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Subtraction is done by adding the second vector in reverse
direction to the first vector.
−b
Vectors and vector calculus
c=a-b
Vectors in two and three dimensions
a
Free vectors: A vector is free if it is considered to be the
same vector under translation. PQ is a free vector because it
can be repositioned as shown in the following diagram.
y
Q
Linear dependence and independence
Vectors x, y and z are linearly dependent if there exist nonzero scalars p, q and r such that
P
px + qy + rz = 0.
x
If px + qy + rz = 0 only when p = q = r = 0, then x, y and z are
linearly independent.
A position vector is not a free vector because it cannot be
translated. It always starts from a reference point O, the origin.
Example 1
OP and OQ shown below are position vectors.
3w
y
x
2y
z
Q
P
O
x
Since 3w + x + z = 2y, i.e. 3w + x – 2y + z = 0,
Therefore, w, x, y and z are linearly dependent.
Most vectors can be added or subtracted if they represent the
same quantity.
Example 2 Three non-coplanar vectors are linearly
independent.
Exceptions:
1. Addition of two position vectors is undefined.
2. A displacement vector can be added to a position vector to
give a new position vector.
3. Subtraction of 2 position vectors is defined as displacement.
A typical example is the case where the three vectors are
perpendicular to each other.
4. Force vectors acting on different objects cannot be added.
Addition of vectors is carried out by putting the head of one
to the tail of the other. The order is irrelevant.
b
a
c=a+b
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Vectors and vector calculus
1
Example 3 Given a = 3 units north and b = 4 units east, find
a vector c which is linearly dependent on a and b. Find
another vector d which is also linearly dependent on a and b.
A possible vector c is c = a + b because c − a − b = 0; a
possible vector d is d = a − 2b because d − a − 2b = 0.
Example 4 Given that vectors p, q and r are linearly
independent, and a 2 (p + q − r) = (a + 6)p + (2a + 3)q − 9r,
find the value(s) of a.
a 2 (p + q − r) = (a + 6)p + (2a + 3)q − 9r,
∴ (a 2 − a − 6) p + (a 2 − 2a + 3) q − (a 2 − 9) r = 0.
Since p, q and r are linearly independent,
∴ a 2 − a − 6 = a 2 − 2a + 3 = −(a 2 − 9) = 0,
Resolution of a vector into rectangular components
A vector is resolved into rectangular components when it is
expressed in terms of i, j and k, e.g. r = li + mj + nk.
The magnitude of r is |r| = r = l 2 + m 2 + n 2 .
Example 1 An aeroplane is 50 km NE of Melbourne Airport
and it is flying at an altitude of 10 km. Resolve its position
vector from Melbourne Airport into i, j and k components.
They are respectively pointing east, north and up. What is the
straight-line distance of the plane from the runway?
y
N
(
P 50 cos 45 o ,50 sin 45 o
i.e. (a + 2)(a − 3) = (a + 1)(a − 3) = −(a + 3)(a − 3) = 0 .
a = 3 is the only value that satisfies all three equations.
)
50
45o
Unit vectors
0
Any vector with a magnitude of 1 is called a unit vector. A
unit vector in the direction of vector s is labelled as ŝ. It is
found by dividing vector s by its magnitude |s|, i.e. ŝ = s/|s|.
There are three unit vectors, i, j and k that are particularly
useful in vector analysis. i, j and k are in the direction of x, y
and z axes respectively. They are perpendicular to each other
and therefore, linearly independent, i.e. if pi + qj + rk = 0,
then p = q = r = 0.
x
OP = 50 cos 45 o i + 50 sin 45 o j +10 k
= 25 2 i + 25 2 j +10 k.
Distance from the runway = OP = 50 2 + 10 2 = 10 26 km
Example 2 Find the magnitude of s = 2i – j + 2.5k, and a
unit vector in the direction of s.
3 5
.
2
Unit vector in the direction of s:
|s| = s = 2 2 + (− 1) + 2.5 2 =
2
z
y
k
ŝ = s/|s| = (2i – j + 2.5k)/
j
3 5
2
( 2i – j + 2.5k).
=
2
3 5
Scalar (dot) product of two vectors
i
x
Any vector can be written in terms of i, j and k, e.g.
s = 2i – j + 2.5k.
s
2.5k
Scalar (dot) product of two vectors is defined to meet the
requirements of many physical situations,
e.g. in physics work done W joules by a force F newtons in
displacing an object s metres is
F
W = Fs cos θ
θ
s
If the scalar product of F and s is defined as F.s = Fs cosθ ,
then W = F.s.
2i
The scalar product of two vectors gives a scalar.
−j
s, i, j and k are linearly dependent.
Note: (1) In a scalar product, θ is the angle between two
vectors that are placed tail to tail.
(2) r.r = r 2 , ∴r = √ r.r
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Vectors and vector calculus
2
Example 1 Find the scalar product of the two vectors.
5
(3i + 2j + 4k).(2i + j – 2k) = 6 + 2 − 8 = 0.
∴the two vectors are parallel.
120 o
5
Example 2 Show that 3i + 2j + 4k is perpendicular to
2 i + j – 2 k.
Place the two vectors tail to tail, θ = 60 o .
60 o
Example 3 Find c such that 2i + 2j – k is perpendicular to
i + cj + 6k.
(2i + 2j – k).(i + cj + 6k) = 0, ∴ 2 + 2c − 6 = 0, c = 2 .
Scalar product = 5 × 5 cos 60 o = 12.5 .
Example 4 Find a vector perpendicular to g = i – 2j + 3k.
Parallel and perpendicular vectors
Two vectors, p and q, are parallel if one equals the other
multiplied by a constant, p = αq.
Two parallel vectors are linearly dependent.
The angle between two parallel vectors is either zero or 180 o .
∴F.s = Fs or −Fs.
Two vectors are perpendicular ( θ = 90 ) if F.s = 0.
Conversely, F.s = 0 if two vectors are perpendicular.
o
Two perpendicular vectors are linearly independent.
Note that i.i = j.j = k.k = 1 and i.j = j.k = k.i = 0.
Scalar product of vectors in i, j, k components
F
θ
s
If F and s are in terms of i, j and k, i.e. F = ai + bj + ck and
s = xi + yj + zk, then
F.s = ( ai + bj + ck ).( xi + yj + zk ) = ax + by + cz.
ax + by + cz
Since F.s = Fs cos θ , ∴ cosθ =
.
Fs
Hence cos θ =
ax + by + cz
a + b + c2 x2 + y2 + z2
2
2
.
Let r = li + mj + nk be a vector perpendicular to g.
r.g = 0. ∴ l − 2m + 3n = 0 .
Let m = 0 and n = 1 , ∴l = −3.
Hence r = −3i + k is a vector (out of an infinite number of
them) perpendicular to g.
Example 5 Find a unit vector perpendicular to p = 2i – 3j
and (a) on the same plane, (b) not on the same plane.
Let u = li + mj + nk be a unit vector ⊥ to p = 2i – 3j,
∴ l 2 + m 2 + n 2 = 1, l 2 + m 2 + n 2 = 1 and 2l − 3m = 0 .
(a) If u is on the same plane as p, then the k component of u is
zero, i.e. n = 0. Hence l 2 + m 2 = 1 ……(1)
and 2l − 3m = 0 ……(2)
Solve the simultaneous equations for l and m:
2
From (2), m = l …...(3). Substitute (3) in (1),
3
2 13
3
3 13
4
13
, ∴m = ±
l 2 + l 2 = 1, l 2 = 1, ∴l = ±
.
=±
9
9
13
13
13
Hence u =
3 13
2 13
3 13
2 13
i+
j or −
i−
j.
13
13
13
13
(b) u (or −u) is a unit vector ⊥ to p, but not on the same plane
as p. It contains i, j and k components.
l 2 + m 2 + n 2 = 1 ……(1) and, 2l − 3m = 0 ……(2)
2
From (2), m = l …...(3). Substitute (3) in (1),
3
13
4
13
l 2 + l 2 + n 2 = 1, ∴ l 2 + n 2 = 1, n = ± 1 − l 2 .
9
9
9
Choose l =
Example 1 Show that p = 3i + 2j + 4k is parallel to
1
1
1
q = i + j + k.
4
6
3
1
1
1
3
2
4
q= i+ j+ k=
i+
j+
k
4
6
3
12
12
12
1
1
=
(3i + 2j + 4k) =
p. ∴p and q are parallel.
12
12
n = ± 1−
1
13
=
13
2
2 13
, then m =
and
=
13
39
3 13
13 2
1
2 2
l = ± 1− = ±
.
9
9
3
13
2 13
2 2
i+
j+
k, or
13
39
3
13
2 13
2 2
u=
i+
j−
k.
13
39
3
Hence u =
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Vectors and vector calculus
3
Direction cosines
Consider vector h = ai + bj + ck, and the unit vector in the
direction of h is:
a
b
c
ĥ=
i +
j+
k,
a2 + b2 + c2
a2 + b2 + c2
a2 + b2 + c2
where
a 2 + b 2 + c 2 = |h|.
Hence α ≈ 66 o , β ≈ 145 o and γ ≈ 114 o.
Let the angles that the unit vector make with the x, y and zaxes be α , β , γ respectively.
z
y
h
α
β
Hence c 2 = a 2 + b 2 + c 2 c cosθ , and
c
∴ cosθ =
.
2
a + b2 + c2
b
0
a
x
and cos γ =
a
a +b +c
c
2
∴a.c + b.c + c.c = √(a + b + c).(a + b + c) c cos θ .
Since a, b and c are orthogonal, ∴a.b = 0, b.c = 0 and c.a = 0.
γ
2
Example 4 a, b and c are three orthogonal (i.e. mutually
perpendicular) vectors. Find an expression for the cosine of
the angle between (a + b + c) and c.
Consider the scalar product of two vectors: p.q = pq cos θ ,
∴(a + b + c).c = |a + b + c| |c| cos θ ,
c
∴ cos α =
Example 3 Find the angles that s = i – 2j − k makes with the
axes.
1
1
2
1
|s| = 6 , ŝ =
(i – 2j − k) =
i−
j−
k.
6
6
6
6
1
2
1
, cos β = −
, and cos γ = −
.
∴ cos α =
6
6
6
b
, cos β =
2
a2 + b2 + c2
Scalar and vector resolutes
a + b2 + c2
2
.
Instead of resolving a vector into i, j and k components, it can
be resolved in other ways, e.g. vector q can be resolved into
two perpendicular components, one component q1 parallel to
another vector s and the other component q2 ⊥ to s.
Hence ĥ = cos α i + cos β j + cos γ k .
q
cos α , cos β , cos γ are called the direction cosines of h.
q1
Example 1 Vector r has a magnitude of 10, and it makes
angles of 30 o , 45 o and 60 o respectively with i, j and k.
Express r in terms of i, j and k.
r = 10 cos 30 o i + 10 cos 45 o j + 10 cos 60 o k
∴r = 5 3 i + 5 2 j +5 k.
q2
s
q1 and q1 are respectively called the scalar and vector resolutes
of q parallel to s (or the scalar and vector projections of q onto
s).
q2 and q2 are respectively called the scalar and vector resolutes
of q perpendicular to s.
q1 = q.ŝ and q1 = q1ŝ = (q.ŝ)ŝ ,
Example 2 Find the magnitude and direction cosines of
3i − 4j + 5k.
q2 = q – q1, and q2 = |q – q1|.
Magnitude = 3 2 + (− 4 ) + 5 2 = 5 2 .
Example 1 Find the projection of 4i + k onto −3i + j – 2k,
i.e. scalar resolute of 4i + k in the direction of −3i + j – 2k.
2
Unit vector =
3
5 2
i−
4
5 2
Direction cosines: cos α =
4
j+
5
5 2
3
5 2
=
k.
3 2
,
10
2 2
5
2
cos β = −
=−
and cos γ =
=
.
5
2
5 2
5 2
Let h = 4i + k and g = −3i + j – 2k.
Unit vector in the direction of g: ĝ =
1
14
(−3i + j – 2k).
The projection of h onto g:
1
1
h.ĝ = (4i + k).
(−3i + j – 2k) =
(−12 − 2) = − 14 .
14
14
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Vectors and vector calculus
4
Example 2 Find the scalar and vector resolutes of
b = i – 2j + k in the direction of a = 2i – 3j and, perpendicular
to a.
1
â =
(2i − 3j).
13
1
8
.
Scalar resolute of b parallel to a: b.â =
(2 + 6) =
13
13
Vector resolute of b parallel to a:
8
1
8
×
(b.â)â =
(2i − 3j) = (2i − 3j).
13
13
13
Example 2 Use vectors to prove Pythagoras’ theorem and
the cosine rule.
B
B
a
c
a
c
θ
C
A
b
C
b
Introduce vectors a, b and c as shown above to simplify the
notations.
Vector resolute of b perpendicular to a:
8
3
2
b − (b.â)â = (i – 2j + k) − (2i − 3j) = − i − j + k.
13
13 13
Pythagoras’ theorem: c = b − a, c.c = (b − a).(b − a),
∴c.c = b.b − b.a − a.b + a.a.
Since a.b = b.a = 0, ∴ c 2 = a 2 + b 2 .
Scalar resolute of b perpendicular to a:
The cosine rule: c = b − a, c.c = (b − a).(b − a),
∴c.c = b.b − b.a − a.b + a.a.
Since a.b = b.a = ab cosθ = ab cos C ,
∴ c 2 = a 2 + b 2 − 2ab cos C .
2
2
182
 3  2
|b − (b.â)â| =  −  +  −  + 12 =
.
13
 13   13 
A
Example 3 Prove that the angle subtended by a diameter in a
circle is a right angle.
A
C
q
O
Vector proofs of simple geometric results
p
B
Example 1 Prove that the diagonals of a rhombus are
perpendicular.
Consider rhombus OPQR, OP = PQ = QR = OR .
Q
Introduce vectors p and q from centre O to B and C
respectively.
AO = p, AC = AO + OC = p + q, CB = OB − OC = p − q,
∴ AC • CB = (p + q).(p − q) = p.p − q.q = p 2 − q 2 = 0,
because p = q = radius of the circle.
R
Hence AC ⊥ CB , i.e. ∠ACB = 90o .
P
O
Find vector OP in terms of m, n, a and b.
OQ = OP + OR, RP = OP − OR,
(
Example 4 Point P divides AB according to the ratio m : n.
)(
OQ • RP = OP + OR • OP − OR
A
)
a
= OP • OP − OP • OR + OR • OP − OR • OR
2
m
P
n
B
b
2
= OP − OR = 0 .
Since OQ and RP are non-zero vectors, and OQ • RP = 0 ,
∴the diagonals are perpendicular.
O
m
AB
m+n
m
1
(b − a) =
(ma + na + mb − ma)
=a+
m+n
m+n
1
=
(na + mb).
m+n
OP = OA + AP = OA +
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Vectors and vector calculus
5
Example 5 Prove that the medians of a triangle are
concurrent, and the intersection trisects each median.
B
Example 1 Sketch the locus of a particle with its position
described by r(t) = t 2 i − (t−1)j, t ≥ 0 .
Vector equation: r(t) = t 2 i − (t−1)j, t ≥ 0 .
Parametric equations: x = t 2 , y = −(t − 1) .
M
b
Cartesian equation:
P
O
c
C
a
N
From the second equation, t = 1 − y , substitute into the first
equation, x = (1 − y ) 2 , ∴ y = ± x + 1
Since t ≥ 0 , ∴ y = −(t − 1) ≤ 1 , ∴ y = − x + 1 .
A
y
AM and BN are medians and CP is a line segment passing
1
through the intersection O of AM and BN .
1
Let OA = a, OB = b and OC = c. ∴ OM = (b + c) and
2
1
ON = (a + c). Let m, n and p be some positive constants
2
such that OM = −m a, ON = −n b and OP = − p c.
1
1
∴ −m a = (b + c) and −n b = (a + c).
2
2
From the last two equations, –2ma – b = –2nb – a,
hence (1 – 2m)a – (1 – 2n)b = 0.
Since a and b are not parallel, ∴ they are independent and
1
hence 1 – 2m = 0 and –(1 – 2n) = 0, i.e. m = n = .
2
1
1
∴ − b = (a + c), ∴ b = –a – c.
2
2
0
x
Example 2 Sketch the locus of a particle with position vector
r(t) = (3 sin t + 1) i + (2 cos t − 1) j, and t ≥ 0 .
Vector equation: r(t) = (3 sin t + 1) i + (2 cos t − 1) j, and t ≥ 0 .
Parametric equations:
AP = AO + OP = –a – pc, and AB = b – a = –2a – c.
x −1
,
3
y +1
y = 2 cos t − 1, ∴ cos t =
.
2
Let AP = k AB where k is a positive constant.
Cartesian equation:
Since a and c are independent, ∴ 2k − 1 = 0 and k − p = 0 ,
1
1
i.e. k = and p = . ∴ P is the mid-point of AB and CP is
2
2
a median. Hence the three medians are concurrent at O and the
intersection O trisects each median.
 x −1  y +1
Since sin 2 t + cos 2 t = 1, ∴ 
 = 1,
 +
 3   2 
∴ –a – pc = k(–2a – c), i.e. (2k – 1)a + (k – p)c = 0.
Vector equations, parametric equations and cartesian
equations (2-dimensional)
x = 3 sin t + 1, ∴ sin t =
2
i.e.
(x − 1)2 + ( y + 1)2
9
4
2
= 1.
The locus is an ellipse centred at (1,−1) . At t = 0 , the particle
π
is at x = 1, y = 1. At t = , it is at x = 2.5, y = 3 − 1 .
6
∴the particle starts at (1,1) and moves clockwise.
As a particle moves its position changes with time. Its position
can be described with a position vector r(t) that is expressed in
terms of i and j components, r(t) = x(t)i + y(t)j.
This equation is called a vector equation and t the parameter.
y
0
1
x
x = x(t) and y = y(t) are two rectangular components in terms
of the parameter t. They are called parametric equations.
When the parameter t is eliminated from the two parametric
equations, we obtain a cartesian equation of the locus (path),
f(x,y) = 0.
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Vectors and vector calculus
6
Example 3 Sketch the locus of a particle with position given
 1
 1
by r(t) =  t +  i +  t −  j, t > 0.
 t
 t
 1
 1
Vector equation: r(t) =  t +  i +  t −  j, t > 0.
 t
 t
Parametric equations:
1
1
x = t + ……(1), y = t − ……(2)
t
t
Cartesian equation:
x+ y
(1) + (2), x + y = 2t , ∴ t =
.
2
x+ y
2
Substitute into (1), x =
+
,
2
x+ y
∴x =
(x + y )2 + 4 ,
2(x + y )
Example 5 Two ships are on a collision course. Their
position vectors are r(t) = (2t + 11)i + (7t +6)j and
r(t) = (5t + 7)i + (4t + 10)j, where t ≥ 0 . When and where do
the ships collide?
The two ships collide when they are at the same place and at
the same time. Let (2t + 11)i + (7t +6)j = (5t + 7)i + (4t + 10)j.
Equate corresponding components:
4
2t + 11 = 5t + 7 and 7t +6 = 4t + 10, ∴t = .
3
4
41
46
At t = , the ships are at the same position r(t) =
i +
j.
3
3
3
Example 6 The position vectors for particles A and B are
rA(t) = (2t 2 + t − 1) i +(2t + 3) j and rB(t) = (t 2 + 5t − 4) i +3tj
respectively. Find an expression for their separation at time
t ≥ 0 . When and where do they collide?
∴ 2 x( x + y ) = (x + y ) + 4 ,
2
2 x 2 + 2 xy = x 2 + 2 xy + y 2 + 4, ∴ x 2 − y 2 = 4,
i.e.
y
x2 y2
−
= 1.
4
4
rB − rA
The locus of the particle is a hyperbola.
For t > 0 , x > 0, ∴ the locus consists of the right hand branch
only. At t → 0 , x → +∞, y → −∞, ∴ the particle moves
upwards along the hyperbola.
y
0
2
x
rA
1
1


Vector equation: r(t) =  t 2 + 2  i +  t 2 − 2  j, t > 0.
t
t




Parametric equations:
1
1
x = t 2 + 2 ……(1), y = t 2 − 2 ……(2)
t
t
Cartesian equation:
x+ y
(1) + (2), x + y = 2t 2 , ∴ t 2 =
.
2
x+ y
2
x2 y2
Substitute into (1), x =
+
, ∴ −
= 1.
2
x+ y
4
4
= −(t 2 − 4t + 3) i + (t − 3) j
(t
2
)
− 4t + 3 + (t − 3) .
2
2
The particles collide when their separation is zero,
i.e.
(t
2
)
(
)
− 4t + 3 + (t − 3) = 0, t 2 − 4t + 3 + (t − 3) = 0 ,
2
2
2
2
(t − 3)2 (t − 1)2 + (t − 3)2 = 0, (t − 3)2 ((t − 1)2 + 1) = 0,
∴t = 3 and hence r = 20i + 9j is where they collide.
Differentiation of a vector with respect to time
Given vector r(t) = x(t)i + y(t)j, its first and second
derivatives with respect to t are respectively
d
dx
dy
d2
d 2x
d2y
r=
i+
j and
r=
i+
j.
2
2
dt
dt
dt
dt
dt
dt 2
Derivatives of position and velocity vectors
y
0
x
rB − rA = (t 2 + 5t − 4) i +3tj − [ (2t 2 + t − 1) i +(2t + 3) j]
Separation = |rB − rA| =
Example 4 A particle has position vector
1 
1

r(t) =  t 2 + 2  i +  t 2 − 2  j, t > 0 . Find the cartesian
t
t

 

equation and sketch the graph of its locus. Compare the
motion of this particle with that in example 3.
rB
0
2
The particle has the same locus but different speed.
x
If r(t) is a position vector, its first derivative is a velocity
vector v(t), and the second derivative is an acceleration vector
a(t).
v(t) =
d
d
r , a(t) =
v or a(t) =
dt
dt
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d2
r.
dt 2
Vectors and vector calculus
7
Example 1 The position vector of an object moving in a
plane is given by r(t) = t 3 i + t 2 j, t ≥ 0 . (a) Find its velocity,
speed and acceleration when t = 1 . (b) Draw the graph of the
locus, and the velocity and acceleration vectors at t = 1 .
(a) r(t) = t 3 i + t 2 j, t ≥ 0 , v(t) =
v=
(3t ) + (2t )
a(t) =
d
v = 6t i + 2j.
dt
2 2
2
d
r = 3t 2 i +2t j,
dt
= 9t 4 + 4t 2 ,
(a)
At t = 1, v = 3i + 2j, v = 13 , a = 6i + 2j.
2
2
 1
(b) x = t , y = t , ∴ y =  x 3  = x 3 . Since t ≥ 0 , ∴ x ≥ 0 .
 
 
y
2
3
3
v
a
1
4
r(t) = et i + e −t j, v(t) =
(e ) + (− e )
t 2
−t 2
a = −12 cos(2t ) i − 12 sin( 2t ) j = −4(3 cos(2t ) i + 3 sin(2t ) j),
∴ a = −4 r. Hence a is opposite in direction to r.
x = 3 cos(2t ), ∴
x
y
= cos(2t ). y = 3 sin(2t ), ∴ = sin(2t ).
3
3
7
x
= e 2t + e − 2t , a(t) =
y
a
1
v
1
At t = 0.1, x > 0 and y > 0. ∴The particle starts from (3,0)
and moves anticlockwise.
y
v
a
d
v = e t i + e − t j.
dt
Parametric equations: x = e t , y = e −t .
1
1
Cartesian equation: y = e −t = t , ∴ y = . When t = 0, x = 1
x
e
and y = 1. ∴ the particle starts from (1,1) .
x
2
x  y
Use sin 2 A + cos 2 A = 1 to eliminate t:   +   = 1,
3  3
2
2
∴ x + y = 9. The locus is a circle of radius 3 and centre
(0,0) . At t = 0, x = 3 cos 0 = 3, y = 3 sin 0 = 0.
d
r = e t i − e − t j,
dt
At t = 0, v = i − j, a = i + j.
0
(b) v.r = −18 sin(2t ) cos(2t ) + 18 sin( 2t ) cos(2t ) = 0, ∴ v ⊥ r.
2
Example 2 Find the velocity, acceleration and speed of a
particle whose position vector is given by r(t) = et i + e −t j,
t ≥ 0. Sketch the path of the particle and draw the velocity and
acceleration vectors at t = 0 .
v=
r = 3cos(2t)i +3sin(2t)j,
d
v=
r = −6 sin(2t ) i + 6 cos(2t ) j,
dt
d
a = v = −12 cos(2t ) i − 12 sin( 2t ) j.
dt
(c)
1
0
Example 3 A particle moves so that its position vector at
time t is given by r = 3cos(2t)i + 3sin(2t)j for t ≥ 0 .
(a) Find the velocity and acceleration at time t.
(b) Show that in this case, v is perpendicular to r, and a is
opposite in direction to r.
(c) Sketch the locus of the particle and show the velocity and
acceleration vectors at time t.
0
x
Example 4 An object moves in a path with position vector
r(t) = cos t i + 2 sin t j, t ≥ 0 . (a) Show that the path is
elliptical. (b) Show that the acceleration points towards the
origin.
y
(a) x = cost , y = 2 sin t , ∴ = sin t.
2
2
y
Eliminate t, x 2 +
= 1. Path is elliptical.
4
(b) r(t) = cos t i + 2 sin t j, v = − sin t i +2 cos t j,
a = − cos t i −2 sin t j = −(cos t i +2 sin t j) = − r. ∴a is opposite
to r and hence it is towards the origin.
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Vectors and vector calculus
8
Antidifferentiation of a vector with respect to time
Given acceleration a(t) = p(t)i+ q(t)j,
∫
position is r(t) = ∫
velocity is v(t) =
a(t) dt = ∫ p(t ) i dt + ∫ q (t ) j dt , and
v(t) dt.
Example 1 Given the acceleration a = 2 cos t i + 2 sin t j, and
when t = 0, r = −2i and v = −2j, of a particle.
(a) Show that the speed of the particle is constant.
(b) Show that the velocity and the acceleration are always
perpendicular.
(c) Show that the acceleration is always towards the centre
of the circular path.
(d) Show that the path is circular.
(e) Sketch the path, and show the direction of motion of the
particle at time t.
∫
∫
∴v = 2 sin t i −2 cos t j + c, where c is a constant vector.
When t = 0, the velocity is v = 7i + 10j, ∴c = 7i + 10j.
Hence v = −10t j + 7i + 10j = 7i +(10 − 10t ) j.
∴r = ∫ 7 i dt + ∫ (10 − 10t ) j dt = 7t i + (10t − 5t 2 ) j + d, where d
is a constant vector.
When t = 0, the position r = 0, ∴ d = 0.
Hence r = 7t i + (10t − 5t 2 ) j.
(b) x = 7t , y = 10t − 5t 2 .
10
5 2
x x
x ,
Eliminate t, y = 10  − 5  , y = x −
7
49
7 7
y=−
When t = 0, v = −2j, ∴c = 0, ∴ v = 2 sin t i −2 cos t j.
(2 sin t )2 + (− 2 cos t )2
(a) a = −10j, v = ∫ − 10 j dt = −10t j + c, c is a constant vector.
2
(a) a = 2 cos t i + 2 sin t j, v = 2 cos t i dt + 2 sin t j dt ,
Hence v =
Example 2 Given the acceleration a = −10j, and when t = 0,
the position is r = 0 and the velocity is v = 7i + 10j.
(a) Find the position vector at time t ≥ 0 .
(b) Describe the path.
(c) At what time is the y-coordinate the same as that at t = 0?
(d) What is the x-coordinate at that time?
(
)
(
)
5
5 2
5 2
2
x − 14 x = −
x − 14 x + 49 + 5 = − (x − 7 ) + 5 .
49
49
49
Path is a parabola, vertex at (7,5).
y
5
= 4 = 2 , a constant.
(b) v.a = 4 sin t cos t − 4 cos t sin t = 0, ∴v is ⊥ a.
(c)
∫
∫
0
r = 2 sin t i dt − 2 cos t j dt = −2 cos t i −2 sin t j + d,
7
14
x
where d is a constant vector.
(c) At t = 0, y = 0. When y = 0, 10t − 5t 2 = 0, ∴t = 2.
When t = 0, r = −2i, ∴d = 0, ∴r = −2 cos t i −2 sin t j.
∴a = −r. Hence a is always towards the origin, i.e. the centre
of the circular path.
(d)
x
y
x = −2 cos t , ∴ − = cos t. y = −2 sin t , ∴ − = sin t.
2
2
2
2
 x  y
Use sin A + cos A = 1 to eliminate t:  −  +  −  = 1,
 2  2
2
2
∴ x + y = 4. The path is a circle of radius 2 and centre (0,0 ) .
2
2
(e)
y
v
0
Example 3 A particle is projected at the origin with a
velocity given by v = i + j + 10k, and it has a constant
acceleration a = −10k. The unit vector k is pointing vertically
upward.
(a) Find the velocity vector of the particle at time t.
(b) Find the position vector of the particle at time t.
(c) Find the maximum height above the origin.
(d) Find the distance of the particle from the origin when it
returns to the same level as the origin.
(a) a = −10k, v = ∫ − 10 k dt = −10t k + c, where c is a
r
−2
(d) At t = 2, x = 2t = 14.
2
x
constant vector.
When t = 0, the velocity is v = i + j +10k, ∴c = i + j + 10k.
Hence v = −10t k + i + j + 10k = i + j + (10 − 10t ) k.
(b) ∴r =
∫
i dt +
∫
∫
j dt + (10 − 10t ) k dt
= t i +t j + (10t − 5t ) k + d, where d is a constant vector.
2
When t = 0, the position r = 0, ∴ d = 0.
Hence r = t i +t j + (10t − 5t 2 ) k.
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Vectors and vector calculus
9
(c) At the highest point, the particle moves horizontally, i.e.
the k component of v is zero. ∴10 − 10t = 0, t = 1.
At t = 1, the k component of r is 10t − 5t 2 = 10 − 5 = 5, ∴the
maximum height is 5.
(d) When the particle returns to the starting level, the k
component of r is zero. ∴10t − 5t 2 = 0, t = 2.
At t = 2, r = t i +t j + (10t − 5t 2 ) k = 2i + 2j.
∴distance from origin = |r(2)| = 2 2 + 2 2 = 2 2 .
The last two examples are projectile motions under constant
acceleration due to gravity, g.
A general consideration of projectile motion under
constant acceleration due to gravity (2 D)
Eliminating t from the last two equations:
x
x = (v0 cos θ )t , ∴ t =
. Substitute into
v0 cos θ
1
y = (v0 sin θ )t − gt 2 ,
2
2
x 
1 
x
 ,
− g
v0 cos θ 2  v0 cos θ 
g
∴ y = x tan θ −
x 2 . This is the equation of the
2(v0 cos θ ) 2
locus of the projectile satisfying the conditions stated in the
opening sentence.
y = (v0 sin θ )
y
0
x
Let a = −gj and the initial position and velocity are
respectively r = 0 and v = v0 cos θ i + v0 sin θ j.
Range
v
Initial speed v0
a
θ
v0 sin θ
j
i
v0 cos θ
the j component of r is y = (v0 sin θ )t −
∫
a = −gj, v = − g j dt = − gt j + c, where c is a constant vector.
When t = 0, the velocity is v = v0 cos θ i + v0 sin θ j,
∴c = v0 cos θ i + v0 sin θ j.
Hence v = − gt j + v0 cos θ i + v0 sin θ j
= v0 cos θ i + (v0 sin θ − gt ) j.
∫
The maximum height is reached when the j component of v is
v sin θ
.
zero, i.e. v0 sin θ − gt = 0, ∴ t = 0
g
v sin θ
At t = 0
,
g
∫
∴r = v0 cos θ i dt + (v0 sin θ − gt ) j dt
= (v0 cos θ )t i + ((v0 sin θ )t −
1 2
gt ) j + d, where d is a constant
2
vector.
When t = 0, the position r = 0, ∴ d = 0.
1
∴ r = (v0 cos θ )t i + ((v0 sin θ )t − gt 2 ) j.
2
Hence at time t, the horizontal and vertical components of the
velocity vector are v = v0 cos θ and v = v0 sin θ − gt
respectively.
Also, at time t, the horizontal and vertical components of the
1
position vector are x = (v0 cos θ )t and y = (v0 sin θ )t − gt 2
2
respectively.
1 2 (v0 sin θ ) 2
.
gt =
2
2g
(v0 sin θ ) 2
.
2g
Also, maximum height is reached at mid-range where
2
2
v sin θ v0 sin θ cos θ v0 sin 2θ
=
=
.
x = (v0 cos θ )t = (v0 cos θ ) 0
g
g
2g
∴ maximum height =
v0 sin 2θ
.
g
2
∴the range of the projectile is
Example 1 A particle is projected with a speed of 20 ms-1 at
60o with the horizontal from the origin. Take g = 10 ms-2.
(a) Find the equation of the path of the particle.
(b) Find the maximum height reached.
(c) Find the range of the particle.
(a) y = x tan 60 o −
x2
10
x 2 , ∴ y = 3x − .
o 2
20
2(20 cos 60 )
(b) Maximum height =
(v0 sin θ ) 2 (20 sin 60 o ) 2
=
= 15 m.
2g
2 × 10
v0 sin 2θ 20 2 sin 120 o
=
= 20 3 m.
10
g
2
(c) Range =
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Vectors and vector calculus
10