & 2, 1, 3"# $%

22+ 02

+ (–2)2

=2

8

! 0.71

The distance is approximately 0.71 units. Chapter 8 Practice Test Question 21 Page 505 Choose three coplanar normal vectors.

[ ]1 2, 3, 5n = !

!. Choose any non-parallel vector for n

!

, Choose [1, 1, 1].

For a third normal, choose a coplanar vector. Choose [ ]1 2 3, 2, 6n n+ = !

! !.

Possible equations for the other planes are 2 and 3 2 6 2x y z x y z+ + = ! + = . To be sure that the planes do not intersect in a common line, check that + ≠ (i.e.

1 2 3D D D+ ! ).

Chapter 8 Practice Test Question 22 Page 505 a) Substitute the parametric equations in the scalar equation.

2(3r)+ 3(2r)+ 4(6 ! 3r) = 24

0r = 0

2(14 + s)+ 3(!12 ! 2s)+ 4(8+ s) = 24

0s = 0

2(9 + t)+ 3(–10 + 6t)+ 4(9 ! 5t) = 24

0t = 0

Since each of the equations is true for all values of r, s, and t respectively, all three lines are contained in the plane.


b) To find the first vertex, equate like coordinates for lines1 2 and ! ! .

3r = 14 + s !

2r = !12 ! 2s "

6 ! 3r = 8+ s #


3r ! s = 14 !

r + s = !6 0.5"

4r = 8 !+0.5"

r = 2

Substitute r = 2 in ! .

2 + s = !6 0.5!

s = !8


L.S.= 6 ! 3(2)

= 0

R.S.= 8+ (!8)

= 0

L.S. = R.S. Substitute 2 or 8 r s= = ! to find the point of intersection. The two lines intersect at the point (6, 4, 0). To find the second vertex, equate like coordinates for lines

2 3 and ! ! .

14 + s = 9 + t !

!12 ! 2s = !10 + 6t "

8+ s = 9 ! 5t #


s ! t = !5 !

s + 3t = !1 0.5"

! 4t = !4 !!0.5"

t = 1


s !1= !5 !

s = !4


L.S.= 8+ (!4)

= 4

R.S.= 9 ! 5(1)

= 4

L.S. = R.S. Substitute 1 or 4 t s= = ! to find the point of intersection. The two lines intersect at the point (10, –4, 4).


To find the third vertex, equate like coordinates for lines1 3 and ! ! .

3r = 9 + t !

2r = !10 + 6t "

6 ! 3r = 9 ! 5t #


3r ! t = 9 !

3r ! 5t = !3 "

4t = 12 !!"

t = 3


3r ! 3= 9 !

r = 4


L.S.= 2(4)

= 8

R.S.= !10 + 6(3)

= 8

L.S. = R.S. Substitute 3 or 4t r= = to find the point of intersection. The two lines intersect at the point (12, 8, –6). The vertices are A(12, 8, –6), B(10, –4, 4), and C(6, 4, 0).

c)

AB = (10 !12)2+ (!4 ! 8)2

+ (4 ! (–6))2

= 248

AC = (6 !12)2+ (4 ! 8)2

+ (0 ! (–6))2

= 88

BC = (10 ! 6)2+ (!4 ! 4)2

+ (4 – 0)2

= 96

Perimeter = 248 + 88 + 96 ! 34.9 The perimeter is approximately 34.9 units.


d)

Area =1

2AB! "!!

! BC! "!!

=1

2"2, "12, 10#$ %& ! "4, 8, " 4#$ %&

=1

2"32, " 48, " 64#$ %&

=1

2(–32)2

+ (–48)2+ (–64)2

=1

27424

# 43.1

The area of the triangle is approximately 43.1 units2.


Chapters 6 to 8 Review Chapter 6 to 8 Review Question 1 Page 506 a) 340º b) 144º c) 260º Chapter 6 to 8 Review Question 2 Page 506 a) S50ºE b) N70ºW c) S86ºE Chapter 6 to 8 Review Question 3 Page 506 a) DB

! "!!

or EA

! "!!

b) No. The magnitudes of the vectors are marked equal in the diagram. The vectors look parallel in the

diagram but the angles in the isosceles triangles involved may be different in magnitude and so we cannot be sure the vectors are parallel. (e.g., ∠FAE could equal 45º while ∠BDC could equal 46º.)

c) The two isosceles triangles involved need to be congruent. This requires either ∠FAE = ∠BDC or

FE = BC . Both conditions would lead to AF and CD having equal lengths and being parallel.

d)

AB

! "!!

= EB

! "!!

! EA

! "!!

= AE

! "!!

! BE

! "!!

= DB

! "!!

! DA

! "!!

= AD

! "!!

! BD

! "!!


Chapter 6 to 8 Review Question 4 Page 506 The first two vectors are opposites of each other and have a vector sum of 0

!. The resultant force is 50 N

in the northwest direction.

Chapter 6 to 8 Review Question 5 Page 506


Chapter 6 to 8 Review Question 6 Page 506 Draw a scale diagram using the vectors from question 5.

Chapter 6 to 8 Review Question 7 Page 506 a) Normal acceleration due to gravity on Earth is 9.8 m/s2. On a roller coaster at an amusement park the

acceleration felt by a rider is twice as great at the bottom of a short dip. b) A ball hits a racquet with a force of 100 N. The force of the racquet on the ball is ten times greater and

in the opposite direction. Chapter 6 to 8 Review Question 8 Page 506 a) Draw a diagram. Use the Pythagorean theorem and trigonometry.

R

!"

= 102+ 8

2

# 12.8

tan ! =8

10

! = tan"1

8

10

#

$%&

'(

! ! 38.7o

The resultant is approximately 12.8 N in a direction N38.7ºW.


b) Draw a diagram. Use the Pythagorean theorem and trigonometry.

R

!"

= 802+12

2

# 80.9

tan ! =8

10

! = tan"1

12

80

#

$%&

'(

! ! 8.5o

The resultant is approximately 80.9 m/s, 8.5º up from the horizontal. c) Draw a diagram of the situation.

Use the cosine law to find the magnitude of the resultant displacement. The angle between the displacements in the diagram is 170º.

R

!" 2

= 302+ 252

! 2(30)(25)cos 1700

R

!" 2

# 3002.2

R

!"

# 54.8


To find the direction of the resultant, use the sine law. Let ! represent the angle of R!"

to the horizontal direction.

sin !25

=sin 170

o

54.8

sin ! =25sin 170

o

54.8

! = sin"1

25sin 170o

54.8

#

$%&

'(

! ! 4.5o

The resultant force has a magnitude of approximately 54.8 N, 4.5! from the horizontal Chapter 6 to 8 Review Question 9 Page 506

F

!"

ramp = 75cos 70o

# 25.7

F

!"

perp. = 75sin 70o

# 70.5

The component parallel to the ramp is 25.7 N. The component perpendicular to the ramp is 70.5 N. Chapter 6 to 8 Review Question 10 Page 506

a) Unit vectors parallel to u!

are of the form 1u

u

±

!

! .

u

!

= 22+ 3

2

= 13

.

The required unit vectors are 2 3 2 3

, and ,13 13 13 13

! " ! "# #$ % $ %

& ' & '.

b) 13


c) Let Q(x, y) be the point. [ ]7, 5PQ x y= + !

!!!"

7 2

5

x

x

+ =

= !

5 3

8

y

y

! =

=

The point is Q(–5, 8). Chapter 6 to 8 Review Question 11 Page 506 Answers may vary. For example: For k = 2 and m = 3 and [ ]5, 6u =

!.

L.S.= (k + m)u!

= (2 + 3) 5, 6!" #$

= 5 5, 6!" #$

= 25, 30!" #$

R.S.= ku

!

+ mu

!

= 2 5, 6!" #$ + 3 5, 6!" #$

= 10, 12!" #$ + 15, 18!" #$

= 25,30!" #$

L.S. = R.S. Chapter 6 to 8 Review Question 12 Page 506

The boat’s vector is

18cos 216

o, 18sin 216

o!"

#$ = %14.5623, %10.5801!" #$ .

The current’s vector is

8cos 146o, 8sin 146

o!"

#$ = %6.6323, 4.4735!" #$ .

The resultant vector is [ ]21.1946, 6.1066! ! .

The magnitude of the resultant is ( ) ( )2 2

21.1946 6.1066 22.05! + ! ! . The angle of the resultant with the east direction is:

! = tan"1

"6.1066

"21.1946

#

$%&

'(

! 16.1o+180

o

= 196.1o

(The resultant is pointing SW or at a bearing of 253.9º.)



a

!

!b

!

= a

!

b

!

cos "

= (90)(108)cos 40o

" 7446.0

The force is approximately 7446.0 N. Chapter 6 to 8 Review Question 14 Page 506 a)

m

!"

!n"

= "2, " 8#$ %& ! 9, 0#$ %&

= "2(9)+ (–8)(0)

= "18

b)

p!"

!q"

= 4, 5, "1#$ %& ! 6, " 2, 7#$ %&

= 4(6)+ 5(–2)+ (–1)(7)

= 7

Chapter 6 to 8 Review Question 15 Page 506 Choose any vector the makes a dot product of zero with u

!.

[ ]5,6v =!


cos ! =u

!

"v!

u

!

v

!

cos ! =5, 7, #1$% &' " 8, 7, 8$% &'

52+ 72

+ (–1)2 82+ 72

+ 82

cos ! =81

75 177

cos ! " 0.7030

! = cos#1 0.7030( )! = 45.3o

The angle between the vectors is approximately 45º.



W = F

!"

! s

"

= F

!"

s

"

cos "

= (80)(20)cos 10o

# 1575.7

Tyler does 1575.7 J of mechanical work. Chapter 6 to 8 Review Question 18 Page 507

v

!

= OB" !""

!OA" !""

v

!

= 9, 1, 5"# $% ! 3, 7, ! 2"# $%

v

!

= 6,!6,7"# $%

v

!

= 62+ (–6)2

+ 72

v

!

= 11

Chapter 6 to 8 Review Question 19 Page 507 a)

b

!

!a!

" c

!

= 7, 3, 4#$ %& ! 2, ' 4, 5#$ %& " '3, 7, 1#$ %&

= 7, 3, 4#$ %& ! '4(1)' 7(5), 5(–3)'1(2), 2(7)' (–3)(–4)#$ %&

= 7, 3, 4#$ %& ! '39, '17, 2#$ %&

= 7(–39)+ 3(–17)+ 4(2)

= '316

b) [ ] [ ] [ ] [ ]

[ ] [ ]

[ ]

2, 4, 5 7, 3, 4 7, 3, 4 3, 7, 1

31, 27, 34 25, 19, 58

6, 46, 24

a b b c! " ! = " ! " ! "

= " " " "

= " "

! ! ! !


Chapter 6 to 8 Review Question 20 Page 507 No, u v v u! = " !

! ! ! !.

i j k

j i k

! =

! = "

! ! !

! ! !


c

!

! d

"!

= 6,3, " 2#$ %& ! 4,5, " 7#$ %&

= 3(–7)" 5(–2), " 2(4)" (–7)(6), 6(5)" 4(3)#$ %&

= "11, 34, 18#$ %&

d

"!

! c

!

= "c

!

! d

"!

= 11, " 34, "18#$ %&

Two possible orthogonal vectors are (–11, 34, 18) and (11, –34, –18). Chapter 6 to 8 Review Question 22 Page 507

A = u

!

! v

!

= u

!

v

!

sin "

sin " =95

(12)(10)

" = sin#1 95

(12)(10)

$

%&'

()

" " 52.3o

Adjacent interior angles in a parallelogram add to 180º. The interior angles are o o

52.3 and 127.7 .


Chapter 6 to 8 Review Question 23 Page 507 Assume the force acts at right angles to the shaft.

!

!

= r

!

F

"!

sin"

= (0.18)(40)sin90o

# 7.2

The magnitude of the torque is 7.2 N·m. Chapter 6 to 8 Review Question 24 Page 507 a)

PQ! "!!

= !4, 7"# $% ! 3, 5"# $%

= !7, 2"# $%


x, y!" #$ = 3, 5!" #$ + t %7, 2!" #$ , t &! .

Parametric equations are:

x = 3! 7t

y = 5+ 2t, t "!

b)

AB! "!!

= !2, ! 3, 6"# $% ! 6, !1, 5"# $%

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


x, y, z!" #$ = 6, %1, 5!" #$ + t %8, % 2, 1!" #$ , t &! .

Parametric equations are:

x = 6 ! 8t

y = !1! 2t

z = 5+ t, t "!

Chapter 6 to 8 Review Question 25 Page 507 The direction of the line is [ ]7, 3m = !

!".

A normal is any perpendicular vector (i.e., a vector having a dot product of zero with m!"

). Use [ ]3, 7n =!

. The scalar equation is of the form 3 7 0x y C+ + = . The point (6, –2) is on the line and leads to C = –4. The equation of the line is 3x + 7y – 4 = 0.


Chapter 6 to 8 Review Question 26 Page 507 The scalar equation is of the form 5 0x y C+ + = . Substitute (–4, 6).

5(!4)+ 6 +C = 0

C = 14

The scalar equation is 5x + y + 14 = 0. A direction vector is any vector perpendicular to the normal. Use[ ]1, 5! .


x, y!" #$ = %4, 6!" #$ + t %1, 5!" #$ , t &! . Chapter 6 to 8 Review Question 27 Page 507 a) Lines parallel to the x- and y-axes have scalar equations of the form z = k. A vector equation is

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

A scalar equation is z = –1. b) A vector equation is

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

For a scalar equation,

n

!

= BA" !""

!CA" !""

= 1, 1, 6"# $% ! 1, & 5, 5"# $%

= 35, 1, & 6"# $%

The scalar equation is of the form 35 6 0x y z D+ ! + = . Substitute (6, 2, 3).

35(6)+ 2 ! 6(3)+ D = 0

D = !194

A scalar equation is 35x + y – 6z – 194 = 0. c) Parallel planes have equal normals. The scalar equation is of the form 2 6 4 0x y z D+ + + = . Substitute (3, 2, 1).

( ) ( ) ( )2 3 6 2 4 1 0

22

D

D

+ + + =

= !

A scalar equation is 2x + 6y + 4z – 22 = 0 or 3 2 11 0x y z+ + ! = .


Chapter 6 to 8 Review Question 28 Page 507 If three direction vectors were mutually perpendicular, they would define a three-dimensional space, not a plane. Two perpendicular vectors can be used to form a plane, and any other vectors in the plane would be a linear combination of these two vectors and hence coplanar with them. Chapter 6 to 8 Review Question 29 Page 507 a)

n

!

= m

"!

1 ! m

"!

2

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

= "10, 22, "19#$ %&

The scalar equation is of the form10 22 19 0x y z D! + + = . Substitute (2, 3, 5).

10(2)! 22(3)+19(5)+ D = 0

D = !49

A scalar equation is10 22 19 49 0x y z! + ! = .

b) The scalar equation is of the form 2 5 0x y z D+ ! + = .

Substitute (6, –2, 3).

6 + 2(–2)! 5(3)+ D = 0

D = 13

A scalar equation is 2 5 13 0x y z+ ! + = .

Chapter 6 to 8 Review Question 30 Page 507 The angle between planes is defined as the angle between their normal vectors.

cos! =n

!

1 "n!

2

n

!

1 n

!

2

=

2, 2, 7#$ %& " 3, ' 4, 4#$ %&

22+ 22

+ 72 32+ (–4)2

+ 42

=26

57 41

! = cos'1 26

57 41

(

)*+

,-

" 57.5o

The angle is about 57.5º.


Chapter 6 to 8 Review Question 31 Page 507 The direction vectors are not scalar multiples of each other. The lines must intersect. To find the intersection point, equate like coordinates.

2 ! 5s = !11+ 2t !

1+ 3s = 7 ! 3t "

!4 + s = 2 + 3t #


3s + 3t = 6 !

s ! 3t = 6 "

4s = 12 !+"

s = 3

Substitute s = 3 in ! .

3! 3t = 6 !

t = !1


L.S.= 2 ! 5(3) R.S.= !11+ 2(1)

= !13 = !13

L.S.= R.S.

Substitute 3 or 1 s t= = ! to find the point of intersection. The two lines intersect at the point (–13, 10, –1). Chapter 6 to 8 Review Question 32 Page 507

For the distance between skew lines,

d =

P1P

2

! "!!!

!n

"

n

" where P

1!!

1, P

2!!

2, and n

"

= m

#"

1 " m

#"

2 .

Let P1(4, 2, –3), P2(–6, –2, 3), [ ] [ ]1 21, 2, 2 , and 2, 2, 1m m= ! ! = !

!" !".

P1P

2

! "!!!

= !10, ! 4,6"# $% and n"

= m

!"

1 & m

!"

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

d =

!10, ! 4,6"# $% ' !6, ! 3, ! 6"# $%

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

d =36

81

d = 4

The distance between the skew lines is 4 units.


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.


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.


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


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).


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.

b) equate the corresponding coordinates and...

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