Exercise Solutions for Chapter I.
§I.1
Points and Vectors
1)
a)
b)
c)
d)
x = y is the straight line through the origin that makes an angle 45◦ with the x and y–axes.
x + y = 1 is the straight line through the points (1, 0) and (0, 1).
x2 + y 2 = 4 is the circle with centre (0, 0) and radius 2.
x2 + y 2 = 2y, or equivalently, x2 + (y − 1)2 = 1 is the circle with centre (0, 1) and radius 1.
x2 + y 2 = 2y
x2 + y 2 = 4
x=y
x+y =1
2)
a) For each fixed y0 , z = x, y = y0 is a straight line that lies in the plane, y = y0 (which is parallel to the
plane containing the x and z axes and is a distance y0 from it). This line passes through x = z = 0 and
makes an angle 45◦ with the xy–plane. The set z = x is the union of all the lines z = x, y = y0 with
all values of y0 . As y0 varies z = x, y = y0 sweeps out the plane which contains the y–axis and which
makes an angle 45◦ with the xy–plane.
b) x + y + z = 1 is the plane through the points (1, 0, 0), (0, 1, 0) and (0, 0, 1).
b) x + y + z = 1 is the plane through the points (1, 0, 0), (0, 1, 0) and (0, 0, 1).
c) x2 + y 2 + z 2 = 4 is the sphere with centre (0, 0, 0) and radius 2.
d) x2 + y 2 + z 2 = 4, z = 1 or equivalently x2 + y 2 = 3, z = 1, is the intersection of the plane z = 1 with
the sphere
√ of centre (0, 0, 0) and radius 2. It is a circle in the plane z = 1 that has centre (0, 0, 1) and
radius 3.
e) For each fixed z0 , x2 + y 2 = 4, z = z0 is a circle in the plane z = z0 with centre (0, 0, z0 ) and radius 2.
So x2 + y 2 = 4 is the union of x2 + y 2 = 4, z = z0 for all possible values of z0 . It is a vertical stack of
horizontal circles. It is the cylinder of radius 2 centered on the z–axis.
z
z
z
z>
x+y+z =1
z=x
y
y
p
x2 + y 2
y
x
x
y = y0
x
p
is a circle in the plane z = z0 with centre
f) For each fixed z0 ≥ 0, the p
curve z = x2 + y 2 , z = z0p
(0, 0, z0 ) and radius z0 . As x2 + y 2 = z is the union of x2 + y 2 = z, z = z0 for all possible values
1
of z0 ≥ 0, it is a vertical stack
p of horizontal circles whose radii increase linearly with z. It is a cone
centered on the z–axis. z > x2 + y 2 is the region above this cone.
3) For each fixed c, the isobar p(x, y) = c is the √
curve x2 − 2cx + y 2 = 1, or equivalently, (x − c)2 + y 2 = 1 + c2 .
This is a circle with centre (c, 0) and radius 1 + c2 , which for large c is just a bit bigger than c.
4) Call the centre of the circumscribing circle (x̄, ȳ). This centre must be equidistant from the three vertices.
So
x̄2 + ȳ 2 = (x̄ − a)2 + ȳ 2 = (x̄ − b)2 + (ȳ − c)2
0 = a2 − 2ax̄
which implies
a
2
2
2
b +c −ab 2
ȳ =
x̄ =
The radius is
§I.2
q
a 2
2
+
2c
= b2 − 2bx̄ + c2 − 2cȳ
b2 + c2 − 2bx̄
b2 + c2 − ab
=
2c
2c
.
Areas of Parallelograms
1) When (a, b) and (c, d) lie in the first and second quadrants respectively, the desired parallelogram is gotten
by deleting from the rectangle with vertices (a, 0), (a, b+d), (c, b+d) and (c, 0) (and hence area (a−c)(b+d))
two triangles of area 21 ab and two triangles of area 12 (−c)d. So the area of the parallelogram is
(a − c)(b + d) − 2 × 21 ab − 2 × 12 (−c)d = ad − bc
(a + c, b + d)
(c, b + d)
(a, b + d)
(c, d)
(a, b)
2a)
a = r cos θ
b = r sin θ
2b)
A = r cos(θ + ϕ) = r cos θ cos ϕ − r sin θ sin ϕ = a cos ϕ − b sin ϕ
B = r sin(θ + ϕ) = r sin θ cos ϕ + r cos θ sin ϕ = b cos ϕ + a sin ϕ
3)
A
det
C
B
D
=AD − BC
=[a cos ϕ − b sin ϕ][d cos ϕ + c sin ϕ]
− [b cos ϕ + a sin ϕ][c cos ϕ − d sin ϕ]
=ac sin ϕ cos ϕ − sin ϕ cos ϕ + ad cos2 ϕ + sin2 ϕ
− bc sin2 ϕ + cos2 ϕ − bd sin ϕ cos ϕ − cos ϕ sin ϕ
=ad − bc
2
§I.3
Addition of Vectors and Multiplication of a Vector by a Number
1) ~a + ~b = [3, 1], ~a + 2~b = [4, 2], 2~a − ~b = [3, −1]
2) The center of the sphere is 12 (2, 1, 4) + (4, 3, 10) = (3, 2, 7). The radius is k(2, 1, 4) − (4, 3, 10)k =
√
k(−2, −2, −6)k = 2k(1, 1, 3)k = 2 11. The equation of the sphere is
(x − 3)2 + (y − 2)2 + (z − 7)2 = 44
3a) The vectors [0, 3, 7] − [1, 2, 3] = [−1, 1, 4] and [3, 5, 11] − [1, 2, 3] = [2, 3, 8] are not parallel (i.e. multiples of
each other), so the three points are not on the same line.
3b) The vectors [1, 2, −2] − [0, 3, −5] = [1, −1, 3] and [3, 0, 4] − [0, 3, −5] = [3, −3, 9] are parallel (i.e. multiples
of each other), so the three points are on the same line.
4) Let (x, y, z) be a point in P . The distances from (x, y, z) to (3, −2, 3) and (3/2, 1, 0) are
p
p
(x − 3)2 + (y + 2)2 + (z − 3)2 and (x − 3/2)2 + (y − 1)2 + z 2
respectively. To be in P , (x, y, z) must obey
p
p
(x − 3)2 + (y + 2)2 + (z − 3)2 = 2 (x − 3/2)2 + (y − 1)2 + z 2
(x − 3)2 + (y + 2)2 + (z − 3)2 = 4(x − 3/2)2 + 4(y − 1)2 + 4z 2
x2 − 6x + 9 + y 2 + 4y + 4 + z 2 − 6z + 9 = 4x2 − 12x + 9 + 4y 2 − 8y + 4 + 4z 2
3x2 − 6x + 3y 2 − 12y + 3z 2 + 6z − 9 = 0
x2 − 2x + y 2 − 4y + z 2 + 2z − 3 = 0
(x − 1)2 + (y − 2)2 + (z + 1)2 = 9
This is a sphere of radius 3 centered on (1, 2, −1).
5) The parallelogram determined by the vectors ~a and ~b has vertices ~0, ~a, ~b and ~a + ~b. As t varies from 0 to
1, t(~a + ~b) traverses the diagonal from ~0 to ~a + ~b. As s varies from 0 to 1, ~a + s(~b −~a) traverses the diagonal
from ~a to ~b. These two straight lines meet when s and t are such that
t(~a + ~b) = ~a + s(~b − ~a)
or
(t + s − 1)~a = (s − t)~b
Assuming that ~a and ~b are not parallel (i.e. the parallelogram has not degenerated to a line segment), this
is the case only when t + s − 1 = 0 and s − t = 0. That is, s = t = 21 . So the two lines meet at their
midpoints.
§I.4
(1)
The Dot Product
4
cos θ = √ √ = .4961
5 13
0
b) ~a · ~b = (−1, 1) · (1, 1) = 0
cos θ = √ √ = 0
2 2
4
c) ~a · ~b = (1, 1) · (2, 2) = 4
cos θ = √ √ = 1
2 8
2
d) ~a · ~b = (1, 2, 1) · (−1, 1, 1) = 2 cos θ = √ √ = .4714
6 3
0
e) ~a · ~b = (−1, 2, 3) · (3, 0, 1) = 0 cos θ = √ √ = 0
14 10
a) ~a · ~b = (1, 2) · (−2, 3) = 4
3
θ = 60.25◦
θ = 90◦
θ = 0◦
θ = 61.87◦
θ = 90◦
2) projı̂~a = (~a · ı̂)ı̂ = a1 ı̂. proj̂~a = (~a · ̂)̂ = a2 1̂.
3) The vector from (1, 2, 3) to (4, 0, 5) is [3, −2, 2]. The vector from (1, 2, 3) to (3, 6, 4) is [2, 4, 1]. The dot
product between these two vectors is [3, −2, 2]·[2, 4, 1] = 0, so the vectors are perpendicular and the triangle
does contain a right angle.
−−→
−−→ −→
−−→
4) The vectors OU , AU and BU are all of the same length. So, by high school geometry, proj−
→ OU ought to
OA
−−→
be the midpoint of the line from O to A, that is, (a/2, 0), and proj−
→ OU ought to be the midpoint of the
OB
line from O to B, that is, (b/2, c/2). As
B
−−→
proj−
→ OU
OB
O
−−→ −−→
OB · OU = (b, c) ·
−→ −−→
OA · OU = (a, 0) ·
U
−−→
proj−
→ OU
OA
a b2 +c2 −ab
2,
2c
a b2 +c2 −ab
2,
2c
=
ab
2
=
a2
2
+
b2 +c2 −ab
2
−→
= 12 kOAk2
=
b2 +c2
2
−−→
= 12 kOBk2
A
the formula agrees with the guess in both cases.
§I.5
The Pendulum
(1) Denote by x(t), y(t) the position of the skier at time t. As long as the skier remains on the surface of the
hill
y(t) = h x(t)
=⇒ y 0 (t) = h0 x(t) x0 (t)
=⇒ y 00 (t) = h00 x(t) x0 (t)2 + h0 x(t) x00 (t)
So the velocity and acceleration vectors of the skier are
h
i
~v (t) = 1, h0 x(t) x0 (t)
h
h
i
i
~a(t) = 1, h0 x(t) x00 (t) + 0, h00 x(t) x0 (t)2
The skier is subject to two forces. One is gravity. The other acts perpendicularly to the hill and has a
magnitude such that the skier remains on the surface of the hill. From the velocity vector of the skier
(which remain tangential to the hill as long as the skier remains of the surface of the hill),we see that one
vector normal to the hill at x(t), y(t) is
h
i
~n(t) = − h0 x(t) , 1
This vector is not a unit vector, but that’s ok. By Newton’s law of motion
m~a = −mg̂ + p(t)~n(t)
Dot both sides of this equation with ~n(t).
m~a(t) · ~n(t) = −mg̂ · ~n(t) + p(t)k~n(t)k2
Subbing in
=⇒
h
2 i
mh00 x(t) x0 (t)2 = −mg + p(t) 1 + h0 x(t)
h
2 i
p(t) 1 + h0 x(t)
= m g + h00 x(t) x0 (t)2
4
As long as p(t) ≥ 0, the hill is pushing up in order to keep the skier on the surface. When p(t) becomes
negative, the hill has to pull on the skier in order to keep her on the surface. But the hill can’t pull, so the
skier becomes airborne instead. This happens when
g + h00 x(t) x0 (t)2 = 0
That is when x0 (t) =
q
−g/h00 x(t) . At this time x(t) = x0 , y(t) = y0 and the speed of the skier is
q
p
2 q
x0 (t)2 + y 0 (t)2 =
1 + h 0 x0
−g/h00 x0
2) The marble is subject to two forces. The first, gravity, is −mg k̂ with m being the mass of the marble. The
second is the normal force imposed by the plane. This forces acts in a direction perpendicular to the plane.
One vector normal to the plane is aı̂ + b̂ + ck̂. So the force due to the plane is T [a, b, c] with T determined
by the property that the net force perpendicular to the plane must be exactly zero, so that the marble
remains on the plane, neither digging into nor flying off of it. The projection of the gravitational force onto
the normal vector [a, b, c] is
−mg[0,0,1]·[a,b,c]
[a, b, c] = a2−mgc
k[a,b,c]k2
+b2 +c2 [a, b, c]
The condition that determines T is thus
T [a, b, c] +
−mgc
a2 +b2 +c2 [a, b, c]
=0
=⇒
T =
mgc
a2 +b2 +c2
The total force on the marble is then (ignoring friction – which will have no effect on the direction of motion)
T [a, b, c] − mg[0, 0, 1] =
=
mgc
a2 +b2 +c2 [a, b, c]
− mg[0, 0, 1]
2
+b2 +c2 ]
mg c[a,b,c]−[0,0,a
a2 +b2 +c2
2
−b
= mg [ac,bc,−a
a2 +b2 +c2
2
]
The direction of motion [ac, bc, −a2 − b2 ] . If you want to turn this into a unit vector, just divide by
p
(a2 + b2 )(a2 + b2 + c2 ). Note that the direction vector in perpendicular [a, b, c] and hence is parallel to
the plane. If c = 0, the plane is vertical. In this case, the marble doesn’t roll – it falls straight down. If
a = b = 0, the plane is horizontal. In this case, the marble doesn’t roll – it remains stationary.
§I.6
The Cross Product
1)

ı̂ ̂
(1, 2, 3) × (4, 5, 6) = det  1 2
4 5

k̂
3  = ı̂(2 × 6 − 3 × 5) − ̂(1 × 6 − 3 × 4) + k̂(1 × 5 − 2 × 4) = −3ı̂ + 6̂ − 3k̂
6
2)
~b × ~c = (b2 c3 − b3 c2 )ı̂ − (b1 c3 − b3 c1 )̂ + (b1 c2 − b2 c1 )k̂


ı̂
̂
k̂

~a × (~b × ~c) = det 
a1
a2
a3
b2 c3 − b3 c2 −b1 c3 + b3 c1 b1 c2 − b2 c1
= ı̂[a2 (b1 c2 − b2 c1 ) − a3 (−b1 c3 + b3 c1 )]
−̂[a1 (b1 c2 − b2 c1 ) − a3 (b2 c3 − b3 c2 )]
5
+k̂[a1 (−b1 c3 + b3 c1 ) − a2 (b2 c3 − b3 c2 )]
(lhs)
~
~
(~a · ~c)b − (~a · b)~c = (a1 c1 + a2 c2 + a3 c3 )(b1 ı̂ + b2 ̂ + b3 k̂) − (a1 b1 + a2 b2 + a3 b3 )(c1 ı̂ + c2 ̂ + c3 k̂)
= ı̂[a1 b1 c1 + a2 b1 c2 + a3 b1 c3 − a1 b1 c1 − a2 b2 c1 − a3 b3 c1 ]
+̂[a1 b2 c1 + a2 b2 c2 + a3 b2 c3 − a1 b1 c2 − a2 b2 c2 − a3 b3 c2 ]
+k̂[a1 b3 c1 + a2 b3 c2 + a3 b3 c3 − a1 b1 c3 − a2 b2 c3 − a3 b3 c3 ]
= ı̂[a2 b1 c2 + a3 b1 c3 − a2 b2 c1 − a3 b3 c1 ]
+̂[a1 b2 c1 + a3 b2 c3 − a1 b1 c2 − a3 b3 c2 ]
+k̂[a1 b3 c1 + a2 b3 c2 − a1 b1 c3 − a2 b2 c3 ]
(rhs)
(lhs) and (rhs) are the same.
3)
~ = ~a · [~b × (~c × d)]
~
(~a × ~b) · (~c × d)
~ c − (~b · ~c)d]
~
= ~a · [(~b · d)~
~
(by property 9 with ~c → (~c × d))
(by property 10)
~ − (~a · d)(
~ ~b · ~c)
= (~a · ~c)(~b · d)
So
~ = (~a · ~c)(~b · d)
~ − (~a · d)(
~ ~b · ~c)
(~a × ~b) · (~c × d)
§I.7
1)
Application of Cross Products to Rotational Motion
√
â = [2 − 1, −3 − 1, 1 − (−1)]/k[2 − 1, −3 − 1, 1 − (−1)]k = [1, −4, 2]/ 21
Ω = 10
~ = √10 [1, −4, 2]
Ω
21
~r = [1 − 1, 2 − 1, 3 − (−1)] = [0, 1, 4]
~ × ~r = √10 [1, −4, 2] × [0, 1, 4] = √10 [−18, −4, 1]
~v = Ω
21
21
2)
[x(t), y(t), z(t)] = [r cos θ(t), r sin θ(t), 0]
=⇒
~v = [−rθ 0 (t) sin θ(t), rθ0 (t) cos θ(t), 0]
The plate is rotating about the z–axis (i.e. â = k̂) at Ω = θ0 (t) radians per second. So
~v = Ωk̂ × [x(t), y(t), 0] = θ 0 (t)k̂ × [r cos θ(t), r sin θ(t), 0] = θ 0 (t)[−r sin θ(t), r cos θ(t), 0]
§I.8
Equations of Lines in Two Dimensions
1) (0, 1) is one point on the line 3x − 4y = −4. So [−2 − 0, 3 − 1] = [−2, 2] is a vector whose tail is on the
line and whose head is at (−2, 3). [3, −4] is a vector perpendicular to the line, so 51 [3, −4] is a unit vector
perpendicular to the line. The distance from (−2, 3) to the line is the length of the projection of [−2, 2] on
1
1
5 [3, −4], which is the magnitude of 5 [3, −4] · [−2, 2]. So distance=14/5
2a) The midpoint of the side opposite ~a is 21 (~b + ~c). The vector joining ~a to that midpoint is 12~b + 21 ~c − ~a. The
vector parametric equation of the line through ~a and 12 (~b + ~c) is
~x(t) = ~a + t 21~b + 12 ~c − ~a
6
Similarly, for the other two medians (but using s and u as parameters, rather than t)
~x(s) = ~b + s
~x(u) = ~c + u
1
a+
2~
1
a+
2~
1
c − ~b
2~
1~
c
2b −~
2b) The three medians meet at a common point if there are values of s, t and u such that
~a + t 12~b + 12 ~c − ~a = ~b + s 12 ~a + 21 ~c − ~b = ~c + u 12 ~a + 21~b − ~c
(1 − t)~a + 2t ~b + 2t ~c = s2 ~a + (1 − s)~b + s2 ~c = u2 ~a + u2 ~b + (1 − u)~c
Assuming that the triangle has not degenerated to a line segment, this is the case if and only if the coefficients
of ~a, ~b and ~c match
= u2
1 − t = 2s
t
2
t
2
or
= 1 − s = u2
= 2s
=1−u
s = t = u, 1 − t =
t
2
=⇒
s=t=u=
2
3
The medians meet at 31 (~a + ~b + ~c).
§I.9
Equations of Planes in Three Dimensions
1) Two vectors in the plane are [1, 0, 1] − [0, 1, 1] = [1, −1, 0] and [1, 1, 0] − [0, 1, 1] = [1, 0, −1]. So, one vector
perpendicular to the plane is [1, −1, 0] × [1, 0, −1] = [1, 1, 1]. An equation of the plane is
[x − 0, y − 1, z − 1] · [1, 1, 1] = 0
or
x+y+z =2
Actually, you should have just guessed that this was the answer.
2) The two planes x + y + z = 3 and x + y + z = 9 are parallel. The centre must be on the plane x + y + z = 6
half way between them. So, the center is on x + y + z = 6, 2x − y = 0 and 3x − z = 0. Solving these
three equations gives (1, 2, 3) as the centre. (1, 1, 1) is a point on x + y + z = 3. (3, 3, 3) is a point on
x + y + z = 9. So [2, 2, 2] is a vector with tail on x + y + x = 3 and head on x + y + z = 9. Furthermore
√
[2, 2, 2] is perpendicular√to the two planes. So the distance between the planes is k[2, 2, 2]k = 2 3 and the
radius of the sphere is 3. The sphere is
(x − 1)2 + (y − 2)2 + (z − 3)2 = 3
3) Set y = 0 and then solve 2x + 3y − z = 0, x − 4y + 2z = −5. The result, (−1, 0, −2), is one point on the
plane. Set y = 5 and then solve 2x + 3y − z = 0, x − 4y + 2z = −5. The result, (−3, 5, 9), is another
point on the plane. So three points on the plane are (−2, 0, 1), (−1, 0, −2) and (−3, 5, 9). [−1, 0, 3] and
[1, −5, −8] are two vectors having both head and tail in the plane. [3, −1, 1] is a vector perpendicular to
the plane. The plane is
3x − y + z = −5
4) All three points (1, 2, 3), (2, 3, 4) and (3, 4, 5) are on the line ~x(t) = (1, 2, 3) + t(1, 1, 1). There are many
planes through that line.
5) The vector ~n is perpendicular to the plane ~n · ~x = c. So the line
~x(t) = p
~ + t~n
passes through p~ and is perpendicular to the plane. It crosses the plane at the value of t which obeys
~n · ~x(t) = c
or
7
~n · [~
p + t~n] = c
namely
t = [c − ~n · p~]/k~nk2
The vector
~x(t) − p~ = ~n[c − ~n · p
~]/k~nk2
has head on the plane ~n · ~x = c, tail at p~ and is perpendicular to the plane. So the distance is the length of
that vector, which is
distance = |c − ~n · p
~|/k~nk
§I.10
Equations of Lines in Three Dimensions
1) [1, 1, 0] is perpendicular to x + y = 0 and hence to the line. [1, −1, 2] is perpendicular to x − y + 2z = 0
and hence to the line. [1, −1, −1] is perpendicular to both [1, 1, 0] and [1, −1, 2] and hence is parallel to the
line. The line is
[x−2, y +1, z +1] = t[1, −1, −1]
or
x−2 = t, y +1 = −t, z +1 = −t
8
or
x−2 = −y −1 = −z −1