Table Of ContentInstructors Solution Manual for
Introduction to Linear Algebra
Geza Schay and Dennis Wortman
1.1.2.r = (5,5), p q = ( 3,1), q p = (3, 1).
− − − −
1.1.4. The midpoint of the edge from P(0,1,1) to Q(1,1,1) has posi-
tion vector m = (p+q)/2 = (1/2,1,1). The center of the face through
(0,0,1),(0,1,1),(1,1,1) and (1,0,1) has position vector f = [(0,0,1) +
(1,1,1)]/2 = (1/2,1/2,1). Similarly, the center of the whole cube has posi-
tion vectorc = [(0,0,0)+(1,1,1)]/2 = (1/2,1/2,1/2).
1.1.6.Let p = (p ,p ) and q = (q ,q ).
1 2 1 2
ProofofProperty 4:
cp = 0isequivalentto(cp ,cp ) = (0,0),thatis,tocp = 0andcp = 0.
1 2 1 2
Now, theselast two relations hold ifandonlyifeitherc = 0with arbitrary p
1
and p orc = 0 but p = p = 0,thatis,p = 0.
2 1 2
(cid:2)
ProofofProperty 5:
( c)p = ( cp , cp ) = c( p , p ) = c( p) and
1 2 1 2
− − − − − −
( c)p = ( cp , cp ) = (cp ,cp ) = (cp).
1 2 1 2
− − − − −
ProofofProperty 6:
c(p q) = c(p q ,p q ) = (cp cq ,cp cq ) = (cp ,cp )
1 1 2 2 1 1 2 2 1 2
− − − − − −
(cq ,cq ) = cp cq.
1 2
−
ProofofProperty 7:
(c d)p = ((c d)p ,(c d)p ) = (cp ,cp ) (dp ,dp ) = c(p ,p )
1 2 1 2 1 2 1 2
− − − − −
d(p ,p ) = cp dp.
1 2
−
1.1.8.Letp = −O−O→. Thenforeachi, r = r p. Ifr denotestheposition
(cid:1) (cid:1)i i− (cid:1)
vectorofthecenterofmass relativeto O ,then
(cid:1)
n n n
1 1 1
r = r p = m r p = m r m p
(cid:1) i i i i i
− M − M − M
i=1 i=1 i=1
(cid:1) (cid:1) (cid:1)
n n
1 1
= m (r p) = m r .
M i i− M i (cid:1)i
i=1 i=1
(cid:1) (cid:1)
1.1.10. By the solution of Exercise 1.1.9, the centroid P of the face
(cid:1)
formedbyB, C,andD haspositionvectorp = 1(b+c+d). Theposition
(cid:1) 3
vectorofthepointthat is onefourthoftheway fromP toAis
(cid:1)
1 3 1 1 1 1
p + (a p) = p + a = (b+c+d)+ a = (a+b+c+d) = p.
(cid:1) (cid:1) (cid:1)
4 − 4 4 4 4 4
1.2.2. If p = (2, 2,1) and q = (2,3,2), then p+q = (4,1,3) and
−
2
p q = (0, 5, 1). Thus p+q 2 = p q 2 = 26, p 2 = 9, q 2 = 17.
He−nce p+q−2 =−p 2+ q 2|. Thev|ecto|rsp−an|dqspan|ap|arallelo|gr|amwith
| | | | | |
diagonals of equal length, that is, they span a rectangle. Also, the equation
p+q 2 = p 2 + q 2 expresses the Pythagorean Theorem for the triangle
| | | | | |
withedgesp,qandp+q,whichshowsagainthatpandqareperpendicular
to each other.
1.2.4.
p q 26
cosθ = · = = 0.997; θ 3.065 radians.
p q −√20 √34 − ≈
| || | ·
1.2.6.Betweentwo spacediagonals,p = (1,1,1)andq = (1,1, 1):
−
p q 1+1 1 1
cosθ = · = − = , θ 70.5 .
◦
p q √3 √3 3 ≈
| || | ·
Betweentwo facediagonals, p = (1,0, 1) andq = (0,1, 1):
− −
p q 0+0+1 1
cosθ = · = = , θ = 60 .
◦
p q √2 √2 2
| || | ·
Betweenaspace diagonal andanedge, p = (1,1,1) and q = (1,0,0):
p q 1+0+0 1
cosθ = · = = , θ 54.7 .
◦
p q √3 √1 √3 ≈
| || | ·
Betweenaspacediagonalandafacediagonal,p = (1,1,1)andq = (1,1,0):
p q 1+1+0 2
cosθ = · = = , θ 35.3 .
◦
p q √3 √2 √6 ≈
| || | ·
The othertypesofangles areobvious: 45 or90 .
◦ ◦
1.2.8. a) Let p = −A→B = (1,1) and q = −B−→C = (2, 5). The projection
−
ofponto qisthen
p q 2 5 6 15
p = · q = − (2, 5) = , .
1 q 2 4+25 − −29 29
| | (cid:2) (cid:3)
3
b)IfD denotes thefoot oftheprojectionofpontoq,then
6 15 35 14
−A−→D = p p = (1,1) , = , ,
− 1 − −29 29 29 29
(cid:2) (cid:3) (cid:2) (cid:3)
and
1 7
dist Afrom BC = −A−→D = √352 +142 = .
| | 29 √29
(cid:4) (cid:5)
c)
1 1 7 7
Areaof∆ = −B−→C −A−→D = √22 +52 = .
2| || | 2 · √29 2
1.2.10.
p+q 2 = (p+q) (p+q) = p2 +2p q+q2,
| | · ·
and
p q 2 = (p q) (p q) = p2 2p q+q2.
| − | − · − − ·
Thus
p+q 2 + p q 2 = 2p2 +2q2 = 2 p 2 +2 q 2.
| | | − | | | | |
Geometrically, this result says that the sum of the squares of the lengths of
the diagonals of a parallelogram equals the sum of the squares of the lengths
ofthefoursides.
1.2.12.Theresultistriviallytrueifq = 0,sowemayassumethatq = 0.
(cid:2)
Followingthehint,letf(λ) = p λq 2 = p 2 (2p q)λ+λ2 q 2 forany
| − | | | − · | |
scalarλ.Now,f isaquadraticfunctionofλandf(λ) 0forallλ. Hencethe
≥
graph of f is a parabola opening upwards and the minimum value will occur
at the vertex, where f (λ) = 2 q 2λ 2(p q) = 0. Thus λ = pq yields
(cid:1) | | − · 0 q·2
the minimum value of f, which is f(λ ) = p 2 pq. Cauchy’s i|n|equality
0 | | − q·2
follows from the fact that f(λ ) 0. Finally, equa|lit|y occurs if and only if
0
≥
q = 0or p λq = 0,thatis,ifandonlyifq = 0orp = λq,foranyλ,and
| − |
thus, ifand onlyifpand qare parallel.
1.2.14.
a. Substituting r = p q into the Triangle Inequality, r+q r + q
− | | ≤ | | | |
yields the inequality p q p q . Similarly, q p q p =
| | − | | ≤ | − | | | − | | ≤ | − |
p q .Combining thesetwo inequalities yields thedesired result.
| − |
4
b. Equalityoccurswhenthevectorspandqareparallelandpointinthesame
direction; that is, p = λq, with λ 0, or q = λp, with λ 0. This follows
≥ ≥
from the fact that equality occurs in the Triangle Inequality if and only if the
vectors areparalleland point in thesamedirection.
1.2.16.
a. From Figure 1.12 weseethat
p
3
cosα = .
3 p
| |
Ontheotherhand,
p k (p ,p ,p ) (0,0,1) p
u k = · = 1 2 3 · = 3 .
p · p p p
| | | | | |
and so u k = cosα . Theothertwo relationscanbeobtainedsimilarly.
P 3
·
b.. Byexpressingeachofthecosinesasinthefirstequationabove, weobtain
p2 p2 p2 p2 +p2 +p2
cos2α +cos2α +cos2α = 1 + 2 + 3 = 1 2 3 = 1.
1 2 3 p 2 p 2 p 2 p 2
| | | | | | | |
This is thethree-dimensional analog oftheformulasin2α+cos2α = 1.
c.
p = (p ,p ,p ) = ( p cosα , p cosα , p cosα )
1 2 3 1 2 3
| | | | | |
= p (cosα ,cosα ,cosα ).
1 2 3
| |
1.2.18.Theanglebetween e = (cosα,sinα) and e = (cosβ,sinβ) is
a b
α β andthus
−
e e = e e cos(α β) = cos(α β)
a b a b
· | || | − −
Also
e e = cosαcosβ +sinαsinβ.
a b
·
The result follows from combining thetwo equations.
1.2.20.
a. Let α denote the angle between the axes labelled ξ and η. Then p =
1
p1 +p2cosαand p = p2 +p1cosα,fromwhich itfollowsthat
2
p p cosα
p1 = 1 − 2 ,
1 cos2α
−
5
and
p p cosα
p2 = 2 − 1 .
1 cos2α
−
If u and u denote the unit vectors in the directions of the axes, then p =
ξ η
p1u +p2u and q = q1u +q2u and thus
ξ η ξ η
p q = p1q1 +p2q2 +p1q2cosα+p2q1cosα.
·
Substituting the above expressions for p1 and p2 into the last equation and
simplifying yields p q = p q1 + p q2 By a symmetric argument, it also
1 2
·
follows thatp q = p1q +p2q .
1 2
·
b. Fromthefirstexpressionforp qabovewecanreadoffg = g = 1and
11 22
·
g = g = cosα.
12 21
1.3.2. The scalar parametric form is x = 1, y = 2 + t, z = 4 and the
−
nonparametricequations are: x = 1and z = 4 (y isarbitrary).
1.3.4.v = −P−P→ = (0,8, 7), p = p +tv = (1, 2,4)+t(0,8, 7). In
0 1 − 0 − −
scalar form: x = 1, y = 2+8t, z = 4 7t. The nonparametric equations
− −
are: x = 1and y+2 = z 4.
−
8 7
−
1.3.6.v = (3, 4,3)andthevectorparametricformoftheequationofthe
−
line is p = p +tv = (5,4, 8)+t(3, 4,3) and thus the scalar parametric
0 − −
formis x = 5+3t, y = 4 4t, z = 8+3t, andthenonparametricformis
− −
x 5 = y 4 = z+8.
− −
3 4 3
−
1.3.8. The direction vector n of the desired line is orthogonal to the nor-
mal vectors u = (3, 4,3) and v = ( 1,3,4) of the two planes. Proceeding
− −
as in Example 1.3.7 in the text, we can express the equation of the plane
throughtheorigindeterminedbythevectorsuandvintheformp = su+tv,
where s and t are parameters, or, in scalar form, as x = 3s t, y =
−
4s + 3t, z = 3s + 4t. Eliminating the parameters results in the equation
−
5x+ 3y z = 0, and thus n = (5,3, 1). Therefore, the vector parametric
− −
formoftheequationofthelineisp = p +tn = (1, 2,4)+t(5,3, 1),and
0 − −
thus the scalar parametric form is x = 1+5t, y = 2+3t, z = 4 t, and
− −
thenonparametricformis x 1 = y+2 = z 4
− −
5 3 1
−
1.3.10.Theequationofthelinecanbewrittenasp = b+t(a b),with
−
p = band v = a b. Thissimplifies to p = ta+(1 t)b.
0
− −
For any point P between A and B, the parameter t is between 0 and 1
and means the fraction dist(P,B)/dist(A,B), since dist(P,B) = p b =
| − |
6
t(a b) = tdist(A,B). Similarly, 1 tmeans thefraction
| − | −
dist(P,A)/dist(A,B).
If P is outside the interval [A,B] on the side of A, then t is greater
than 1 and is still dist(P,B)/dist(A,B). On the other hand, if P is out-
side the interval [A,B] on the side of B, then t is less than 0 and equals
dist(P,B)/dist(A,B).
−
1.3.12.BothpointsP (1, 2,4)andP (3, 2,1)mustlieintherequired
0 1
− −
plane,andsowemaychooseu = p p = (2,0, 3). Forvwemaychoose
1− 0 −
the same v that we have for the given line v = (2,1, 3), and for the fixed
−
point of the plane we may choose either P or P . Thus a parametric vector
0 1
equation can be written as p = (1, 2,4) + s(2,0, 3) + t(2,1, 3) or as
− − −
p = (3, 2,1)+s(2,0, 3)+t(2,1, 3).
− − −
The nonparametric equation can be obtained by eliminating s and t from
x = 1+2s+2t, y = 2+t, z = 4 3s 3t. This elimination results in
− − −
3x+2z = 11.
1.3.14.In this caseitis easiertostart with thenonparametricform.
Since the required plane is to be orthogonal to the line given by p =
(3, 2,1) + t(2,1, 3), we may take its normal vector to be the direction
− −
vector of the line, that is, let n = (2,1, 3). We may take p = 0, and so
0
−
n (p p ) = 0becomes (2,1, 3) (x,y,z) = 0,thatis, 2x+y 3z = 0.
· − 0 − · −
To write parametric equations, we may choose any two nonparallel vec-
tors (x,y,z) whose components satisfy the above equation to be u and v.
For instance, we may take u = (1,1,1) and v = (0,3,1), and obtain p =
s(1,1,1)+t(0,3,1) as aparametricvectorequation oftheplane.
1.3.16.In this caseitis easiertostart with thenonparametricform.
Since the normal vectors of parallel planes must be parallel, we may take
the n of the two planes to be the same, and read it off the given equation as
n = (3, 4,3). BecauseP (1, 2,4)istobeapointoftherequiredplane,we
0
− −
may take p = (1, 2,4), and so n (p p ) = 0 becomes (3, 4,3)(x
0 − · − 0 − −
1,y +2,z 4) = 0, thatis, 3x 4y +3z = 23.
− −
To write parametric equations, choose any two distinct vectors a and b,
otherthanp ,whosecomponentssatisfytheaboveequation,andletu = a p
0 − 0
and v = b p , because such vectors a and b represent points of the plane
− 0
and so u and v are then nonzero vectors lying in the plane. For instance,
take a = (9,1,0) and b = (0,1,9). Then u = (9,1,0) (1, 2,4) =
− −
7
(8,3, 4) and v = (0,1,9) (1, 2,4) = ( 1,3,5). Thus we obtain
− − − −
p = (1, 2,4)+s(8,3, 4)+t( 1,3,5) as a parametric vector equation of
− − −
theplane. (Asacheck,noticethats = 1, t = 0givesp = a,ands = 0, t = 1
gives p = b.)
1.3.18.Thevectorsu = p 0 = (1,6, 3)andv = p 0 = (7, 2,5)
1− − 2− −
are two nonparallel vectors lying in the required plane. Since a vector para-
metric equation of the plane is p = s(1,6, 3) + t(7, 2,5), the corre-
− −
sponding scalar form of the equation is x = s + 7t, y = 6s 2t, z =
−
3s + 5t. Eliminating the parameters results in the nonparametric equation
−
12x 13y 22z = 0.
− −
1.3.20.We can proceed as in Example 1.3.3 in the text to decompose the
vectorequation
(5,1,1)+s( 2,1,6) = (3, 2,1)+t(2,1, 3)
− − −
into threescalarequations:
5 2s = 3+2t,1+s = 2+t,1+6s = 1 3t
− − −
Solving this system yields s = 1 and t = 2. Hencethelines intersect at the
−
point(7,0, 5).
−
1.3.22.In terms ofscalarcomponents,theequation ofthegiven lineis
x = 5 2s, y = 1+s, z = 1+6s.
−
Substitutingtheseequationsintothatofthegivenplaneresultsin7(5 2s)+
−
(1+s)+2(1+6s) = 8;thesolutioniss = 30. Hencethepointofintersection
is ( 55,31,181).
−
1.3.24. Rewriting the equation of the given line in component form and
replacingsby r yield
x = 3 3r, y = 2+5r, z = 6+7r
− −
and rewriting theequation ofthegiven planein componentformgives
x = 4 2s+t, y = 2+s+3t, z = 1+3s+2t.
− −
8
Combining thosesets of equationsand simplifyingresult in thesystem
3r 2s+t = 1
− −
5r s 3t = 0
− −
7r 3s 2t = 5
− − −
The solution of this system is r = 4 (and s = 59/7 and t = 27/7), so the
pointofintersection is( 9,18,34).
−
1.3.26.PickapointP ontheplane,say,P(1,0,0). Then−P−P→ = (0,2, 4).
0
−
From thegivenequation,n = (3,2, 2). Thus, as in Example1.3.6,
−
−P−P→ n 0 3+2 2+( 4) ( 2) 12
0
D = | · | = | · · − · − | = .
n √9+4+4 √17
| |
1.3.28. First find the plane through 0 that is parallel to the direction vec-
tors of the given lines, as in Example 1.3.6: p = s( 3,5,7) + t( 2,1,6),
− −
or, in scalar parametric form, x = 3s 2t, y = 5s + t, z = 7s + 6t.
− −
Eliminating the parameters yields the equation of the plane in scalar form:
23x+4y +7z = 0, fromwhichweobtainanormalvector,n = (23,4,7), to
theplane, and hencealsoto thetwo given lines.
ThepointP(3, 2,6)liesonthefirstlineandQ(5,1,1)liesonthesecond
−
one. Thus −P→Q = (2,3, 5),and as inExample 1.3.6,
−
−P→Q n 2 23+3 4+( 5) 7 23
D = | · | = | · · − · | = .
n √529+16+49 √594
| |
1.3.30.Following thehint, let Q be the point (3, 2,6) on theline L, so
−
−Q−P→ = (0,6, 6). The direction of the line L is u = ( 3,5,7), and thus the
0
− −
component of −Q−P→ parallel to L is given by −Q−P→ u u = 12( 3,5,7).
0 0 · u u 83 −
| | | |
Thenthecomponent of −Q−P→orthogonal to Li(cid:6)s (cid:7)
0
12 1
(0, 6,6) ( 3,5,7) = (36, 558,414),
− − 83 − 83 −
9
and thereforethe distancefrom P to P is
0
1
d = (36, 558,414) 8.38.
83 − ≈
(cid:8) (cid:8)
(cid:8) (cid:8)
1.3.32. The vector n is(cid:8)orthogonal to the l(cid:8)ine containing P and P, since
(cid:8) (cid:8) 0
−P−P→ = p p and so Equation 1.83 is an equation of the line through the
0 0
−
pointP that isorthogonal ton.
0
1.3.34.ApplyingtheresultofExercise1.3.33,withq = −O−P→ = (1, 2,4),
0
−
n = (3,2,−2), p = (1,0,0) and d = n p = 3 ,wehavef(q) = n q d =
√17 · √17 · −
12 . HenceD = 12 .
−√17 √17
1.3.36.Let q = −O→Q. Then −P→Q = q p = q p tv, and so
− − 0 −
2
−P→Q = −P→Q −P→Q = [(q p ) tv] [(q p ) tv]
· − 0 − · − 0 −
(cid:8) (cid:8) = (q p )2 2t(q p ) v+t2v2
(cid:8)(cid:8) (cid:8)(cid:8) − 0 − − 0 ·
(q p ) v 2 ((q p ) v)2
= v2 t − 0 · + (q p )2 − 0 · ,
− v2 − 0 − v2
(cid:9) (cid:10)
(cid:2) (cid:3)
and −P→Q v = (q p ) v tv2. Thus −P→Q is orthogonal to v if and only if
· − 0 · −
2
(q p ) v tv2 = 0, and also −P→Q (and hence −P→Q ,too) is minimized
− 0 · −
ifandonlyift− (q−vp20)·v = 0. He(cid:8)(cid:8)nce−P(cid:8)(cid:8)→Qisorthogo(cid:8)(cid:8)nalto(cid:8)(cid:8) vifandonlyifthe
distancebetween P and Qis mini(cid:8)mize(cid:8)d. (cid:8) (cid:8)
2.1.2.The samerowoperations asin Exercise2.1.1 give
2 2 3 0 2 2 3 0
− −
1 5 2 0 0 4 7/2 0 .
(cid:8) → (cid:8)
4 0 6 (cid:8) 0 0 0 7/2 (cid:8) 0
− (cid:8) − (cid:8)
(cid:8) (cid:8)
The last row corresponds(cid:8)(cid:8)to the equation −72x3 = 0, s(cid:8)(cid:8)o x3 = 0. The
secondrowcorrespondstotheequation4x +7x = 0andthusx = 7 0 =
2 2 3 2 −8·
0. Finally, the first rowcorresponds to 2x +2x 3x = 0, from which we
1 2 3
−
obtain x = x + 3x = 0. Thus, in vectorform thesolutionis x = 0.
1 − 2 2 3
2.1.4. This equation constitutes a system, which is already in echelon
form. The variables x and x are free, and so we set x = s and x = t.
2 3 2 3
10
