Vector Geometric Interpretation

1
Geometric Interpretation
Vector
Let R3 denote the set of all triples (a1 , a2 , a3 ) with a1 , a2 , a3 ∈ R. Elements in R3 are
called vectors. The sum of vectors
a = (a1 , a2 , a3 ) and b = (b1 , b2 , c3 )
is
a + b = (a1 + b1 , a2 + b2 , a3 + b3 ).
The scalar multiplication of α ∈ R with the vector a = (a1 , a2 , a3 ) is
αa = (αa1 , αa2 , αa3 ).
The zero vector is denoted 0 = (0, 0, 0). For a ∈ R3 , −a is defined by −a = (−1)a.
These rules have properties
• (a + b) + c = a + (b + c) for a, b, c ∈ R3
• α(a + b) = αa + αb for α ∈ R,a, b ∈ R3
• a + (−a) = 0 for a ∈ R3
• 0 + a = a = a + 0 for a ∈ R3
• 0a = 0 for a ∈ R3
• a + b = b + a for a, b ∈ R3
• (α + β)a = αa + βa for α, β∈ R, for a ∈ R3
• α(βa) = (αβ)a for α, β∈ R, for a ∈ R3
Geometric Interpretation
translation: x → x + a
magnification: x → αx, α > 1
contraction: x → αx, 0 < α < 1
parallelogram construction
March 17, 2001
2
Cauchy-Schwarz Inequality
Dot Product
The dot product of two vectors a = (a1 , a2 , a3 ) and b = (b1 , b2 , c3 ) is
a · b = a1 b1 + a2 b2 + a3 b3
Proposition
• a · 0 = 0 · a = 0 for a ∈ R3
• a · b = b · a for a, b ∈ R3
• a · (b + c) = a · b + a · c for a, b, c ∈ R3
• (b + c) · a = b · a + c · a for a, b, c ∈ R3
• α(a · b) = (αa) · b = a · (αb) for α, β ∈ R,a, b ∈ R3
• (αa) · (βb) = αβ(a · b) for α, β ∈ R,a, b ∈ R3
• a · a > 0 unless a = 0 for a ∈ R3 (Proof. a · a = a21 + a22 + a23. )
Cauchy-Schwarz Inequality
|a · b|2 ≤ (a · a)(b · b) for a, b ∈ R3
The equality holds if and only if there exist α, β ∈ R\{0} with αa + βb = 0.
Proof. Case 1. a = 0. Then a · b = 0, a · a = 0 so the equality holds.
Case 2. a 6= 0. Then a · a > 0. Consider the vector
c=b−
a·b
a.
a·a
Then
0 ≤ c · c = (b −
a·b
a·b
a, b −
a)
a·a
a·a
a·b
a·b
a·b
(a · b) −
(b · a) +
= (b · b) −
a·a
a·a
a·a
= (b · b) −
!2
(a · a)
a·b
(a · b).
a·a
March 17, 2001
3
Properties of the Norm
Therefore (a · a)(b · b) ≥ |a · b|2 .
a·b
If (a · a)(b · b) = |a · b|2 , and a 6= 0, then (b · b) − a·a
(a · b) = 0 so c = 0; i.e. b =
If (a · a)(b · b) = |a · b|2 , and b 6= 0, then a =
a·b
a.
a·a
a·b
b.
b·b
The norm of the vector a = (a1 , a2 , a3 ) is
kak =
q
√
a · a = a21 + a22 + a32
It follows from Cauchy-Schwarz inequality that
−1 ≤
a·b
≤1
kak kbk
if a =
6 0, b 6= 0. The unique angle θ ∈ [0, π] such that
cos θ =
a·b
kak kbk
is called the angle between the vectors a and b. This is consistent with the Law of
Cosine:
ka − bk2 = kak2 + kbk2 − 2 kak kbk cos θ
ka − bk2 − kak2 − kbk2 = (a1 − b1 )2 + (a2 − b2 )2 + (a3 − b3 )2
−(a12 + a22 + a32 + b12 + b22 + b32 ) = −2(a1 b1 + a2 b2 + a3 b3 )
= −2a · b = −2 kak kbk cos θ.
Properties of the Norm
• kak ≥ 0 for all a ∈ R3 , kak = 0 if and only if a = 0.
• kαak = |α| kak for α ∈ R,a ∈ R3
• Triangle Inequality: ka + bk ≤ kak + kbk for a, b ∈ R3
Proof.
ka + bk2 = (a + b) · (a + b)
=a·a+b·b+a·b+b·a
= kak2 + kbk2 + 2(a · b)
≤ kak2 + kbk2 + 2 kak kbk
= (kak + kbk)2 .
March 17, 2001
4
Direction Cosine
• ka − bk ≤ kb − ck + kc − ak for a, b, c ∈ R3
• Parallelogram Law: ka + bk2 + ka − bk2 = 2 kak2 + 2 kbk2 for a, b ∈ R3 .
Proof. Add the identities
ka + bk2 = (a + b) · (a + b) = a · a + b · b + 2a · b
ka − bk2 = (a − b) · (a − b) = a · a + b · b − 2a · b
Two vectors a, b are perpendicular (a ⊥ b) if the angle between them equals π/2 :
a ⊥ b if and only if a · b = 0
if and only if a1 b1 + a2 b2 + a3 b3 = 0.
Projections and Components
Theorem. Let a 6= 0. Each b ∈ R3 can be uniquely expressed as
b = bk + b⊥
where bk is parallel to a (i.e., there exists λ ∈ R with bk = λa) and b⊥ is perpendicular
to a (i.e., b⊥ · a = 0).
Proof. Set bk =
a·b
a
a·a
and b⊥ = b − bk . Then
b⊥ · a = (b − bk ) · a = b · a −
a·b
a · a = 0.
a·a
If b = b1 + b2 with b1 = λa, b2 · a = 0, then
b · a = (b1 + b2 ) · a = λa · a
so λ =
a·b
;
a·a
this shows b1 =
Definition bk =
a·b
a
a·a
a·b
a
a·a
= bk and b2 = b −
a·b
a
a·a
= b⊥ .
is called the projection of the vector b on the vector a.
Direction Cosine
Notation: i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1).
March 17, 2001
5
Properties of Cross Product
If a =
6 0, then the vector u =
a
kak
satisfies kuk = 1, i.e., u is a unit vector. The
components of u in the directions i, j, k
u · i = cos α
u · j = cos β
u · k = cos γ
are called the direction cosines of a.
Cross Product
Let a, b ∈ R3 . Their cross product (also called vector product) is given by
a×b =

i
j
a1 a2
b1 b2

b3
= (a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 )
= (a2 b3 − a3 b2 )i + (a3 b1 − a1 b3 )j + (a1 b2 − a2 b1 )k.
Properties of Cross Product
• a × a = 0 for a ∈ R3
• a × b = −b × a for a, b ∈ R3
• a × (b + c) = a × b + a × c for a, b, c ∈ R3
• (a + b) × c = a × c + b × c for a, b, c ∈ R3
• αa × βb = αβ(a × b) for α, β∈ R, for a, b ∈ R3
• i × i = j × j = k × k = 0, i × j = k, j × k = i, k × i = j
• Scalar Triple Product:
(a × b) · c =

a1 a2
b1 b2
c1 c2

c3
March 17, 2001
6
Properties of Cross Product
Proof.
(a × b) · i = a2 b3 − a3 b2
(a × b) · j = a3 b1 − a1 b3
(a × b) · k = a1 b2 − a2 b1
Therefore
(a × b) · (c1 i + c2 j + c3 k)
= c1 (a × b) · i + c2 (a × b) · j + c3 (a × b) · k
=

a1 a2
b1 b2
c1 c2

b3 .

c3
• (a × b) · c = (b × c) · a = (c × a) · b for a, b, c ∈ R3
• (a × b) · c = a · (b × c)
• (a × b) · a = 0, (a × b) · b = 0. Hence a × b is perpendicular to both a and b.
• Lagrange Identity: ka × bk2 + (a · b)2 = kak2 kbk2 for a, b ∈ R3
Proof.
ka × bk2 = (a2 b3 − a3 b2 )2 + (a3 b1 − a1 b3 )2 + (a1 b2 − a2 b1 )2
(a · b)2 = (a1 b1 + a2 b2 + a3 b3 )2
kak2 = a21 + a22 + a23
kbk2 = b21 + b22 + b23
• ka × bk is the area of the parallelogram formed with sides a and b.
6 0. From the identity
Proof. We may assume that a 6= 0 and b =
(a · b)2
ka × bk2
+
=1
kak2 kbk2 kak2 kbk2
we see that
ka × bk2
2
2
2 + cos θ = 1,
kak kbk
March 17, 2001
7
Properties of Cross Product
where θ is the angle between a and b. It follows that
sin θ =
ka × bk
kak kbk
therefore the required area equals
kak kbk sin θ = ka × bk2 .
• The absolute value of the triple product|(a · b) × c| represents the volume of the
parallelepiped with edges a, b, c.
Proof. Write

a×b

ka × bk
ka × bk
|(a · b) × c| =
The quantity
a×b
·c
ka × bk
is the component of the vector c along the direction

· c .
a×b
;
ka×bk

a×b
parallelepiped with base formed by the vectors a, b is ka×bk
from the fact
so the height of the

· c . The result follows
volume=base area × height.
• CAB Minus BAC Identity a × (b × c) = (c · a)b − (b · a)c
The result holds for a, b, c ∈ {i, j, k} (there are altogether 27 different cases.)
Write
e1 = i, e2 = j, e3 = k.
The general case follows from
3
X
p=1
=

ap ep × 
3 X
3 X
3
X
p=1 q=1 r=1
=
3 X
3
3 X
X
p=1 q=1 r=1
=(
3
X
r=1
cr er ·
3
X
p=1
3
X
q=1
bq eq ×
3
X
r=1

cr er 
ap bq cr ep × (eq × er )
ap bq cr [(er · ep )eq − (eq · ep )er ]
ap ep )
3
X
q=1
bq eq − (
3
X
q=1
bq eq ·
3
X
p=1
ap ep )
3
X
cr er .
r=1
March 17, 2001
8
Cramer’s Rule
• Extended Lagrange Identity: (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c).
Proof.
(a × b) · (c × d) = a · [b × (c × d)]
It follows from CAB minus BAC identity that this equals
a · [(b · d)c − (b · c)d]
= (a · c)(b · d) − (a · d)(b · c).
Note that the Lagrange identity may be obtained from the extended Lagrange
identity by setting c = a and d = b.
• Jacobi Identity: a × (b × c) + b × (c × a) + c × (a × b) = 0
Proof. Add the identities
a × (b × c) = (c · a)b − (b · a)c
b × (c × a) = (a · b)c − (c · b)a
c × (a × b) = (b · c)a − (a · c)b
Cramer’s Rule
Consider the simultaneous set of equations
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
(1)
a3 x + b3 y + c3 z = d3
From the basic properties of the determinant we have
x

a 1 b1
a 2 b2
a3 b3

a1 x b1
= a2 x b2
a3 x b3

a1 x + b1 y + c1 z b1
= a2 x + b2 y + c2 z b2
a3 x + b3 y + c3 z b3

c2 .

c3
March 17, 2001
9
Cramer’s Rule
Hence if x is to satisfy equations (1) then
x
Write

c3
a1 b 1
a2 b2
a3 b3

= d 2 b2
a1 b1
∆ = a 2 b2
If ∆ 6= 0, then
x=
6 0, then
Similarly, if ∆ =

y=
a1 d 1
a2 d2
a3 d3
∆
a 3 b3

d1 b1
d2 b2
d3 b3
∆

c3
d 1 b1
d3 b3

c2 .

c3
z=
,

c2 .

a1 b1
a2 b2
a3 b3
∆

d3
.
On the other hand, the following identity always holds:

ar br cr dr
a1 b1 c1 d1
a2 b2 c2 d2
a3 b3 c3 d3

= 0 for r = 1, 2, 3.
This four-by-four determinant can be expanded along the first row:

d1 b1
d2 b2
d3 b3

c2 + br

a1 d1
a2 d2
a3 d3

−dr

a1 b1
a2 b2
a3 b3

c2 + cr

c3
a 1 b1
a2 b2
a3 b3

d3
March 17, 2001
10
Straight Line in R3
Hence if ∆ 6= 0, then
x=

d1 b1
d2 b2
d3 b3
∆

c3
,y =

a1 d1
a2 d 2
a3 d3
∆

c3
,z =

a1 b1
a 2 b2
a3 b3
∆

d3
is indeed the (unique) solution to the system (1).
The method can be extended to n linear equations in n unknowns.
Straight Line in R3
A straight line in R3 can be given by the parametric equations:
x = x0 + at
y = y0 + bt
z = z0 + ct
t∈R
A point (x, y, z) belong to this line if and only if
(x, y, z) = (x0 + at, y0 + bt, z0 + ct)
for some t ∈ R. This can therefore be written in vector equations as:
p + tq
p = (x0 , y0 z0 ), q = (a, b, c), t ∈ R.
After eliminating the parameter t from the parametric equation, we obtain the symmetric equation
x − x0
y − y0
z − z0
=
=
a
b
c
if abc =
6 0.
Example Find the equations of the straight line passing through p = (−1, 4, 3)
and q = (2, −1, 5).
First we determine the direction: q − p = (3, −5, 2); from this the vector equation
is obtained:
(x, y, z) = (−1, 4, 3) + t(3, −5, 2)
March 17, 2001
11
Straight Line in R3
= (−1 + 3t, 4 − 5t, 3 + 2t), t ∈ R.
Equating the components we obtain the parametric equation:
x = −1 + 3t
y = 4 − 5t,
z = 3 + 2t,
t∈R
and the symmetric equations:
z−3
y−4
x+1
=
=
.
3
−5
2
Example Find the distance between the skew lines L1 given by
x=t−1
y = −t + 2
z = −2t − 1
t∈R
and L2 given by
x = −2t + 1
y = 2t + 2
z = −2t + 2
t∈R
We note that L1 has direction u = (1, −1, −2) and L2 has direction v = (−2, 2, −2).
The vector
a=

i
j

−1 −2 = 6i + 6j

−2 2 −2
1
is perpendicular to both u and v. Choose any point p ∈ L1 and any point q ∈ L2 .
(There are infinitely many ways to do so.) The distance between L1 and L2 is the
norm of the projection of p − q on a. Take p = (−1, 2, −1) and q = (1, 2, 2). Then
p − q = (−2, 0, −3) and so the required distance is

(−2, 0, −3) · a

a·a

|(−2, 0, −3) · (6, 6, 0)| −12 √
√
=
= √ = 2.
6 2
6 2
March 17, 2001
12
Plane in R3
Plane in R3
Let (x0 , y0 , z0 ) be any point of the plane. The plane consists of all the points (x, y, z) ∈
R3 satisfying the vector equation
0 = (x − x0 , y − y0 , z − z0 ) · n
= a(x − x0 ) + b(y − y0 ) + c(z − z0 )
for a nonzero vector n = (a, b, c). This vector n is called the normal of the plane. Thus
the equation of the plane takes the form
ax + by + cz = d
where
d = ax0 + by0 + cz0 .
Example Find the distance from p = (−1, 4, 5) to the plane 3x − 2y + 4z = 12.
The vector n = (3, −2, 4) is normal to the plane. Choose a point on the plane:
(0, 0, 3). the distance from p to the plane is the norm of the projection of the vector
(−1, 4, 5) − (0, 0, 3) on n =
|(−1, 4, 2) · (3, −2, 4)|
q
32 + (−2)2 + 42
3
=√ .
29
Example Show that the line
x−1
y−5
z+2
=
=
2
4
1
can be expressed as the intersection of two planes, each of which is parallel to a coordinate axis.
Consider the plane P1 with equation obtained from the first equality
P1 :
x−1
y−5
=
2
4
P1 can be expressed by the equation
4x − 2y + 6 = 0
March 17, 2001
13
Intersecting Planes
which has (4, −2, 0) as normal. P1 is therefore parallel to the z-axis since (−4, −2, 0)
is perpendicular to k. Similarly, the plane satisfying the equation
P2 :
z+2
y−5
=
,
4
1
which is the same as the equation
y − 4z − 7 = 0
has (0, 1, −4) as normal, must be parallel to the x-axis since (0, 1, −4) is perpendicular
to i. The given line is the intersection of P1 and P2 .
Intersecting Planes
The angle between two intersecting planes is the same as the angle between their
normals n1 , n2 . Depending on the choices of the normals, there are two such angles,
each the supplement of the other. We choose the smaller angle, the one with the
nonnegative cosine:
cos θ =
|n1 · n2 |
kn1 k kn2 k
Example Find the cosine of the angle between the planes with equations
2(x − 1) − 3y + 5(z − 2) = 0
and
−4x + 6y + 10z + 24 = 0.
Solution: n1 = (2, −3, 5), n2 = (−4, 6, 10); so
|(2, −3, 5) · (−4, 6, 10)|
6
q
cos θ = q
= .
19
22 + (−3)2 + 52 (−4)2 + 62 + 102
Example The planes
a1 x + b1 y + c1 z + d1 = 0
and
a2 x + b2 y + c2 z + d2 = 0
intersect to form a line. Find the vector equation of this line.
March 17, 2001
14
Distance from a Point to a Plane
Solution: The direction vector of the intersecting line must be perpendicular to both
normal vectors (a1 , b1 , c1 ) and (a2 , b2 , c2 ). Hence it is parallel to (a1 , b1 , c1 ) × (a2 , b2 , c2 ).
Let (x0 , y0 , z0 ) be any point on the intersecting line. The required vector equation takes
the form
(x0 , y0 , z0 ) + t(a1 , b1 , c1 ) × (a2 , b2 , c2 ), t ∈ R.
Plane Determined by Three Noncollinear Points
Suppose that the points
a = (a1 , a2 , a3 ), b = (b1 , b2 , b3 ), c = (c1 , c2 , c3 )
do not lie on a straight line. A point x = (x1 , x2 , x3 ) lies on the plane determined
by these three points if and only if the volume of the parallelepiped with a, b, c, x as
vertices is zero. In view of the vector triple product identity, the condition is given by
(x − a) · (b − a) × (c − a) = 0.
Example Find the equation for the plane passing through
(0, 1, 1), (1, 0, 1), (1, 1, 0).
Solution: Expanding the determinant
we have

x1 − 0 x2 − 1 x3 − 1
1−0
0−1
1−1
1−0
1−1
0−1

=0
x1 + x2 + x3 − 2 = 0
Distance from a Point to a Plane
Suppose that p = (x1 , y1 , z1 ) does not lie on the plane
P : ax + by + cz + d = 0.
March 17, 2001
15
Exercise
Choose any point q = (x0 , y0 , z0 ) on the plane. The distance from p to the plane is the
norm of the projection of p − q on the normal (a, b, c) of the plane:
|(x1 − x0 , y1 − y0 , z1 − z0 ) · (a, b, c)|
√
.
a2 + b2 + c2
Since (x0 , y0 , z0 ) lies on the plane, it follows that
ax0 + by0 + cz0 + d = 0.
Therefore
(x1 − x0 , y1 − y0 , z1 − z0 ) · (a, b, c) = ax1 + by1 + cz1 + d;
consequently the required distance is given by
|ax1 + by1 + cz1 + d|
√
.
a2 + b2 + c2
Exercise
1. Find the equation for the plane
(a) through (1, 2, −1) perpendicular to i + j.
(b) through (1, 2, −1) perpendicular to i + 2j − k.
(c) through (1, 0, 1) parallel to x + 2y + z = 0.
(d) through (x0 , y0 , z0 ) parallel to x + y + z = 1.
2. Explain why a plane cannot
(a) contain (1, 2, 3) and (2, 3, 4) and be perpendicular to i + j
(b) contain (1, 0, 0), (0, 1, 0), (0, 0, 1), and (1, 1, 1)
(c) go through the origin and have the equation ax + by + cy = 1
3. Find the projection P of B along A and also find kP k
(a) A = (4, 2, 4), B = (1, −1, 0)
(b) A = (1, −1, 0), B = (4, 2, 4)
March 17, 2001
16
Exercise
(c) B =unit vector at 60◦ angle with A
(d) B =vector of length 2 at 60◦ angle with A
(e) B = −A
(f) A = i + j, B = i + k
(g) A is perpendicular to x − y + z = 0, B = i + j
(h) A is perpendicular to x − y + z = 5, B = i + j + 5k
4. True or false:
(a) A × B never equals A · B.
(b) If A × B = 0 and A · B = 0, then either A = 0 or B = 0.
(c) If A × B = A × C and A 6= 0, then B = C.
(d) A × (A × B) = 0.
(e) A · (B × C) = C · (B × A).
(f) (A − B) × (A + B) = 2(A × B).
5. Which of the following equals to A × B?
(a) (A + B) × B
(b) (−B) × (−A)
(c) kAk kBk |sin θ|
(d) (A + C) × (B − C)
(e)
1
(A
2
− B) × (A + B).
March 17, 2001

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Vector Geometric Interpretation

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