Preliminary Questions
1. What is the terminal point of the vectorv= 3,2,1based at the pointP =(1,1,1)?
solution We denote the terminal point byQ=(a, b, c). Then by the definition of components of a vector, we have 3,2,1 = a−1, b−1, c−1
Equivalent vectors have equal components respectively, thus,
3=a−1 a=4
2=b−1 ⇒ b=3
1=c−1 c=2
The terminal point ofvis thusQ=(4,3,2).
2. What are the components of the vectorv= 3,2,1based at the pointP =(1,1,1)?
solution The component ofv= 3,2,1are3,2,1regardless of the base point. The component ofvand the base pointP =(1,1,1)determine the headQ=(a, b, c)of the vector, as found in the previous exercise.
3. Ifv= −3w, then (choose the correct answer):
(a) vandware parallel.
(b) vandwpoint in the same direction.
solution The vectorsvandwlie on parallel lines, hence these vectors are parallel. Sincevis a scalar multiple ofw by a negative scalar,vandwpoint in opposite directions. Thus, (a) is correct and (b) is not.
4. Which of the following is a direction vector for the line throughP =(3,2,1)andQ=(1,1,1)?
(a) 3,2,1 (b) 1,1,1 (c) 2,1,0
solution Any vector that is parallel to the vector−→
P Qis a direction vector for the line throughPandQ. We compute the vector−→
P Q:
−→P Q= 1−3,1−2,1−1 = −2,−1,0. The vectors3,2,1and1,1,1are not constant multiples of−→
P Q, hence they are not parallel to−→
P Q. However2,1,0 =
−1−2,−1,0 = −−→
P Q, hence the vector2,1,0is parallel to−→
P Q. Therefore, the vector2,1,0is a direction vector for the line throughP andQ.
5. How many different direction vectors does a line have?
solution All the vectors that are parallel to a line are also direction vectors for that line. Therefore, there are infinitely many direction vectors for a line.
6. True or false? Ifvis a direction vector for a lineL, then−vis also a direction vector forL.
solution True. Every vector that is parallel tovis a direction vector for the lineL. Since−vis parallel tov, it is also a direction vector forL.
Exercises
1. Sketch the vectorv= 1,3,2and compute its length.
solution The vectorv= 1,3,2is shown in the following figure:
1 3
2
y x
z
v = 〈1, 3, 2〉
The length ofvis
v =
12+32+22=√ 14
Letv=−−−→
P0Q0, whereP0 =(1,−2,5)andQ0= (0,1,−4). Which of the following vectors (with tailP and headQ) are equivalent tov?
v1 v2 v3 v4
P (1,2,4) (1,5,4) (0,0,0) (2,4,5) Q (0,5,−5) (0,−8,13) (−1,3,−9) (1,7,4) 3. Sketch the vectorv= 1,1,0based atP =(0,1,1). Describe this vector in the form−→
P Qfor some pointQ, and sketch the vectorv0based at the origin equivalent tov.
solution The vectorv= 1,1,0based atP =(0,1,1)is shown in the figure:
v
v0 x y
Q = (1, 2, 1) P = (0, 1, 1) z
May 16, 2011 The headQof the vectorv=−→
P Qis at the pointQ=(0+1,1+1,1+0)=(1,2,1).
v y
x S = (1, 1, 0) O
z
The vectorv0based at the origin and equivalent tovis
v0= 1,1,0 =−→OS, whereS=(1,1,0).
Determine whether the coordinate systems (A)–(C) in Figure 17 satisfy the right-hand rule.
In Exercises 5–8, find the components of the vector−→
P Q.
5. P =(1,0,1), Q=(2,1,0)
solution By the definition of the vector components we have
−→P Q= 2−1,1−0,0−1 = 1,1,−1
P =(−3,−4,2), Q=(1,−4,3) 7. P =(4,6,0), Q=
−12,92,1
solution Using the definition of vector components we have
−→P Q=
−1 2−4,9
2−6,1−0
=
−9 2,−3
2,1
P =
−12,92,1, Q=(4,6,0) In Exercises 9–12, letR=(1,4,3).
9. Calculate the length of−→
OR.
solution The length of−→ORis the distance fromR=(1,4,3)to the origin. That is, −→OR =
(1−0)2+(4−0)2+(3−0)2=√
26≈5.1.
Find the pointQsuch thatv=−→
RQhas components4,1,1, and sketchv.
11. Find the pointP such thatw=−→
P Rhas components3,−2,3, and sketchw.
solution DenotingP =(x0, y0, z0)we get
−→P R= 1−x0,4−y0,3−z0 = 3,−2,3 Equating corresponding components gives
1−x0=3 4−y0= −2 3−z0=3
⇒ x0= −2, y0=6, z0=0
The pointP is, thus,P =(−2,6,0).
w z
x P = (−2, 6, 0) y
R = (1, 4, 3)
(0, 0, 3)
(−2, 0, 0)
(0, 6, 0)
Find the components ofu=−→
P R, whereP =(1,2,2).
13. Letv= 4,8,12. Which of the following vectors is parallel tov? Which point in the same direction?
(a) 2,4,6 (b) −1,−2,3
(c) −7,−14,−21 (d) 6,10,14
solution A vector is parallel tovif it is a scalar multiple ofv. It points in the same direction if the multiplying scalar is positive. Using these properties we obtain the following answer:
(a) 2,4,6 =12v⇒The vectors are parallel and point in the same direction.
(b) −1,−2,3is not a scalar multiple ofv, hence these vectors are not parallel.
(c) −7,−14,−21 = −74v⇒The vectors are parallel but point in opposite directions.
(d) 6,10,14is not a constant multiple ofv, hence these vectors are not parallel.
In Exercises 14–17, determine whether−→
ABis equivalent to−→
P Q.
A=(1,1,1) B=(3,3,3) P =(1,4,5) Q=(3,6,7) 15. A=(1,4,1) B=(−2,2,0)
P =(2,5,7) Q=(−3,2,1) solution We compute the two vectors:
−→AB= −2−1,2−4,0−1 = −3,−2,−1
−→P Q= −3−2,2−5,1−7 = −5,−3,−6 The components of−→
ABand−→
P Qare not equal, hence they are not a translate of each other, that is, the vectors are not equivalent.
A=(0,0,0) B=(−4,2,3) P =(4,−2,−3) Q=(0,0,0) 17. A=(1,1,0) B=(3,3,5)
P =(2,−9,7) Q=(4,−7,13) solution The vectors−→
ABand−→
P Qare the following vectors:
−→AB= 3−1,3−1,5−0 = 2,2,5
−→P Q= 4−2,−7−(−9),13−7 = 2,2,6 Thez-coordinates of the vectors are not equal, hence the vectors are not equivalent.
In Exercises 18–23, calculate the linear combinations.
52,2,−3 +31,7,2 19. −28,11,3 +42,1,1
solution Using the operations of vector addition and scalar multiplication we have
−28,11,3 +42,1,1 = −16,−22,−6 + 8,4,4 = −8,−18,−2.
6(4j+2k)−3(2i+7k) 21. 124,−2,8 − 1312,3,3
solution Using the operations on vectors we have 1
24,−2,8 −1
312,3,3 = 2,−1,4 − 4,1,1 = −2,−2,3. 5(i+2j)−3(2j+k)+7(2k−i)
23. 46,−1,1 −21,0,−1 +3−2,1,1
solution Using the operations of vector addition and scalar multiplication we have
46,−1,1 −21,0,−1 +3−2,1,1 = 24,−4,4 + −2,0,2 + −6,3,3
= 16,−1,9. In Exercises 24–27, find the given vector.
ev, wherev= 1,1,2 25. ew, wherew= 4,−2,−1
solution We first find the length ofw:
w =
42+(−2)2+12=√ 21 Hence,
ew= 1 ww=
4
√21, −2
√21, −1
√21
Unit vector in the direction ofu= 1,0,7 27. Unit vector in the direction opposite tov= −4,4,2
solution A unit vector in the direction opposite tov= −4,4,2is the following vector:
−ev= − 1 vv We compute the length ofv:
v =
(−4)2+42+22=6 The desired vector is, thus,
−ev= −1
6−4,4,2 = −4
−6, 4
−6, 2
−6
= 2
3,−2 3,−1
3
May 16, 2011
Sketch the following vectors, and find their components and lengths.
(a) 4i+3j−2k (b) i+j+k
(c) 4j+3k (d) 12i+8j−k
In Exercises 29–36, find a vector parametrization for the line with the given description.
29. Passes throughP =(1,2,−8), direction vectorv= 2,1,3 solution The vector parametrization for the line is
r(t)=−→
OP+tv Inserting the given data we get
r(t)= 1,2,−8 +t2,1,3 = 1+2t,2+t,−8+3t
Passes throughP =(4,0,8), direction vectorv= 1,0,1 31. Passes throughP =(4,0,8), direction vectorv=7i+4k
solution Sincev=7i+4k= 7,0,4we obtain the following parametrization:
r(t)=−→
OP+tv= 4,0,8 +t7,0,4 = 4+7t,0,8+4t
Passes throughO, direction vectorv= 3,−1,−4 33. Passes through(1,1,1)and(3,−5,2)
solution We use the equation of the line through two pointsP andQ:
r(t)=(1−t)−→
OP+t−−→
OQ Since−→
OP = 1,1,1and−−→
OQ= 3,−5,2we obtain
r(t)=(1−t)1,1,1 +t3,−5,2 = 1−t,1−t,1−t + 3t,−5t,2t = 1+2t,1−6t,1+t
Passes through(−2,0,−2)and(4,3,7) 35. Passes throughOand(4,1,1)
solution By the equation of the line through two points we get
r(t)=(1−t)0,0,0 +t4,1,1 = 0,0,0 + 4t, t, t = 4t, t, t
Passes through(1,1,1)parallel to the line through(2,0,−1)and(4,1,3) In Exercises 37–40, find parametric equations for the lines with the given description.
37. Perpendicular to thexy-plane, passes through the origin
solution A direction vector for the line is a vector parallel to thez-axis, for instance, we may choosev= 0,0,1.
The line passes through the origin(0,0,0), hence we obtain the following parametrization:
r(t)= 0,0,0 +t0,0,1 = 0,0, t orx=0,y=0,z=t.
Perpendicular to theyz-plane, passes through(0,0,2)
39. Parallel to the line through(1,1,0)and(0,−1,−2), passes through(0,0,4)
solution The direction vector isv= 0−1,−1−1,−2−0 = −1,−2,−2. Hence, using the equation of a line we obtain
r(t)= 0,0,4 +t−1,−2,−2 = −t,−2t,4−2t
Passes through(1,−1,0)and(0,−1,2)
41. Which of the following is a parametrization of the line throughP =(4,9,8)perpendicular to thexz-plane (Figure 18)?
(a) r(t)= 4,9,8 +t1,0,1 (b) r(t)= 4,9,8 +t0,0,1 (c) r(t)= 4,9,8 +t0,1,0 (d) r(t)= 4,9,8 +t1,1,0
y P = (4, 9, 8) z
x
FIGURE 18
solution Since the direction vector must be perpendicular to thexz-plane, then the direction vector for the line must be parallel toj, which is only satisfied by solution (c).
Find a parametrization of the line throughP =(4,9,8)perpendicular to theyz-plane.
In Exercises 43–46, letP =(2,1,−1)andQ=(4,7,7). Find the coordinates of each of the following.
43. The midpoint ofP Q
solution We first parametrize the line throughP =(2,1,−1)andQ=(4,7,7):
r(t)=(1−t)2,1,−1 +t4,7,7 = 2+2t,1+6t,−1+8t The midpoint ofP Qoccurs att= 12, that is,
midpoint=r 1
2
=
2+2ã1
2,1+6ã1
2,−1+8ã1 2
= 3,4,3
The midpoint ofP Qis the terminal point of the vectorr(t), that is,(3,4,3). (One could also use the midpoint formula to arrive at the same solution.)
The point onP Qlying two-thirds of the way fromP toQ 45. The pointRsuch thatQis the midpoint ofP R
solution We denoteR=(x0, y0, z0). By the formula for the midpoint of a segment we have 4,7,7 =
2+x0 2 ,1+y0
2 ,−1+z0 2
Equating corresponding components we get 4= 2+x0
2 7= 1+y0
2 7= −1+z0
2
⇒ x0=6, y0=13, z0=15 ⇒ R=(6,13,15)
The two points on the line throughP Qwhose distance fromP is twice its distance fromQ 47. Show thatr1(t)andr2(t)define the same line, where
r1(t)= 3,−1,4 +t8,12,−6 r2(t)= 11,11,−2 +t4,6,−3
Hint:Show thatr2passes through(3,−1,4)and that the direction vectors forr1andr2are parallel.
solution We observe first that the direction vectors ofr1(t)andr2(t)are multiples of each other:
8,12,−6 =24,6,−3
Thereforer1(t)andr2(t)are parallel. To show they coincide, it suffices to prove that they share a point in common, so we verify thatr1(0)= 3,−1,4lies onr2(t)by solving fort:
3,−1,4 = 11,11,−2 +t4,6,−3 3,−1,4 − 11,11,−2 =t4,6,−3
−8,−12,6 =t4,6,−3 This equation is satisfied fort= −2, sor1andr2coincide.
Show thatr1(t)andr2(t)define the same line, where
r1(t)=t2,1,3, r2(t)= −6,−3,−9 +t8,4,12
49. Find two different vector parametrizations of the line throughP =(5,5,2)with direction vectorv= 0,−2,1. solution Two different parameterizations are
r1(t)= 5,5,2 +t0,−2,1 r2(t)= 5,5,2 +t0,−20,10
Find the point of intersection of the linesr(t)= 1,0,0 +t−3,1,0ands(t)= 0,1,1 +t2,0,1.
51. Show that the linesr1(t)= −1,2,2 +t4,−2,1andr2(t)= 0,1,1 +t2,0,1do not intersect.
solution The two lines intersect if there exist parameter valuest1andt2such that
−1,2,2 +t14,−2,1 = 0,1,1 +t22,0,1
−1+4t1,2−2t1,2+t1 = 2t2,1,1+t2
May 16, 2011
Equating corresponding components yields
−1+4t1=2t2 2−2t1=1
2+t1=1+t2
The second equation impliest1= 12. Substituting into the first and third equations we get
−1+4ã1
2=2t2 ⇒ t2= 1 2 2+1
2=1+t2 ⇒ t2= 3 2
We conclude that the equations do not have solutions, which means that the two lines do not intersect.
Determine whether the linesr1(t)= 2,1,1 +t−4,0,1andr2(s)= −4,1,5 +s2,1,−2intersect, and if so, find the point of intersection.
53. Determine whether the linesr1(t)= 0,1,1 +t1,1,2andr2(s)= 2,0,3 +s1,4,4intersect, and if so, find the point of intersection.
solution The lines intersect if there exist parameter valuestandssuch that 0,1,1 +t1,1,2 = 2,0,3 +s1,4,4
t,1+t,1+2t = 2+s,4s,3+4s (1)
Equating corresponding components we get
t=2+s 1+t=4s 1+2t=3+4s Substitutingtfrom the first equation into the second equation we get
1+2+s=4s
3s=3 ⇒ s=1, t=2+s=3 We now check whethers=1,t=3 satisfy the third equation:
1+2ã3=3+4ã1 7=7
We conclude thats= 1,t=3 is the solution of (1), hence the two lines intersect. To find the point of intersection we substitutes=1 in the right-hand side of (1) to obtain
2+1,4ã1,3+4ã1 = 3,4,7 The point of intersection is the terminal point of this vector, that is,(3,4,7).
Find the intersection of the linesr1(t)= −1,1 +t2,4andr2(s)= 2,1 +s−1,6inR2.
55. Find the components of the vectorvwhose tail and head are the midpoints of segmentsACandBCin Figure 19.
B=(1, 1, 0) C =(0, 1, 1) A=(1, 0, 1)
y x
z
FIGURE 19
solution We denote byP andQthe midpoints of the segmentsACandBCrespectively. Thus, v=−→
P Q (1)
y x
A = (1, 0, 1) C = (0, 1, 1)
B = (1, 1, 0) P
Q z
We use the formula for the midpoint of a segment to find the coordinates of the pointsP andQ. This gives P =
1+0 2 ,0+1
2 ,1+1 2
= 1
2,1 2,1
Q= 1+0
2 ,1+1 2 ,0+1
2
= 1
2,1,1 2
Substituting in (1) yields the following vector:
v=−→
P Q= 1
2−1 2,1−1
2,1 2−1
=
0,1 2,−1
2
.
Find the components of the vectorwwhose tail isCand head is the midpoint ofABin Figure 19.
Further Insights and Challenges
In Exercises 57–63, we consider the equations of a line insymmetric form, whena=0,b=0,c=0.
x−x0
a = y−y0
b = z−z0
c 12
57. LetLbe the line throughP0 = (x0, y0, c0)with direction vectorv = a, b, c. Show thatLis defined by the symmetric Eq. (12).Hint:Use the vector parametrization to show that every point onLsatisfies Eq. (12).
solution Lis given by vector parametrization
r(t)= x0, y0, z0 +ta, b, c which gives us the equations
x=x0+at y=y0+bt z=z0+ct.
Solving fortgives
t= x−x0 a t= y−y0
b t= z−z0
c Setting each equation equal to the other gives Eq. (12).
Find the symmetric equations of the line throughP0=(−2,3,3)with direction vectorv= 2,4,3. 59. Find the symmetric equations of the line throughP =(1,1,2)andQ=(−2,4,0).
solution This line has direction vector−→
P Q= −3,3,−2. Using(x0, y0, z0)=P =(1,1,2)anda, b, c =−→
P Q=
−3,3,−2in Eq. (12) gives
x−1
−3 = y−1
3 = z−2
−2 Find the symmetric equations of the line
x=3+2t, y=4−9t, z=12t 61. Find a vector parametrization for the line
x−5
9 = y+3
7 =z−10 solution Using(x0, y0, z0)=(5,−3,10)anda, b, c = 9,7,1gives
r(t)= 5,−3,10 +t9,7,1
v
w 4 1 2
3
u
May 16, 2011
Find a vector parametrization for the line x 2 = y
7 = z 8.
63. Show that the line in the plane through(x0, y0)of slopemhas symmetric equations x−x0= y−y0
m
solution The line through(x0, y0)of slopemhas equationy−y0=m(x−x0), which becomesx−x0= m1(y−y0), which becomes
x−x0
1 = y−y0 m
A median of a triangle is a segment joining a vertex to the midpoint of the opposite side. Referring to Figure 20(A), prove that three medians of triangleABCintersect at the terminal pointP of the vector13(u+v+w). The pointP is thecentroidof the triangle.Hint:Show, by parametrizing the segmentAA, thatPlies two-thirds of the way from AtoA. It will follow similarly thatP lies on the other two medians.
65. A median of a tetrahedron is a segment joining a vertex to the centroid of the opposite face. The tetrahedron in Figure 20(B) has vertices at the origin and at the terminal points of vectorsu,v, andw. Show that the medians intersect at the terminal point of14(u+v+w).
solution We first find vectors from the origin to the centroids of the four faces (labelled 1,2,3,4 after their opposite vertices, also labelled 1,2,3,4). Now, by the previous problem (Exercise 64), a vector from the origin (vertex 1) to the centroid of the opposite face (face 1) is13(u+v+w). As for face 2, a vector from vertex 2 to the centroid of face 2 is
1
3(−u+(v−u)+(w−u)), but since vertex 2 is at the head of vectoru, then a vector from the origin to the centroid of face 2 isu+ 13(−u+(v−u)+(w−u))= 13(v+w). Similarly, a vector from the origin to the centroid of face 3 is v+13(−v+(u−v)+(w−v))= 13(u+w), and from the origin to the centroid of face 4 is13(u+v).
We now find the paramentric equations of four lines1, . . . , 4, each from vertexito the centroid of the (opposite) facei.
1(t)=t0+(1−t)1
3(u+v+w) 2(t)=tu+(1−t)1
3(v+w) 3(t)=tv+(1−t)1
3(u+w) 4(t)=tw+(1−t)1
3(u+v)
By substitutingt=1/4 into each line, we find that they all intersect in the same point:
1(1/4)=1/40+(1−1/4)1
3(u+v+w)=1/4(u+v+w) 2(1/4)=1/4u+(1−1/4)1
3(v+w)=1/4(u+v+w) 3(1/4)=1/4v+(1−1/4)1
3(u+w)=1/4(u+v+w) 4(1/4)=1/4w+(1−1/4)1
3(u+v)=1/4(u+v+w)
We conclude that all four lines intersect at the terminal point of the vector 1/4(u+v+w), as desired.