Download The geometry of Euclidean Space

Chapter 1
The geometry of Euclidean
Space
We consider the basic operations on vectors in 3 and 3 dim. space: vector
addition, scalar multiplication, dot product and cross product. In section ??
we generalize these notions to n dim’l space.
1.1
Vectors in 2, 3 dim space
Definition 1.1.1. A vector in Rn , n = 2, 3 is an ordered pair(triple) of real
numbers, such as
(a1 , a2 ), or (a1 , a2 , a3 ).
Here the numbers a1 , a2 are called x coordinate, y coordinate or x component, y component of (a1 , a2 ). The point (0, 0) is called the origin and
denoted by O.
a = (a1 , a2 ) or a = (a1 , a2 , a3 ) will be standard notation for vectors. A
point P is represented by ordered pairs of real numbers (a1 , a3 ) called Cartesian coordinate) of P.
Vectors are identified with points in the plane or space.
R2 = {(a1 , a2 ) | a1 ∈ R, a2 ∈ R}
1
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CHAPTER 1. THE GEOMETRY OF EUCLIDEAN SPACE
y
✻
a2
b1
P q(a1 , a2 )
O
q
b2
a1
✲x
Q(b1 , b2 )
Figure 1.1: Coordinate plane
Vector addition and scalar multiplication-algebraic view
The operation of addition can be extended to R3 . Given two triples, a =
(a1 , a2 , a3 ), b = (b1 , b2 , b3 ), we define
a + b = (a1 , a2 , a3 ) + (b1 , b2 , b3 ) = (a1 + b1 , a2 + b2 , a3 + b3 ) = b + a
to be the sum of (a1 , a2 , a3 ) and (b1 , b2 , b3 ). The vector 0 = (0, 0, 0) is the
zero element.
Scalar multiple of a vector. For real s and vector v, sv is the vector having
magnitude |s|, having same direction as v when s > 0, opposite direction when
s < 0. Here s is called scalar sv is the scalar multiple of v. Componentwise,
it is written as
s(v1 , v2 , v3 ) = (sv1 , sv2 , sv3 ).
Example 1.1.2. Show that (−s)v = −(sv) for any scalar s and vector v.
Commutative law and associate law for additions:
(i) u + v = v + u
(ii) (u + v) + w = u + (v + w)
(iii) −(u1 , u2 , u3 ) = (−u1 , −u2 , −u3 )
(commutative law)
(associate law)
(additive inverse)
The difference is defined as
(a1 , a2 , a3 ) − (b1 , b2 , b3 ) = (a1 − b1 , a2 − b2 , a3 − b3 ).
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1.1. VECTORS IN 2, 3 DIM SPACE
For any real α, and u = (u1 , u2 , u3 ) in R3 , scalar multiple αu is defined
as
αu = (αu1 , αu2 , αu3 ).
Additions and scalar multiplication has the following properties:
(i) (αβ)u = α(βu)
(associate law)
(ii) (α + β)u = αu + βu
(distributive law)
(iii) α(u + v) = αu + αv
(distributive law)
(iv) α(0, 0, 0) = (0, 0, 0)
(property of 0 )
(v) 0u = 0
(property of 0 )
(vi) 1u = u
(property of 1 )
Vectors-Geometric view
We associate a vector a with a point (a1 , a2 , a3 ) in the space. Thus, we can
visualize it with a position vector a = (a1 , a2 , a3 ). One can also interpret a
vector as directed line segment having an initial point (Usually the
initial point is at the origin), i.e, a line segment with specified magnitude
and direction, and initial point at the origin. Vectors are usually denoted by
boldface such as a or ~a.
The elements in R3 are not only ordered triple of numbers, but are also
regarded as vectors. We call a1 , a2 and a3 the components of a. The triple
(0, 0, 0) is called (zero vector) denoted by 0 or ~0.
Two vectors a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) are equal if a1 = b1 , a2 = b2
and a3 = b3 . Geometrically, this means they have the same direction and
magnitude.
Geomteric representation of vectors
−→
See Figure 1.3. The directed line segment P Q from P to Q is denoted by P Q.
P and Q are called initial position and terminal position respectively.
The vector with tail at origin is called position vector. If we move the
vector in parallel, we regard it as the same vector. In other words, a vector is
determined by direction and magnitude. Referring the parallelogram ABDC
−→
−→
−→
−→
in Figure 1.3, we see AB = CD and AC = BD.
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CHAPTER 1. THE GEOMETRY OF EUCLIDEAN SPACE
z
✻
a3
❍❍
❍❍r
✑P
✑ ❅
O ✑
❅ ✲y
✑
a2
✑
✑
x
✠
a1
Figure 1.2: point P (a1 , a2 , a3 )
✯Q
✟
✟
✟
✟
P
✶D
✏✏
❖❈
✏
C ✏✏
❈
❖❈
❈
❈
✶❈
✏
❈
✏✏ B
✏
❈✏
A
Figure 1.3: vector
✑❅
✸
✑
❅
✑
❅
✑
u+v
✲
❅
P ✑
✑
❅
✑
❅
✑
v❅
✑
✑
❘✑
❅
u ✑
✑❅
✸
✑
❅v
✑
❅
✑
u+v
✲
✑
❘
❅
u ✑
(1)
Figure 1.4: sum of two vectors
(2)
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1.1. VECTORS IN 2, 3 DIM SPACE
✸
✑
v✑
✑
q✑
✑
−v
✑
✑
✰
✑
Figure 1.5: v and −v
2v✂
✂
✂
✂
✂
✂
✂✂✍
3
v
2
✂
✂✂✍
✂✍
✂ − 12 v
✂
✂
✂
✂✌
✂ v✂
✂
Figure 1.6: scalar multiple of v
See figure 1.4 (1). If two vectors u, v have same tail P , we define the sum
of u and v as the vector u + v having position at the opposite vertex of the
parallelogram.
Alternative notation
v = (v1 , v2 , · · · , vn ) = [v1 , v2 , · · · , vn ]
z
✻
A(a
r 1 , a2 , a3 )
❳❳
❆ ❳❳❳❳Pr (a1 + b1 , a2 + b2 , a3 + b3 )
❆
❆
O❆❳
✲y
r ❳
❳❳❳❆
❳❆r
B(b1 , b2 , b3 )
x
✠
Figure 1.7: Addition
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CHAPTER 1. THE GEOMETRY OF EUCLIDEAN SPACE
z
k
j
y
O
i
x
Figure 1.8: standard basis vector
is called a row vector, while it can also be written as a column vector:

v1

 
 v2 
.
.
.
vn
(1.1)
Standard basis vectors
Definition 1.1.3. The following vectors i, j, k are called (standard basis
vector) of R3 (Figure 1.13).
i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)
Remark 1.1.4. For a given v = (a1 , a2 , a3 ),
(a1 , a2 , a3 ) = a1 (1, 0, 0) + a2 (0, 1, 0) + a3 (0, 0, 1)
we write v = a1 i + a2 j + a3 k.
1.2
Inner product, length, distance
Dot product-Inner product
Definition 1.2.1. Given two vectors a = a1 i+a2 j+a3 k and b = b1 i+b2 j+b3 k
we define
a1 b1 + a2 b2 + a3 b3
1.2. INNER PRODUCT, LENGTH, DISTANCE
7
to be the dot product or (inner product) of a and b and write a · b.
Definition 1.2.2. If u = (u1 , u2 , · · · , un ) is a vector in Rn , then the length
or the norm of u is given by
kuk =
q
u21 + · · · u2n =
√
u · u.
to be the dot product or (inner product) of a and b and write a · b.
Example 1.2.3. Let a = 2i − 3j + k, b = i + 2j − k. Find
(1) a · a
(2) a · b
(3) a · (a − 3b)
(4) (3a + 2b) · (a − b)
Proposition 1.2.4 (Properties of Inner Product). For vectors a, b, c and
scalar α, the following hold:
(1) a · a ≥ 0 (equality holds only when a = 0)
(2) a · b = b · a
(3) (a + b) · c = a · c + b · c
(4) (αa) · b = α(a · b)
(5) kak =
√
a·a
Example 1.2.5. For a, b, c we have the following.
(1) (a − b) · c = a · c − b · c
(2) a · (b + c) = a · b + a · c
(3) a · (b − c) = a · b − a · c
(4) a · b = 12 (kak2 + kbk2 − ka − bk2 )
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CHAPTER 1. THE GEOMETRY OF EUCLIDEAN SPACE
❅
■
a❅
✶
b ✏✏
❅ θ
✏✏
✏
❅✏
Figure 1.9: Angle between two vectors
Length of vectors
The length, norm of a vector a = (a1 , a2 , a3 ) is
p
(a1 − 0)2 + (a2 − 0)2 + (a3 − 0)2 =
q
a21 + a22 + a23
denoted by kak. Also we note that
kak = (a · a)1/2 .
Example 1.2.6. Find the lengths of the following vectors.
(1) a = (3, 2, 1)
(2) 3i − 4j + k
Unit vectors
u=
u
kuk
Definition 1.2.7. A vector with norm 1 is called a unit vector. Any nonzero
vector u can be made into a unit vector by setting u/kuk. This process is called
a normalization.
Example 1.2.8. Normalize the followings.
(1) i + j + k
(2) 3i + 4k
(3) ai − j + ck
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1.2. INNER PRODUCT, LENGTH, DISTANCE
Distance between two points
Definition 1.2.9. If P = (u1 , u2 , u3 ) and Q = (v1 , v2 , v3 ) then
d(u, v) := ku − vk =
p
(u1 − v1 )2 + (u2 − v2 )2 + (u3 − v3 )2
is called the distance between P and Q.
Angle between two vectors
Proposition 1.2.10. Let u, v be two nonzero vectors in R2 or R3 and let θ
be the angle between them. Then
u · v = kuk kvk cos θ.
Hence
θ = cos−1
−→
u·v
kuk kvk
−→
−→
Proof. Let u = AB, v = AC. Then u − v = CB. Let∠CAB = θ. Then by
C
✑
✑ ❆
✑
❆
v ✑
❆
✑
✑
❆
✑
❆
✑ θ
❆
A ✑
u
B
|BC|2 = |AB|2 + |AC|2 − 2|AB| |AC| cos θ
Figure 1.10: law of cosine
the law of cosine (figure 1.10) we have
kv − uk2 = kuk2 + kvk2 − 2kuk kvk cos θ.
The left hand side is
ku − vk2 = (u − v) · (u − v)
= u·u−u·v−v·u+v·v
= kuk2 − 2 u · v + kvk2 .
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CHAPTER 1. THE GEOMETRY OF EUCLIDEAN SPACE
Hence we obtain
kuk kvk cos θ = u · v.
Corollary 1.2.11. Two nonzero vector u and v are perpendicular, orthogonal if and only if u · v = 0.
Example 1.2.12. Find the angle between i + j + 2k and −i + 2j + k.
sol.
By proposition 1.2.10,
(i + j + 2k) · (−i + 2j + k)
−1 + 2 + 2
3
1
√
=√
= =
ki + j + 2kk k − i + 2j + kk
6
2
1+1+4 1+4+1
Hence the angle is cos−1 (1/2) = π/3.
Corollary 1.2.13. Given two points A(u1 , u2 , u3 ), B(v1 , v2 , v3 ), the area of
the triangle OAB is
1p
(u2 v3 − u3 v2 )2 + (u3 v1 − u1 v3 )2 + (u1 v2 − u2 v1 )2
2
−→
−→
Proof. Let OA = u, OB = v, ∠BOA = θ. Then the area of △OAB is
=
=
=
=
1
|OA| |OB| sin θ
2
p
1
kuk kvk 1 − cos2 θ
2
1p
kuk2 kvk2 − (u · v)2
2q
1
(u21 + u22 + u23 )(v12 + v22 + v32 ) − (u1 v1 + u2 v2 + u3 v3 )2
2
1p
(u2 v3 − u3 v2 )2 + (u3 v1 − u1 v3 )2 + (u1 v2 − u2 v1 )2
2
Theorem 1.2.14 (Cauchy-Schwarz inequality). For any two vectors u, v
|u · v| ≤ kuk kvk
holds, and the equality holds iff u and v are parallel.
1.2. INNER PRODUCT, LENGTH, DISTANCE
11
Proof. We may assume u, v are nonzero. Let θ be the angle between u and
v. Then by prop 1.2.10
|u · v| = kuk kvk | cos θ| ≤ kuk kvk
holds. Since kuk kvk =
6 0, if equality holds | cos θ| = 1 i.e, θ = 0 or π. Hence
a and b are parallel.
Remark 1.2.15. The Cauchy-Schwarz inequality reads, componentwise, as
(ax + by + cz)2 ≤ (a2 + b2 + c2 )(x2 + y 2 + z 2 )
Theorem 1.2.16 (Triangle inequality). For any two vector u, v it holds that
ku + vk ≤ kuk + kvk
and equality holds when u, v are parallel and having same direction.
Proof.
ku + vk2 = (u + v) · (u + v) = kuk2 + 2u · v + kvk2
By C-S
ku + vk2 ≤ kuk2 + 2kuk kvk + kvk2 = (kuk + kvk)2
Equality holds iff
u · v = kuk kvk
i.e, the angle is 0.
Definition 1.2.17. If two vectors u, v satisfy u · vb = 0 then we say they are
orthogonal(perpendicular).
Example 1.2.18. for any real θ two vectors are iθ = (cos θ)i + (sin θ)j, jθ =
−(sin θ)i + (cos θ)j orthogonal.
Example 1.2.19. Find a unit vector orthogonal to 2i − j + 3k and i + 2j + 9k.
sol.
Let ai + bj + ck be the desired vector. Then a, b, c are determined by
2a − b + 3c = 0 orthogonality
a + 2b + 9c = 0 orthogonality
a2 + b2 + c2 = 1 unicity
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CHAPTER 1. THE GEOMETRY OF EUCLIDEAN SPACE
y
iθ
θ
jθ
θ
x
O
Figure 1.11: iθ and jθ
Hence the desired vector is
1
± √ (3i + 3j − k) .
19
Triangle inequality
Theorem 1.2.20. For any vectors a, b, we have
ka + bk ≤ kak + kbk.
Use C-S.
1.3
Vector equations of lines and planes
Parametric equation of lines(Point-Direction form)
z
P = (x, y, z)
P0
b
v
~
OP
~0
OP
y
O
x
Figure 1.12: A line is determined by a point and a vector
1.3. VECTOR EQUATIONS OF LINES AND PLANES
13
~ 0 and pointing in the
The equation of the line ℓ through the tip of OP
direction of P~0 P is
→
→
ℓ(t) = OP0 + tP0 P = x0 + tv
where t takes all real values. In coordinate form, we have
x = x0 + at,
y = y0 + bt,
z = z0 + ct,
where x0 = (x0 , y0 , z0 ) and v = (a, b, c).
Example 1.3.1.
(1) Find equation of line through (2, 1, 5) in the direction
of 4i − 2j + 5k.
(2) In what direction, the the line x = 3t − 2, y = t − 1, z = 7t + 4 points ?
sol.
(1) v = (2, 1, 5) + t(4, −2, 5)
(2) (3, 1, 7) = 3i + j + 7k.
Example 1.3.2. Does the two lines (x, y, z) = (t, −6t + 1, 2t − 8) and (3t +
1, 2t, 0) intersect ?
sol.
If two line intersect, we must have
(t1 , −6t1 , 2t1 − 8) = (3t2 + 1, 2t2 , 0)
for some numbers t1 , t2 .(Note: we have used two different parameters t1 and
t2 ). But since the system of equation
t1 = 3t2 + 1
−6t1 = 2t2
2t1 − 8 = 0
has no solution, the lines do not meet.
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CHAPTER 1. THE GEOMETRY OF EUCLIDEAN SPACE
Two point form
We describe the equation of line through two points x0 , x1 .
The direction is given by v = x1 − x0 . So by point -direction form
ℓ(t) = x0 + t(x1 − x0 ).
If P = (x1 , y1 , z1 ) is the tip of x0 and Q = (x2 , y2 , z2 ) is the tip of x1 , then
v = (x2 − x1 , y2 − y1 , z2 − z1 ). Writing it componentwise, we see
x = x1 + (x2 − x1 )t
y = y1 + (y2 − y1 )t
z = z1 + (z2 − z1 )t
Solving these for t, we see
t=
x − x1
y − y1
z − z1
=
=
x2 − x1
y2 − y1
z2 − z1
This is another equation of line.
Example 1.3.3. Find eq. of a line through (2, 1, −3) and (6, −1, −5).
Example 1.3.4. Find eq. of line segment between (1, 1, −3) and (2, −1, 0)
Equation of a plane, Point-Normal form
Let x0 be a point in the plane and n = (A, B, C) be a nonzero normal vector
to the plane. If x is any point in the plane, then
n · (x − x0 ) = 0.
(1.2)
A(x − x0 ) + B(y − y0 ) + C(z − z0 ) = 0.
(1.3)
Or equivalently,
Equation of a plane, vector and parametric form
Let x0 be a point in the plane W . If v1 , v2 are two vectors positioned with
their initial point at x0 in the plane then for any vector x in the plane the
1.3. VECTOR EQUATIONS OF LINES AND PLANES
15
vector x − x0 lies in the plane W so that there exist two scalars t1 , t2 such that
x − x0 = t1 v1 + t2 v2 .
(1.4)
This is a parametric equation of the plane.
If v1 = (a1 , b1 , c1 ), v2 = (a2 , b2 , c2 ) the equation becomes
x = x0 + a1 t1 + a2 t2
y = y0 + b1 t1 + b2 t2 , (−∞ < t1 , t2 < ∞) .
(1.5)
z = z0 + c1 t1 + c2 t2
Example 1.3.5. Find eq. of plane through the origin and is parallel to the
vectors v1 = (1, 1, −3) and v2 = (2, −1, 0)