Anton Linear Algebra 10th edition Solutions Set 3.3

EXERCISE SET 3.1

1.

2.

147

x

(3, 4, 5)

y

z

x

y

z

(3, 4, 5)

x

( 3, 4, 5)

y

z

x

z

y

(3, 0, 3)

(a) (c)

(e) ( j)

v1= (3, 6)

x

y

x

y

v3 = ( 4, 3)

(a) (c)

3. (a) = (3 4, 7 8) = (1, 1)(e) = (2 3, 5 + 7, 4 2) = (5, 12, 6)

5. (a) Let P = (x, y, z) be the initial point of the desired vector and assume that this vector

has the same length as v. Since has the same direction as v = (4, 2, 1), we have

the equation

= (3 x, 0 y, 5 z) = (4, 2, 1)

If we equate components in the above equation, we obtain

x = 1, y = 2, and z = 4

Thus, we have found a vector which satises the given conditions. Any positive

multiple k will also work provided the terminal point remains xed at Q. Thus, P

could be any point (3 4k, 2k, k 5) where k > 0.

(b) Let P = (x, y, z) be the initial point of the desired vector and assume that this vector

has the same length as v. Since is oppositely directed to v = (4, 2, 1), we

have the equation

= (3 x, 0 y, 5 z) = (4, 2, 1)

If we equate components in the above equation, we obtain

x = 7, y = 2, and z = 6

Thus, we have found a vector which satises the given conditions. Any positive

multiple k will also work, provided the terminal point remains xed at Q. Thus,

P could be any point (3 + 4k, 2k, k 5) where k > 0.

PQ

PQ

PQ

PQ

PQ

PQ

PQ

PQ

P P1 2

P P1 2

v7= (3, 4, 5)

x

y

z

v9= (0, 0, 3)x

z

y

(i)(g)

148 Exercise Set 3.1

6. (a) v w = (4 6, 0 (1), 8 (4)) = (2, 1, 4)

(c) v + u = (4 3, 0 + 1, 8 + 2) = (7, 1, 10)

(e) 3(v 8w) = 3(4 48, 0 + 8, 8 + 32) = (132, 24, 72)

7. Let x = (x1, x2, x3). Then

2u v + x = (6, 2, 4) (4, 0, 8) + (x1, x2, x3)

= (10 + x1, 2 + x2, 12 + x3)

On the other hand,

7x + w = 7(x1, x2, x3) + (6, 1, 4)

= (7x1 + 6, 7x2 1, 7 x3 4)

If we equate the components of these two vectors, we obtain

7x1 + 6 = x1 10

7x2 1 = x2 + 2

7x3 4 = x3 + 12

Hence, x = (8/3, 1/2, 8/3).

9. Suppose there are scalars c1, c2, and c3 which satisfy the given equation. If we equatecomponents on both sides, we obtain the following system of equations:

2c1 3c2 + c3 = 0

9c1 + 2c2 + 7c3 = 5

6c1 + c2 + 5c3 = 4

The augmented matrix of this system of equations can be reduced to

The third row of the above matrix implies that 0c1 + 0c2 + 0c3 = 1. Clearly, there do notexist scalars c1, c2, and c3 which satisfy the above equation, and hence the system isinconsistent.

2 3 1 0

0 2 2 1

0 0 0 1

Exercise Set 3.1 149

10. If we equate components on both sides of the given equation, we obtain

c1 + 2c2 = 0

2c1 + c2 + 3c3 = 0

c2 + c3 = 0

The augmented matrix of this system of equations can be reduced to

Thus the only solution is c1 = c2 = c3 = 0.

11. We work in the plane determined by the three points O = (0, 0, 0), P = (2, 3, 2), and Q =

(7, 4, 1). Let X be a point on the line through P and Q and let t (where t is a positive,

real number) be the vector with initial point P and terminal point X. Note that the length

of t is t times the length of . Referring to the gure below, we see that

and

Q

O

PX

tPQ

OP PQ OQ

+ =

OP t PQ OX

+ =

PQ

PQ

PQ

1 0 0 0

0 1 0 0

0 0 1 0


Therefore,

(a) To obtain the midpoint of the line segment connecting P and Q, we set t =1/2. Thisgives

(b) Now set t = 3/4.This gives

12. The relationship between the xy-coordinate system and the xy-coordinate system is givenby

x = x 2, y = y (3) = y + 3

(a) x = 7 2 = 5 and y = 5+ 3 = 8

(b) x = x + 2 = 1 and y = y 3 = 3

OX

= + =

1

42 3 2

3

47 4 1

23

4

9

4

1

4( , , ) ( , , ) , ,

OX OP OQ

= +

= +

1

2

1

21

22 3 2

1

27 4 1( , , ) ( , , )

==

9

2

1

2

1

2, ,

OX OP t OQ OP

t OP t OQ

= +

= +

( )

( )1


(c)

(d) (1, 10)

(e) v1 = v1 2, v2 = v + 3

13. Q = (7, 3, 19)

14. Let (x0, y0, z0) denote the origin in the x y z-coordinate system with respect to the xyz-coordinate system. Suppose that P1 and P2 are the initial and terminal points, respectively,for a vector v. Let (x1, y1, z1) and (x2, y2, z2) be the coordinates of P1 and P2 in the xyz-coordinate system, and let (x1, y1, z1) and (x2, y2, z2) be the coordinates of P1 and P2 in thexyz-coordinate system. Further, let (v1, v2, v3) and (v1, v2, v3) denote the coordinates ofv with respect to the xyz- and xyz-coordinate systems, respectively. Then

v1 = x2 x1, v2 = y2 y1, v3 = z2 z1

and

v1 = x2 x1, v2 = y2 y1, v3 = z2 z1

However,

x1 = x1 x0, y1 = y1 y0, z1 = z1 z0

and

x2 = x2 x0, y2 = y2 y0, zz2 = z2 z0

Q

y y'

x'

x

P85

6

7

53


If we substitute the above expressions for x1 and x2 into the equation for v1, we obtain

v1 = (x2 x0) (x1 x0 ) = x2 x1 = v1

In a similar way, we can show that v2 = v2 and v3 = v3.

16. Let v = (v1, v2) and u = kv = (u1, u2). Let P and Q be the points (v1, v2) and (u1, u2),respectively. Since the triangles OPR and OQS are similar (see the diagram),

we have

Thus, u1 = kv1. Similarly, u2 = kv2.

17. The vector u has terminal point Q which is the midpoint of the line segment connecting P1and P2.

OP1

OP2 OP1

(OP2 OP1)

O

Q P2P1

OP1OP2

12

u OS

OR

1

1= =

length of

length of

length oof

length of

OQ

OPk

=

O R S

Q(u1, u2)

P(v1, v2)

kv

y

x

v


18. We have x + y + z + w = 0 where

19. Geometrically, given 4 nonzero vectors attach the tail of one to the head of another andcontinue until all 4 have been strung together.The vector from the tail of the rst vectorto the head of the last one will be their sum.

x

yz

w

x + y + z + w

x

yz

w


EXERCISE SET 3.2

1. (a) v = (42 + (3)2)1/2 = 5(c) v = [(5)2 + 02]1/2 = 5(e) v = [(7)2 + 22 + (1)2]1/2 = 54

2. (a) d = [(3 5)2 + (4 7)2]1/2 = 13(c) d = [(7 (7))2 + ((5) (2))2 + (1 (1))2]1/2 = 209

3. (a) Since u + v = (3, 5, 7), then

u + v = [32 + (5)2 + 72]1/2 = 83

(c) Since

2u = [(4)2 + 42 + (6)2]1/2 = 2 17

and

2u = 2 [22 + (2)2 + 32]1/2 = 2 17

then

2u + 2u = 4 17

(e) Since w = [32 + 62 + (4)2]1/2 = 61, then

, ,1

ww =

3

61

6

61

4

61

155

4.

1 = v w 5

5. (a) k = 1, l = 3

(b) no possible solution

6. l = 4

7. Since kv = (k, 2k, 5k), then

kv = [k2 + 4k2 + 25k2]1/2 = |k| 30

If kv = 4, it follows that |k| 30 = 4 or k = 4/ 30.

9. (b) From Part (a), we know that the norm of v/v is 1. But if v = (3, 4), then v = 5.Hence u = v/v = (3/5, 4/5) has norm 1 and has the same direction as v.

10. (b) By the result of Part (a), we have 4u = 4(u cos 30, u sin 30) = 4(3( 3/2), 3(1/2))

= (6 3, 6) and 5 v = 5(v cos(135), v sin(135)) = 5(2(1/ 2), 2(1/ 2))

= (10/ 2, 10/ 2) = (5 2, 5 2). Thus 4u 5v = (6 3 + 5 2, 6 5 2).

11. Note that p p0 = 1 if and only if p p02 = 1. Thus

(x x0)2 + (y y0)

2 + (z z0)2 = 1

The points (x, y, z) which satisfy these equations are just the points on the sphere of radius1 with center (x0, y0, z0); that is, they are all the points whose distance from (x0, y0, z0) is 1.

22

33

0

0

w v2 3w

w w

v

v v


12. First, suppose that u and v are neither similarly nor oppositely directed and that neither isthe zero vector.

If we place the initial point of v at the terminal point of u, then the vectors u, v, and u+ v form a triangle, as shown in (i) below.

Since the length of one side of a trianglesay u + vcannot exceed the sum of thelengths of the other two sides, then u + v < u + v.

Now suppose that u and v have the same direction. From diagram (ii), we see that u + v = u + v. If u and v have opposite directions, then (again, see diagram (ii)) u + v < u + v.

Finally, if either vector is zero, then u + v = u + v.

13. These proofs are for vectors in 3-space. To obtain proofs in 2-space, just delete the 3rdcomponent. Let u = (u1, u2, u3) and v = (v1, v2, v3). Then

(a) u + v = (u1 + v1, u2 + v2, u3 + v3)

= (v1 + u1, v2 + u2, v3 + u3) = v + u

(c) u + 0 = (u1 + 0, u2 + 0, u3 + 0)

= (0+ u1, 0 + u2, 0 + u3)

= (u1, u2, u3) = 0 + u = u

(e) k(lu) = k(lu1, lu2, lu3) = (klu1, klu2, klu3) = (kl)u

14. Again, we work in 3-space. Let u = (u1, u2, u3).

(d) u + (u) = (u1 + (u1), u2 + (u2), u3 + (u3)) = (0, 0, 0) = 0

(g) (k + l)u = ((k + l)u1, (k + l)u2, (k + l)u3)

= (ku1 + lu1, ku2 + lu2, ku3 + lu3)

= (ku1, ku2, ku3) + (lu1, lu2, lu3)

= ku + lu

(h) 1u = (1u1, 1u2, 1u3) = (u1, u2, u3) = u

u

uu

u

u + vu + v

u + v

u + v

v

v

vv

(i) (ii)


15. See Exercise 9. Equality occurs only when u and v have the same direction or when one isthe zero vector.

16. (a) For p = (a, b, c) to be equidistant from the origin and the xz-plane, we must have

= |b|. Thus, a2 + b2 + c2 = b2 or a2 + c2 = 0, so that a = c = 0. That is,p must lie on the y-axis.

(b) For p = (a, b, c) to be farther from the origin than the xz-plane, we must have

> |b|. Thus a2 + b2 + c2 > b2 or a2 + c2 > 0, so that a and c are not bothzero. That is, p cannot lie on the y-axis.

17. (a) If x < 1, then the point x lies inside the circle or sphere of radius one with center atthe origin.

(b) Such points x must satisfy the inequality x x0 > 1.

18. The two triangles pictured are similar since u and ku and v and kv are parallel and theirlengths are proportional; i.e., ku = ku and kv = kv. Therefore ku + kv = ku + v.That is, the corresponding sides of the two triangles are all proportional, where k is theconstant of proportionality. Thus, the vectors k(u + v) and ku + kv have the same length.Moreover, u + v and ku + kv have the same direction because corresponding sides of thetwo triangles are parallel. Hence, k(u + v) and ku + kv have the same direction. Therefore,k(u + v) = ku + kv.

a b c2 2 2+ +

a b c2 2 2+ +


EXERCISE SET 3.3

1. (a) u v = (2)(5) + (3)(7) = 11

(c) u v = (1)(3) + (5)(3) + (4)(3) = 0

2. (a) We have u = [22 + 32]1/2 = 13 and v = [52 + (7)2]1/2 = 74. From Problem1(a), we know that u v = 11. Hence,

(c) From Problem 1(c), we know that u v = 0. Since neither u nor v is the zero vector,this implies that cos = 0.

3. (a) u v = (6)(2)+ (1)(0)+ (4)(3) =0. Thus the vectors are orthogonal.

(b) u v = 1 < 0. Thus is obtuse.

4. (a) Since u a = 0 and a 0, then w1 = (0, 0).

(c) Since u a = (3)(1) + (1)(0) + (7)(5) = 32 and a2 = 12 + 02 + 52 = 26, we have

5. (a) From Problem 4(a), we have

w2 = u w1 = u = (6, 2)

w132

261 0 5

16

130

80

13= =

( , , ) , ,

cos

. = 1113 74

110 8

159

5. (c) From Problem 4(c), we have

w2 = (3, 1, 7) (16/13, 0, 80/13) = (55/13, 1, 11/13)

6. (a) Since |u a| = |4 + 6| = |2| = 2 and a = 5, it follows from Formula (10) that projau= 2/5.

(d) Since |u a| = |3 4 42| = |43| and a = 54, Formula (10) implies that projau =43/ 54.

8. (c) From Part (a) it follows that the vectors (4, 3) and (4, 3) are orthogonal to (3, 4).Divide each component by the norm, 5, to obtain the desired unit vectors.

10. We look for vectors (x, y, z) such that (x, y, z) (5, 2, 3) = 0. That is, such that 5x 2y+ 3z = 0. One such vector is (0, 3, 2). Another is (1, 1, 1). To nd others, assign arbitraryvalues, not both zero, to any two of the variables x, y, and z and solve for the remainingvariable.

12. Since the three points are not collinear (Why?), they do form a triangle. Since

= (1, 3, 2) (4, 2, 1) = 0

the right angle is at B.

13. Let w = (x, y, z) be orthogonal to both u and v. Then u w = 0 implies that x + z = 0 andv w = 0 implies that y + z = 0. That is w = (x, x, x). To transform into a unit vector, wedivide each component by w = 3x2. Thus either (1/ 3, 1/ 3, 1/ 3) or (1/ 3,1/ 3, 1/ 3) will work.

16. (a) We look for a value of k which will make p a nonzero multiple of q. If p = cq, then (2, k) = c(3, 5), so that k/5 = 2/3 or k = 10/3.

(c) We want k such that

p q = p q cos 3

BC

AB


or

If we square both sides and use the quadratic formula, we nd that

The minus sign in the above equation is extraneous because it yields an angle of 2/3.

17. (b) Here

18. From the denition of the norm, we have

u + v2 = (u + v) (u + v)

Using Theorem 3.3.2, we obtain

u + v2 = (u u) + (u v) + (v u)+ (v v)

or

(*) u + v2 = u2 + 2(u v) + v2

Similarly,

(**) u v2 = u2 2(u v) + v2

If we add Equations (*)and (**), we obtain the desired result.

19. If we subtract Equation (**) from Equation (*) in the solution to Problem 18, we obtain

u + v2 u v2 = 4(u v)

If we then divide both sides by 4, we obtain the desired result.

D =+

+=

4 2 1 5 2

4 1

1

172 2

( ) ( )

( ) ( )

k = 60 34 3

33

6 5 4 3 51

22 2 2+ = + +

k k


20. Let u1, u2, and u3 be three sides of thecube as shown. The diagonal of the faceof the cube determined by u1 and u2 is b= u1 + u2; the diagonal of the cube itselfis a = u1 + u2 + u3. The angle betweena and b is given by

Because u1, u2, and u3 are mutually orthogonal, we have ui uj = 0 whenever i j. Also ui ui = ui

2 and u1 = u2 = u3. Thus

That is, =arccos(2/ 6).

21. (a) Let i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) denote the unit vectors along the x, y,and z axes, respectively. If v is the arbitrary vector (a, b, c), then we can write v = ai+ bj + ck. Hence, the angle between v and i is given by

since i = 1 and i j = i k = 0.

cos = =+ +

=

vv ii

vv ii vv

a

a b c

a2 2 2

cos =+

+ + +

=

uu uu

uu uu uu uu uu

12

22

12

22

32

12

22

2 uuuu

uu uu

12

12

12

3 2

2

6=

cos

( ) ( )

( ) (

=

=

+ + +

+ +

aa bb

bb

uu uu uu uu uu

uu uu uu uu

aa

1 2 3 1 2

1 2 3 11 2 3 1 2 1 2+ + + +uu uu uu uu uu uu) ( ) ( )


23. By the results of Exercise 21, we have that if vi = (ai, bi, ci) for i = 1 and 2, then cos i =

, cos i = , and cos i = . Now

24. (a) Area = 19 units

(b) B(3, 12); area = 2 19 = 38 units

25. Note that

v (k1w1 + k2w2) = k1(v w1) + k2(v w2) = 0

because, by hypothesis, v w1 = v w2 = 0. Therefore v is orthogonal to k1w1 + k2w2 for anyscalars k1 and k2.

26. Note that w is a multiple of u plus a multiple of v and thus lies in the plane determined byu and v. Let be the angle between u and w and let be the angle between v and w. Then

Similarly, we can show that

Since both and lie between 0 and , then cos = cos implies that = .

27. (a) The inner product x y is dened only if both x and y are vectors, but here v w is ascalar.

(b) We can add two vectors or two scalars, but not one of each.

cos

= = + v wv w

u v

w

kl

cos = =+

=

uu

uu vv

uu vv uu uu

uu

ww

ww

k l

k

k

+

=

+

vv

uu vv

lk

k

lk

2

ww

ww

v v1 2 1 2and are orthogonal =

+

vv vv 0

1 2 1a a b bb c c

a a b b c c2 1 2

1 2

1 2

1 2

1 2

1 2

1

0+ =

+ +vv vv vv vv vv

cos cos cos

vv2

1 2 1 2 1 2

0=

+ + =cos cos cos 00

ci

ivv

bi

ivv

ai

ivv


27. (c) The norm of x is dened only for x a vector, but u v is a scalar.

(d) Again, the dot product of a scalar and a vector is undened.

28. Assume that neither a nor u is the zero vector. If projau = projua, then

Thus, either u a = 0 and the two vectors are orthogonal, or

In this case, a is clearly a scalar multiple of u. If we let a = ku, then

Hence, k = 1, so that a = u.

29. If, for instance, u = (1, 0, 0), v = (0, 1, 0) and w = (0, 0, 1), we have u v = u w = 0, butv w.

27. Suppose that r = c1u + c2v + c3w. Then

u r = c1u u + c2u v + c3u w = c1u2

since u v = u w = 0. Similarly, v r = c2v2 and w r = c3w

2. Since we know u r, v r, and w r, this allows us to solve for c1, c2, and c3. We nd that

31. This is just the Pythagorean Theorem.

u

u + vv

rruu rr

uuuu

vv rr

vvvv

ww rr

wwww= + +

2 2 2

aa

aa

uu

uu

uu

uu2 2 2 2

= =

k

k k

aa

aa

uu

uu2 2

=

uu aa

aaaa

aa uu

uuuu

uu aa

uuuu

2 2 2= =


EXERCISE SET 3.4

1. (a)

(c) Since

we have

(e) Since

v 2w = (0, 2, 3) (4, 12, 14) = (4, 10, 17)

we have

2. (a) By Theorem 3.4.1, u v will be orthogonal to both u and v where

so (1, 2, 1), for instance, is orthogonal to both u and v.

3. (a) Since u v = (7, 1, 3), the area of the parallelogram is u v = 59.

uu vv =

=, , ( ,

4 2

1 5

6 2

3 5

6 4

3 118 366 18, )

uu vv ww( 2 ) =

, ,2 1

10 17

3 1

4 17

3 2

4 11044 55 22

= ( , , )

( )uu vv ww =

=, ,

9 6

6 7

4 6

2 7

4 9

2 6(( , , )27 40 42

uu vv =

= , , ( ,

2 1

2 3

3 1

0 3

3 2

0 24 99 6, )

vv ww =

= , , ( , ,

2 3

6 7

0 3

2 7

0 2

2 632 6 )4

165

3. (c) Since u and v are proportional, they lie on the same line and hence the area of theparallelogram they determine is zero, which is, of course, u v.

4. (a) We have u = = (1, 5, 2) and v = = (2, 0, 3). Thus

so that

u v = [(15)2 + 72 + 102]1/2 = 374

Thus the area of the triangle is 374/2.

6. For Part (a), we have

and

Hence

u v = (v u)

For Part (b), we have v + w = (7, 2, 3). Hence

On the other hand, from Part (a)we have

u v = (2, 22, 6)

Also

uu ww =

= , , (

1 2

2 1

5 2

1 1

5 1

1 23,, , )7 11

uu vv ww( ) + =

=, ,

1 2

2 3

5 2

7 3

5 1

7 2( , , )1 29 17

vv uu =

= , , ( ,

0 2

1 2

6 2

5 2

6 0

5 12 22 6, )

uu vv =

=, , ( ,

1 2

0 2

5 2

6 2

5 1

6 02 222 6, )

uu vv =

= , , ( ,

5 2

0 3

1 2

2 3

1 5

2 015 , )7 10

PR

PQ


Thus

(u v) + (u w) = (2, 22, 6) + (3, 7, 11)

= (1, 29, 17)

and we have that

u (v + w) = (u v) + (u w)

The proof of Part (c) is similar to the proof of Part (b).

For Part (d), we already have that u v = (2, 22, 6). Hence k(u v) = (10, 110, 30).But ku = (25, 5, 10) and kv = (30, 0, 10). Thus

and

Thus

k(u v) = (ku) v = u (kv)

For Part (e), we have

Similarly 0 u = 0.

Finally, for Part (f), we have

uu uu =

=, , ( ,

1 2

1 2

5 2

5 2

5 1

5 10 00 0, ) = 00

uu 00 =

=, , ( ,

1 2

0 0

5 2

0 0

5 1

0 00 00 0, ) = 00

uu vv( ) =

k , ,1 2

0 10

5 2

30 10

( , , )

5 1

30 0

10 110 30

=

( , ,kuu vv) =

5 10

0 2

25 10

6 2

25 5

6 0

=( , , )10 110 30


7. Choose any nonzero vector w which is not parallel to u. For instance, let w = (1, 0, 0) or(0, 1, 0). Then v = u w will be orthogonal to u. Note that if u and w were parallel, thenv = u w would be the zero vector.

Alternatively, let w = (x, y, z). Then w orthogonal to u implies 2x 3y + 5z = 0. Nowassign nonzero values to any two of the variables x, y, and z and solve for the remainingvariable.

8. (a) We have

9. (e) Since (u w) v = v (u w) is a determinant whose rows are the components of v,u, and w, respectively, we interchange Rows 1 and 2 to obtain the determinant whichrepresents u (v w). Since the value of this determinant is 3, we have (u w) v = 3.

10. (a) Call this volume V. Then, since

we have V = 16.

11. (a) Since the determinant

the vectors do not lie in the same plane.

12. For a vector to be parallel to the yz-plane, it must be perpendicular to the vector (1, 0, 0),

so we are looking for a vector which is perpendicular to both (1, 0, 0) and (3, 1, 2). Such

a vector is (1, 0, 0) (3, 1, 2) = (0, 2, 1). Since we want a unit vector, we divide through

by the norm to obtain the vector (0, 2/ 5, 1/ 5). Obviously any vector parallel to this

one, such as (0, 2/ 5, 1/ 5), will also work.

=

1 2 1

3 0 2

5 4 0

16 0

2 6 2

0 4 2

2 2 4

24 2

2 42

6 2

4 216

=

+

=

uu vv ww ( ) ( ) ( =

= +

1 2 4

3 4 2

1 2 5

20 4 2 15 + + = 2 4 6 4 10) ( )


15. By Theorem 3.4.2, we have

(u + v) (u v) = u (u v) + v (u v)

= (u u) + (u (v)) + (v u) + (v (v))

= 0 (u v) (u v) (v v)

= 2(u v)

17. (a) The area of the triangle with sides and is the same as the area of the trianglewith sides (1, 2, 2) and (1, 1, 1) where we have moved A to the origin andtranslated B and C accordingly. This area is 12(1, 2, 2) (1, 1, 1) =

12(4, 1, 3) =

26/2.

18. If the vector u and the given line make an angle , then the distance D between the pointand the line is given by

D = u sin (Why?)

= u v / v (Why?)

19. (a) Let u = = (4, 0, 2) and v = = (3, 2, 4). Then the distance we want is

(4, 0, 2) (3, 2, 4)/(3, 2, 4) = (4, 22, 8)/ 29. = 2 141/ 29

20. We have that u v = u v cos and u v = u v sin . Thus if u v 0, then u, v,and cos are all nonzero, and

21. (b) One vector n which is perpendicular to the plane containing v and w is given by

n = w v = (1, 3, 3) (1, 1, 2) = (3, 1, 2)

Therefore the angle between u and n is given by

= =

cos cos

.

1 1 9

14

0 8726

uu nn

uu nn

rradians or( . )49 99

tansin

cos

= =

=

uu vv

uu vvuu vvuu vv

uu vv

uu vv

AB

AP

AC

AB


Hence the angle between u and the plane is given by

= 2 .6982 radians (or 4019)

If we had interchanged the roles of v and w in the formula for n so that

n = v w = (3, 1, 2), then we would have obtained = cos1 2.269 radians or 130.0052. In this case, = 2.

In either case, note that may be computed using the formula

24. (a) By Theorems 3.4.2 and 3.2.1 and the denition of k0,

(u + kv) v = (u v) + (kv v)

= (u v) + k(v v)

= (u v) + k0

= u v

(b) Let u = (u1, u2, u3), v = (v1, v2, v3), and z = (z1, z2, z3). Then we know from the textthat

Since x y = y x for any vectors x and y, we have

u v z( ) =

u u u

v v v

z z z

1 2 3

1 2 3

1 2 3

cos .=

1 u n

u n

9

14


25. (a) By Theorem 3.4.1, we know that the vector v w is perpendicular to both v and w.Hence v w is perpendicular to every vector in the plane determined by v and w;moreover the only vectors perpendicular to v w which share its initial point must bein this plane. But also by Theorem 3.4.1, u (v w) is perpendicular to v w for anyvector u 0 and hence must lie in the plane determined by v and w.

(b) The argument is completely similar to Part (a), above.

26. Let u = (u1, u2, u3), v = (v1, v2, v3), and w = (w1, w2, w3). Following the hint, we nd

v i = (0, v3, v2)

and

u (v i) = (u2v2 u3v3, u1v2, u1v3)

But

(u i)v (u v)i = u1(v1, v2, v3) (u1v1 + u2v2 + u3v3)(1, 0, 0)

= (u1v1, u1v2, u1v3) (u1v1 + u2v2 + u3v3, 0, 0)

= (u2v2 u3v3 , u1v2, u1v3)

= u (v i)

Similarly,

u (v j) = (u j)v (u v)j

and

u (v k) = (u k)v (u v)k

( ) = ( ) =u z v v u z

v v v1 2 33

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

u u u

u u u

v v v

z z z

z z z

=

= u ( )v z


Now write w = (w1, w2, w3) = w1i + w2j + w3k. Then

u (v w) = u (v (w1i + w2j + w3k))

= w1[u (v i)] + w2[u (v j)] + w3[u (v k)]

= w1[(u i)v (u v)i] + w2[(u j)v (u v)j] + w3[(u k)v (u v)k]

= (u (w1i + w2j + w3k))v (u v)(w1i + w2j + w3k)

= (u w)v (u v)w

29. If a, b, c, and d lie in the same plane, then (a b) and (c d) are both perpendicular tothis plane, and are therefore parallel. Hence, their cross-product is zero.

30. Recall that |a (b c)| represents the volume of the parallelepiped with sides a, b, and c,which is the area of its base times its height. The volume of the tetrahedron is just 13 (areaof its base) times (its height). The two heights are the same, but the area of the base of thetetrahedron is half of the area of the base of the parallelepiped. Hence, the volume of thetetrahedron is 13 (

12|a (b c)|).

31. (a) The required volume is

= 2/3

33. Let u = (u1, u2, u3), v = (v1, v2, v3), and w = (w1, w2, w3).

For Part (c), we have

u w = (u2w3 u3w2, u3w1 u1w3, u1w2 u2w1)

and

v w = (v2w3 v3w2, v3w1 v1w3, v1w2 v2w1)

1

61 3 2 2 0 3 2 3 1 2 3 3 1 3 0 2 1( , , ) (( , , ) ( , ,+ + + ))

( , , ) ( , , )

3

1

64 4 3 6 10 4

( )= ( )


Thus

(u w) + (v w)

= ([u2 + v2]w3 [u3 + v3]w2, [u3 + v3]w1 [u1 + v1]w3, [u1 + v1]w2 [u2 + v2]w1)

But, by denition, this is just (u + v) w.

For Part (d), we have

k(u v) = (k[u2v3 u3v2], k[u3v1 u1v3], k[u1v2 u2v1])

and

(ku) v = (ku2v3 ku3v3, ku3v1 ku1v3, ku1v2 ku2v1)

Thus, k(u v) = (ku) v. The identity k(u v) = u (kv) may be proved in an analogousway.

35. (a) Observe that u v is perpendicular to both u and v, and hence to all vectors in theplane which they determine. Similarly, w = v (u v) is perpendicular to both v andto u v. Hence, it must lie on the line through the origin perpendicular to v and in theplane determined by u and v.

(b) From the above, v w = 0. Applying Part (d) of Theorem 3.7.1, we have

w = v (u v) = (v v)u (v u)v

so that

u w = (v v)(u u) (v u)(u v)

= v2u2 (u v)2

36. It is not valid. For instance, let u = (1, 0, 0), v = (0, 1, 0), and w = (1, 1, 0). Then u v =u w = (0, 0, 1), which is a vector perpendicular to u, v, and w. However, v w.

37. The expression u (v w) is clearly well-dened.

Since the cross product is not associative, the expression u v w is not well-denedbecause the result is dependent upon the order in which we compute the cross products,i.e., upon the way in which we insert the parentheses. For example, (i j) j = k j = ibut i (j j) = i 0 = 0.


The expression u v w may be deemed to be acceptable because there is only onemeaningful way to insert parenthesis, namely, u (v w). The alternative, (u v) w,does not make sense because it is the cross product of a scalar with a vector.

38. If either u or v is the zero vector, then u v = 0. If neither u nor v is the zero vector, thenu v = u v sin where is the angle between u and v. Thus if u v = 0, then sin = 0, so that u and v are parallel.


EXERCISE SET 3.5

4. (a) We have

= (2, 1, 2) and = (3, 1, 2)

Thus = (0, 10, 5) is perpendicular to the plane determined by and

and P, say, is a point in that plane. Hence, an equation for the plane is

0(x + 4)+ 10(y + 1) 5(z + 1) = 0

or

2y z + 1 = 0

5. (a) Normal vectors for the planes are (4, 1, 2) and (7, 3, 4). Since these vectors are notmultiples of one another, the planes are not parallel.

(b) Normal vectors are (1, 4, 3) and (3, 12, 9). Since one vector is three times theother, the planes are parallel.

6. (a) A normal vector for the plane is (1, 2, 3) and a direction vector for the line is (4, 1,2). The inner product of these two vectors is 4 2 + 6 = 0, and therefore they areperpendicular. This guarantees that the line and the plane are parallel.

7. (a) Normal vectors for the planes are (3, 1, 1) and (1, 0, 2). Since the inner product ofthese two vectors is not zero, the planes are not perpendicular.

8. (a) A normal vector for the plane is (2, 1, 1) and a direction vector for the line is (4, 2,2). Since one of these vectors is a multiple of the other, the line and the plane areperpendicular.

PR

PQ

PR

PQ

PR

PQ

175

10. (a) Call the points P and Q and call the line . Then the vector = (2, 4, 8) is parallelto and the point P = (5, 2, 4) lies on . Hence, one set of parametric equations for is: x = 5 + t, y = 2 + 2t, z = 4 4t where t is any real number.

11. (a) As in Example 6, we solve the two equations simultaneously. If we eliminate y, wehave x + 7z + 12 = 0. Let, say, z = t, so that x = 12 7t, and substitute these valuesinto the equation for either plane to get y = 41 23t.

Alternatively, recall that a direction vector for the line is just the cross-product ofthe normal vectors for the two planes, i.e.,

(7, 2, 3) (3, 1, 2) = (7, 23, 1)

Thus if we can nd a point which lies on the line (that is, any point whose coordinatessatisfy the equations for both planes), we are done. If we set z = 0 and solve the twoequations simultaneously, we get x = 12 and y = 41, so that x = 12 7t, y = 41 23t, z = 0 + t is one set of equations for the line (see above).

13. (a) Since the normal vectors (1, 2, 4) and (2, 4, 8) are parallel, so are the planes.

(b) Since the normal vectors (3, 0, 1) and (1, 0, 3) are not parallel, neither are theplanes.

14. (a) Since (2, 1, 4) (1, 2, 1) = 0, the planes are perpendicular.

(b) Since (3, 0, 2) (1, 1, 1) = 1 0, the planes are not perpendicular.

16. (a) Since 6(0) + 4t 4t = 0 for all t, it follows that every point on the line also lies in theplane.

(b) The normal to the plane is n = (5, 3, 3); the line is parallel to v = (0, 1, 1). But n v= 0, so n and v are perpendicular. Thus v, and therefore the line, are parallel to theplane. To conclude that the line lies below the plane, simply note that (0, 0, 1/3) is inthe plane and (0, 0, 0) is on the line.

(c) Here n = (6, 2, 2) and v is the same as before. Again, n v = 0, so the line and theplane are parallel. Since (0, 0, 0) lies on the line and (0, 0, 3/2) lies in the plane,then the line is above the plane.

17. Since the plane is perpendicular to a line with direction (2, 3, 5), we can use that vectoras a normal to the plane. The point-normal form then yields the equation 2(x + 2) + 3(y 1) 5(z 7) = 0, or 2x + 3y 5z + 36 = 0.

PQ


19. (a) Since the vector (0, 0, 1) is perpendicular to the xy-plane, we can use this as thenormal for the plane. The point-normal form then yields the equation z z0 = 0. Thisequation could just as well have been derived by inspection, since it represents the setof all points with xed z and x and y arbitrary.

20. The plane will have normal (7, 4, 2), so the point-normal form yields 7x + 4y 2 z = 0.

21. A normal to the plane is n = (5, 2, 1) and the point (3, 6, 7) is in the desired plane.Hence, an equation for the plane is 5(x 3) 2(y + 6) + (z 7) = 0 or 5x 2y + z 34 = 0.

22. If we substitute x = 9 5t, y = 1 t, and z = 3 + t into the equation for the plane, we ndthat t = 40/3. Substituting this value for t into the parametric equations for the line yieldsx = 173/3, y = 43/3, and z = 49/3.

24. The two planes intersect at points given by (5 2t, 7 6t, t). Two such points are (5, 7,0) and (3, 1, 1). The plane ax + by + cz + d = 0 through these points and also through(2, 4, 1) will satisfy the equations

5a 7b + d = 0

3a b c + d = 0

2a + 4b c + d = 0

This system of equations has the solution a = (1/2)t, b = (1/2)t, c = 2t, d = t where t isarbitrary. Thus, if we let t = 2, we obtain the equation

x y 4z 2 = 0

for the desired plane.

Alternatively, note that the line of intersection of the two planes has direction given bythe vector v = (2, 6, 1). It also contains the point (5, 7, 0). The direction of the lineconnecting this point and (2, 4, 1) is given by w = (7, 11, 1). Thus the vector v w = (5,5, 20) is normal to the desired plane, so that a point-normal form for the plane is

(x 2) (y 4) 4(z + 1) = 0

or

x y 4z 2 = 0


25. Call the points A, B, C, and D, respectively. Since the vectors = (1, 2, 4) and =(2, 1, 2) are not parallel, then the points A, B, and C do determine a plane (and not justa line). A normal to this plane is = (0, 10, 5). Therefore an equation for theplane is

2y z + 1 = 0

Since the coordinates of the point D satisfy this equation, all four points must lie in thesame plane.

Alternatively, it would sufce to show that (for instance) and are parallel, so that the planes determined by A, B, and C and A, D, and C are parallel.Since they have points in common, they must coincide.

27. Normals to the two planes are (4, 2, 2) and (3, 3, 6) or, simplifying, n1 = (2, 1, 1) andn2 = (1, 1, 2). A normal n to a plane which is perpendicular to both of the given planesmust be perpendicular to both n1 and n2. That is, n = n1 n2 = (1, 5, 3). The plane with thisnormal which passes through the point (2, 1, 5) has the equation

(x + 2) + 5(y 1) + 3(z 5) = 0

or

x + 5y + 3z 18 = 0

28. The line of intersection of the given planes has the equations x = 109 t3 , y =

3118

t3,

z = t and hence has direction (1, 1, 3). The plane with this normal vector which passesthrough the point (2, 1, 4) is the one we are looking for. Its equation is

(x 2) + (y + 1) 3(z 4) = 0

Note that the normal vector for the desired plane can also be obtained by computing thecross product of the normal vectors of the two given planes.

30. The directions of the two lines are given by the vectors (2, 1, 1) and (2, 1, 1). Sinceeach is a multiple of the other, they represent the same direction. The rst line passes (forexample) through the point (3, 4, 1) and the second line passes through the points (5, 1,7) and (7, 0, 8), among others. Either of the methods of Example 2 will yield an equationfor the plane determined by these points.

31. If, for instance, we set t = 0 and t = 1 in the line equation, we obtain the points (0, 1, 3)and (1, 0, 5). These, together with the given point and the methods of Example 2, willyield an equation for the desired plane.

DC

AD

BC

AB

BC

AB

BC

AB


33. The plane we are looking for is just the set of all points P = (x, y, z) such that the distancesfrom P to the two xed points are equal. If we equate the squares of these distances, wehave

(x + 1)2 + (y + 4)2 + (z + 2)2 = (x 0)2 + (y + 2)2 + (z 2)2

or

2x + 1 + 8y + 16 + 4z + 4 = 4y + 4 4z + 4

or

2x + 4y + 8z + 13 = 0

34. The vector v = (1, 2, 5) is parallel to the line and the vector n = (3, 1, 1) isperpendicular to the plane. But n v = 0 and the result follows.

35. We change the parameter in the equations for the second line from t to s. The two lines willthen intersect if we can nd values of s and t such that the x, y, and z coordinates for thetwo lines are equal; that is, if there are values for s and t such that

4t + 3 = 12s 1

t + 4 = 6s + 7

1 = 3s + 5

This system of equations has the solution t = 5 and s = 4/3. If we then substitute t = 5into the equations for the rst line or s = 4/3 into the equations for the second line, we ndthat x = 17, y = 1, and z = 1 is the point of intersection.

36. The vector v1 = (4, 1, 0) is parallel to the first line and v2 = (4, 2, 1) is parallel to thesecond line. Hence n = v1 v2 = (1, 4, 4) is perpendicular to both lines and is therefore anormal vector for the plane determined by the lines. If we substitute t = 0 into theparametric equations for the rst line, we see that (3, 4, 1) must lie in the plane. Hence, anequation for the plane is

(x 3) 4(y 4) + 4(z 1) = 0

or

x 4y + 4z + 9 = 0


37. (a) If we set z = t and solve for x and y in terms of z, then we nd that

38. Call the plane Ax + By + Cz + D = 0. Since the points (a, 0, 0), (0, b, 0), and (0, 0, c) liein this plane, we have

aA + D = 0

bB + D = 0

cC + D = 0

Thus A = D/a, B = D/b, and C = D/c and an equation for the plane is

Alternatively, let P, Q, and R denote the points (a, 0, 0), (0, b, 0), and (0, 0, c),respectively. Then

= (a, b, 0) and = (0, b, c)

So

n = = (bc, ac, ab)

is a normal vector for the plane. Since P, say, lies on the plane, we have

bc(x a)+ acy + abz = 0

Dividing the above equation by abc yields the desired result.

39. (b) By Theorem 3.5.2, the distance is

D =( ) + ( ) ( )

+ + ( )=

2 1 3 2 4 1 1

2 3 4

1

292 2 2

QR

PQ

QR

PQ

x

a

y

b

z

c+ + = 1

x t y t z t= + = =11

23

7

23

41

23

1

23, ,


40. (a) The point (0, 0, 1) lies in the first plane. The distance between this point and thesecond plane is

(b) The two planes coincide, so the distance between them is zero.

41. (a)

(b)

(c)

42. First observe that if we substitute the values for x, y, and z from the line equations into thesymmetric equations, we obtain the equations t = t = t, which hold for all values of t. Thusevery point on the line satises the symmetric equations.

It remains to show that every point (x, y, z) which satises the symmetric equationsalso lies on the line. That is, we must nd a value of t for which the line equations hold.Clearly

is the desired value of t.

44. (a) The symmetric equations for the line are

Thus, two of the planes are given by

or, equivalently,

x 2y 17 = and x + 4z 27 = 0

x y x z=

+ =

7

4

5

2

7

4

5

1and

x y z=

+=

7

4

5

2

5

1

tx x

a

y y

b

z z

c=

=

=

0 0 0

d

d

d

=

=

=

30

11

382

110 since point is on the linne

D =

+ ( ) + ( )=

2 3

6 8 2

1

2 262 2 2


45. (a) Normals to the two planes are (1, 0, 0) and (2, 1, 1). The angle between them isgiven by

Thus = cos1 (2/ 6) 351552.

47. If we substitute any value of the parametersay t0into r = r0 + tv and t0 into r = r0 tv,we clearly obtain the same point. Hence, the two lines coincide. They both pass through thepoint r0 and both are parallel to v.

48. Since the vector (a, b, c) is both a normal vector to the plane and a direction vector for theline, the line must be perpendicular to the plane.

49. The equation r = (1 t)r1 + tr2 can be rewritten as r = r1 + t(r2 r1). This represents a linethrough the point P1 with direction r2 r1. If t = 0, we have the point P1. If t = 1, we havethe point P2. If 0 < t < 1, we have a point on the line segment connecting P1 and P2. Hencethe given equation represents this line segment.

cos, , , ,

= ( ) ( )+ +

=

1 0 0 2 1 1

1 4 1 1

2

6


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Anton Linear Algebra 10th edition Solutions Set 3.3