**Unformatted text preview: **Euclidean and Complex Vector Spaces
Introduction to Linear Algebra, Math115
Ghazal Geshnizjani —— Reference for some of the slides: Introduction to Linear Algebra for Science and Engineering
by Norman and Wolczuk
Chapter 1, 9 Euclidean
Vectors by one number
Scalar
Quantity:defined
Example: distance, speed, mass, etc.
Vector quantity: defined by a number and
direction
Example: displacement, velocity, force, etc.
A
P ~
AB
~
PQ B
Q
x y ~ and AB
~ have the same
PQ
magnitude and direction
~ = AB
~
=⇒ PQ
vectors can be translated
we only need one label to
~ and AB;
~ for
identify PQ
example just ~v . Vectors in R2 x2 Q P Definition (R )
2 O x1
x
is the set of all vectors of the form 1 , where x1 , x2 ∈ R are called
x2
the coordinates(or components) of the vector.
x1 2
R =
x ,x ∈ R
x2 1 2
R2 Notation We use the notation
x
~x = 1
x2 to denote vectors in R2 . The first component is referred to as x1 Vectors in R3
In three dimensions, we always pick our coordinate axes so that they form
a right-handed system.
x3 x3 x2
O O
x2 x1 Definition (R3 ) x1 R3 = x2 | x1 , x2 , x3 ∈ R x3 x1 Extending Vectors to n dimensions
Definition (Rn ) x1 .. n
R = . | x1 , . . . , xn ∈ R xn Definition (Zero vector in Rn ) n
Zero vector in Rn is the vector in R that all of its components are zero
0 .. ~
and it is denoted by 0 = . 0 0
0
~0 = ∈ R4 , · · ·
0
0
we have many different zero vectors!
~0 = 0 ∈ R2 ,
0 0
~0 = 0 ∈ R3 ,
0 Vectors
Definition (Equal Vectors) x1
y1 .. .. ~ = . in Rn are equal if and only if
Two vectors, ~v = . and w
xn
yn
( ⇐⇒ ) x1 = y1 , x2 = y2 , · · · , xn = yn
Examples:
3
3
6= 2
2
0
1
a
=⇒ a = 1, b = 3
=
b−a
2 Standard position of Vectors p1 We can identify the vector ~p = ... in Rn , with the point P(p1 , p2 , · · · , pn ).
pn ~ where O is the origin, O(0, · · · , 0) in Rn .
So we have ~p = OP
~ is the standard vector or position vector representation for ~p . When ~p
OP
has its tail at the origin, we say ~p is in standard position.
We represent the standard vector graphically by drawing an arrow from
(0, · · · , 0) to (p1 , · · · , pn ).
Example for R2 :
x2
P = (p1 , p2 )
p
p = 1
p2 Addition in Rn Definition (Vector Addition)
Suppose ~a , ~b ∈ Rn : x1
y1 .. .. ~
~a = . , b = . .
xn
yn ~ = AC
~
~ + BC
AB
A then x1
y1
x1 + y1 ~a + ~b = ... + ... = ... .
xn
yn
xn + yn
Examples: 1
0
1
−4 + 5 = 1 ,
2
−3
−1 C ~
AC ~
AB
~
BC B 1
3
+ 4 = not defined!
−1
−3 Scalar Multiplication in Rn Definition (Scalar Multiplication)
Suppose x1 .. ~x = . ,
xn t ∈ R. The scalar multiplication of ~x vector by the factor of t, called a scalar, is
defined by x1
tx1 .. .. t~x = t . = . .
xn
txn
Examples: 1
2
2 −4 = −8 ,
2
4 0
3
0
=
= ~0
−1
0 −~v o ~v 2~v Definition (Parallel Vectors)
Two non-zero vectors are parallel if and only if ( ⇐⇒ ) one is a scalar
multiple of the other.
Examples:
3
−9
~v =
~ =
, w
2
−6
1
~v = − w
~ =⇒ ~v and w
~ are parallel
3
3
−8
~v =
~ =
, w
2
−6
@k ∈ R such that 3 = −8k and 2 = −6k
~ are not parallel
=⇒ ~v and w Subtraction in Rn Definition (Vector difference)
Suppose ~a , ~b ∈ Rn : x1 .. ~a = . ,
xn ~ = AC
~ − AB
~
BC y1
.
~b = .. .
yn A then x1
y1
x1 − y1 ~a − ~b = ... + (−1) ... = ... .
xn
yn
xn − yn
Examples: 1
0
1
−4 − 5 = −9 ,
2
−3
5 C ~
AC ~
AB
~
BC B 1
3
− 4 = not defined!
−1
−3 Properties of vectors under addition and scalar
multiplication
Theorem ~ , ~x , ~y ∈ Rn and s, t ∈ R, we have the following.
For all w
1 ~
x + ~y ∈ Rn
(closed under addition)
2 ~
x + ~y = ~y + ~x
(addition is commutative)
3 (~
~ = ~x + (~y + w
~)
x + ~y ) + w
(addition is associative)
n
4 There exists a vector ~
0 ∈ R such that ~z + ~0 = ~z for all ~z ∈ Rn .
(zero vector)
5 There exists a vector −~
x ∈ Rn such that ~x + (−~x ) = ~0.
(additive inverses)
6 t~
x ∈ Rn
(closed under scalar multiplication)
7 s(t~
x ) = (st)~x
(scalar multiplication is associative)
8 (s + t)~
x = s~x + t~x
(a distributive law)
9 t(~
x + ~y ) = t~x + t~y
(another distributive law)
10 1~
x = ~x
(scalar multiplicative identity) Linear combination of vectors
Definition (Linear Combination) Given vectors ~v1 , . . . , ~vk ∈ Rn and numbers t1 , . . . , tk ∈ R, we call
t1~v1 + · · · + tk ~vk ,
a linear combination of ~v1 , . . . , ~vk .
~v1 , . . . , ~vk ∈ Rn , property 1 and 6 =⇒ t1~v1 + · · · + tk ~vk ∈ Rn (closed
under linear combination)
Examples: 1 1
0
1 2 −4 − 5 = −7
2
2
−3
4 1 1
0
2
=⇒ −7 is linear combination of −4 and 5 .
2
−3
4 Standard Basis 1
1
0
0
1
Consider −4 ∈ R3 , −4 = 1 0 + (−4) 1 + 2 0
2
0
0
1
2 if we call 1
0
0
1 ~e1 = 0 ,~e2 = 1 ,~e3 = 0 =⇒ −4 = 1~e1 + (−4)~e2 + 2~e3
0
0
1
2 Definition (Standard Basis for Rn )
In Rn , let ~ei be the vector whose i-th component is 1 and all other
components are 0.
The set {~e1 , . . . ,~en } is called the standard basis for Rn . x1
x1 .. .. n
For . ∈ R we get . = x1~e1 + x2~e2 + · · · + xn~en
xn
xn Definition (Norm) x1 .. The norm or magnitude of ~x = . is
xn
v
u n
q
uX
2
2
k~x k = x1 + · · · + xn = t
xi2 .
i=1 1
2
√
3 =⇒ k~
~ =
Examples: ~v =
=⇒ k~v k = 5, w
w
k
=
6
1
4
0 Theorem Let ~x , ~y ∈ Rn and t ∈ R. Then
1 k~
x k ≥ 0, and k~x k = 0 if and only if ~x = ~0,
2 kt~
x k = |t|k~x k,
3 k~
x + ~y k ≤ k~x k + k~y k.Triangle Inequality Example Find the distance from A(1, −1, 2) to B(3, 2, 1)
Show k~x k − k~y k ≤ k~x − ~y k. Definition (Unit Vector)
A vector ~x ∈ Rn such that k~x k = 1 is called a unit vector.
Examples:
~e2 ∈ R2 is a unit vector. ( ∀ ~ei ∈ standard basis of Rn =⇒ ~ei is a
unit vector.) 1 √ 3 ~ = √13 , =⇒ k~
wk =
w
√1
3 q 1
3 + 1
3 + 1
3 =1 Theorem
~ =
Let ~v ∈ Rn with ~v 6= ~0. Then w 1
~
k~v k v Proof: k~
w k = k~v1k ~v = k~v1k k~v k = is a unite vector in direction of ~v . 1
vk
k~v k k~ =1 ~ is a scalar multiplication of ~v by a positive factor
w
in the same direction as ~v . 1
k~v k so it’s parallel and Example: 4 Find the unit vector in direction of ~v = 5.
6
Solution:k~v k =
so √ 42 + 52 + 62 = √ 77 √4 4
77
1 ~ = √ 5 = √577 w
77 6
√6
77 ~
w ~v The Dot Product
Dot Product (or scalar product ) x1
y1 .. .. The dot product of vectors ~x = . , ~y = . in Rn is
xn
yn
~x · ~y = x1 y1 + x2 y2 + · · · + xn yn .
Examples: 1
−3
1 · −4 = 1(−3) + 1(−4) + 2(5) = −3 − 4 + 10 = 3.
2
5
a
−3
·
= −2 =⇒ a(−3) + 5a = −2 =⇒ 2a = −2 =⇒ a = −1
5
a Properties of Dot Product
Theorem Let ~x , ~y ,~z ∈ Rn and t ∈ R. Then
1 ~
x · ~y ∈ R
√
2 ~
x · ~x = k~x k2 (k~x k = ~x · ~x )
3 ~
x · ~x ≥ 0, and ~x · ~x = 0 if and only if ~x = ~0,
4 ~
x · ~0 = 0,
5 ~
x · ~y = ~y · ~x ,
6 ~
x · (~y ± ~z ) = ~x · ~y ± ~x · ~z ,
7 (t~
x ) · ~y = t(~x · ~y ) = ~x · (t~y ).
Exercise: prove above properties.
For (7), 1st part:
(t~x ) · ~y = (tx1 )y1 + (tx2 )y2 + · · · + (txn )yn = t(x1 y1 + x2 y2 + · · · + xn yn ).
= t(~x · ~y ) Dot Product and Angle between Vectors Let ~v , ~u be non-zero vectors in Rn with a common tail. Then they
determine a unique angle, θ with 0 ≤ θ ≤ π. ~u ~u
θ ~v ~v
Acute: 0 ≤ θ < ~u θ π
2 Obtuse: π
2 <θ≤π θ
~v Orthogonal: θ = π
2 Theorem For two non-zero vectors ~v , ~u ∈ Rn and the determining angle, θ, where
0 ≤ θ ≤ π the following relation holds:
~v · ~u = k~v kk~u k cos θ
~u − ~v ~u
θ
~v k~u − ~v k k~u k
θ
k~v k Proof:From Cosine Law we know how to relates the lengths of the sides of
a triangle to the cosine of one of its angles:
k~u − ~v k2 = k~u k2 + k~v k2 − 2k~u kk~v k cos θ, (1)
From properties of dot product we know
k~u −~v k2 = (~u −~v ) · (~u −~v ) = ~u ·~u +~v ·~v − 2~v ·~u = k~u k2 + k~v k2 − 2~v ·~u , (2)
(1) and (2) =⇒ −2k~u kk~v k cos θ = −2~u .~v =⇒ ~v · ~u = k~u kk~v k cos θ Corollaries based on ~v · ~u = k~v kk~u k cos θ:
Cauchy-Schwarz Inequality: For any two vectors ~v , ~u ∈ Rn , the size of dot product can not exceed the
product of their norms:
|~v · ~u | ≤ k~v kk~u k
Proof: ~v · ~u = k~v kk~u k cos θ =⇒ |~v · ~u | = k~v kk~u k| cos θ| we know| cos θ| ≤ 1, =⇒ |~v · ~u | ≤ k~v kk~u k Calculating angles between vectors
For any two non-zero vectors ~v , ~u ∈ Rn , we can calculate the angle θ
between them (0 ≤ θ ≤ π ):
cos θ = ~v · ~u
k~v kk~u k Example: 2
1
~ = −1.
Compute the angle between ~v = 1 and w
−1
−2
Solution: cos θ = ~v · w
~
2(1) − 1 + 2
3
1
√
=√
=√ √ =
k~v kk~
wk
2
4+1+1 1+1+4
6 6
π
1
θ = cos−1 ( ) =
3
2 Shape of the angle 0 ≤ θ ≤ π between vectors:
k~v kk~u k > 0 =⇒ cos θ and ~v · ~u have the same sign.
π
~v · ~u > 0 ⇐⇒ 0 ≤ θ < , (ACUTE)
2
~v · ~u = 0 ⇐⇒ θ =
~v · ~u < 0 ⇐⇒ ~u ~u
θ π
< θ ≤ π, (OBTUSE)
2 ~u
θ
~v ~v
ACUTE: 0 ≤ θ < π
, (ORTHOGONAL)
2 π
2 OBTUSE: π
2 <θ≤π θ
~v ORTHOGONAL:
θ = π2 Examples: Determine type of the angle in between the two vectors:
1
−1
1 For ~
~ =
v=
and w
.
2
2 2 3 ~v · w
~ = −1 + 4 = 3 > 0 =⇒ acute angle. 3
1 ~ = −5.
For ~v = 0 and w
5
−1
~v · w
~ = 3 − 5 = −2 < 0 =⇒ obtuse angle. 3
1
0
−5 ~ =
For ~v = 5 and w
−1.
2
1
~v · w
~ = 3 − 5 + 2 = 0 =⇒ orthogonal. Exmple: Evaluate the angle for number (1) above:
~v · w
~ 3 3 −1 3 Orthogonal Vectors
1
0
~
Example: What is the angle between ~v =
and 0 =
?
2
0
~v · ~0 = 0 + 0 = 0 =⇒ orthogonal? cos θ?
Note: ∀ ~v ∈ Rn =⇒ ~v · ~0 = 0. So we can define ~0 ∈ Rn to be orthogonal
to every ~v ∈ Rn . Orthogonal vectors
Two vectors, ~v , ~u ∈ Rn are orthogonal ⇐⇒ ~v · ~u = 0. Complex Vectors
Definition (Cn )
The vector space Cn is defined to be the set z 1 .. n
C = . | z1 , . . . , zn ∈ C zn
with addition of vectors and scalar multiplication as for Rn . Definition (Complex Conjugate)
The complex conjugate of z1 .. ~z = . ∈ Cn
zn z1 .. is ~z = . .
zn Definition (Addition and Scalar multiplication in Cn )
~ ∈ Cn and k ∈ C:
Suppose ~z , w z1 .. ~z = . ,
zn w1 ~ = ... w
wn then z1
w1
z1 ± w1 ~z ± w
~ = ... ± ... = ... ,
zn
wn
z n ± wn
−1
Example:What is (1 − 3j)
?
1+j kz1 k~z = ... kzn Inner Product in Complex Vector Spaces
How to define norm or magnitude for complex vectors, such that we
preserve k~z k ∈ R, k~z k ≥ 0, etc.?
√
Can we use k~z k = ~z · ~z ?
Example:
z
1+j
~z = 1 =
z2
2−j
~z · ~z = (1 + j)(1 + j) + (2 − j)(2 − j) = 3 − 2j ∈
/R
but
~z · ~z = (1 − j)(1 + j) + (2 + j)(2 − j) = 1 + 1 + 4 + 1 = 7 ∈ R ≥ 0
p
We can use k~z k = ~z · ~z instead! Definition (Standard Inner Product on Cn )
In Cn the standard inner product h , i is defined by
~ i = ~z · w
~ = z1 w1 + · · · + zn wn ,
h~z , w ~ ,~z ∈ Cn .
for w Definition (Norm for vectios in Cn )
For any ~z ∈ Cn the norm k~z k is defined by
p
k~z k = h~z ,~z i. Remarks:
k~z k = √ z1 z1 + · · · + zn zn = p
|z1 |2 + · · · + |zn |2 dot product is not the same as standard inner product.
~ ∈ Rn , h~z , w
~ i = ~z · w
~
For ~z , w Example
2 − 2j
2+j
~ =
~ i and h~
Let ~z =
and w
. Evaluate h~z , w
w ,~z i.
1+j
3
Properties of Complex Inner Product ~ .~z ∈ Cn , α ∈ C
Let ~v , w
1 h~
z ,~z i ≥ 0 and h~z ,~z i = 0 if and only if ~z = ~0,
2 h~
~ i = h~
z, w
w ,~z i,
3 4 5
6 (i)
(ii)
(i)
(ii) ~i
h~v + ~z , w
~ + ~v i
h~z , w
~i =
hα~z , w
h~z , α~
wi = ~ i + h~z , w
~ i,
= h~v , w
~ i + h~z , ~v i,
= h~z , w
~ i,
αh~z , w
~ i.
αh~z , w ~ i| ≤ k~z k k~
|h~z , w
w k,
(Cauchy-Schwarz Inequality)
~ k ≤ k~z k + k~
k~z + w
w k.
(Triangle Inequality) Proof for (4i):
~ = αz1 w1 + · · · + αzn wn
~ i = α~z · w
hα~z , w
~i
= α[z1 w1 + · · · + zn wn ] = αh~z , w Back to real vectors: Cross-Products in R3
Definition (Cross-Product)
Let ~u , ~v ∈ R3 . The cross product of ~u and ~v is u1
v1
u2 v3 − u3 v2
~u × ~v = u2 × v2 = u3 v1 − u1 v3 .
u3
v3
u1 v2 − u2 v1 1
−1
Example: ~u = 6 , ~v = 3 . Find ~u × ~u , ~u × ~v , ~v × ~u , ~u · (~u × ~v )
3
2
and ~v · (~u × ~v ). Properties of Cross Product
Theorem
For ~x , ~y ,~z ∈ R3 and t ∈ R, we have
1 ~
x × ~y ∈ R3
2 ~
x × ~y = −~y × ~x ,
3 ~
x × ~y is orthogonal to both ~x and ~y .
4 ~
x × ~0 = ~0
5 ~
x × ~x = ~0,
6 ~
x × (~y ± ~z ) = ~x × ~y ± ~x × ~z ,
7 (~
x ± ~y ) × ~z = (~x × ~z ) ± (~y × ~z ),
8 (t~
x ) × ~y = t(~x × ~y ).
Exercise for you: prove these properties
:) x2 x3 − x2 x3
0
pause Proof for (5): ~x × ~x = x3 x1 − x1 x3 = 0
x1 x2 − x2 x1
0 Remark:Order of multiplication matters for cross product! 1
0
0
~ = 0, show
Example: ~u = 1 , ~v = 1 , w
0
0
1
~ 6= ~u × (~v × w
~)
(~u × ~v ) × w
Remark:We can use cross product to find a vector orthogonal to two
non-zero vectors. 1
3 Example: Find a vector orthogonal to ~u = −2 and ~v = 0
1
1 Lagrange Identity
∀ ~u , ~v ∈ R3 =⇒ k~u × ~v k2 = k~u k2 k~v k2 − (~u .~v )2 . Theorem
If θ is the angle between two vectors ~u , ~v ∈ R3 where 0 ≤ θ ≤ π, then
k~u × ~v k = k~u k k~v k sin θ.
Proof:
k~u × ~v k2 = k~u k2 k~v k2 − (~u .~v )2 = k~u k2 k~v k2 − k~u k2 k~v k2 cos2 θ
= k~u k2 k~v k2 (1 − cos2 θ) = k~u k2 k~v k2 sin2 θ
We also know 0 ≤ θ ≤ π =⇒ sin θ ≥ 0
if we take the the square root we get:
k~u × ~v k = k~u k k~v k sin θ Applictions of Cross Product: Area of a Parallelogram
h k~u k ~u θ
θ k~v k = b ~v h = height, b = base, h = k~u k sin θ Area of parallelogram:
A = b · h = k~v kk~u k sin θ = k~v × ~u k = k~u × ~v k The area of the parallelogram formed by ~v , ~u ∈ R3 is
A = k~v × ~u k = k~u × ~v k. Example 3
1
Let ~u = 0 and ~v = −1.
5
1
find the area of the parallelogram determined by ~u and ~v . 5
3
1
~u × ~v = 0 × −1 = 2 −3
5
1
√
√
k~u × ~v k = 25 + 4 + 9 = 38
find the area of the triangle determined by ~u and ~v .
1
1√
A = k~u × ~v k =
38
2
2 ~u
θ
~v A Line in R2 Back in high school: representing a line in R2 :
y =m x +b
slope= m y-intercept= b In Math 115: We use the notation, x1 , x2 ! , We can also determine all the point on a Straight Line, if we have
~ = ~p = p1
one of its points: P(p1 , p2 ) =⇒ OP
p2
the orientation or the direction, ~d
Q(q1 , q2 )
We can use vectors to
represent a line! ~d
P(p1 , p2 )
q1
p
= 1 + 2~d
q2
p2 The Vector Equation of a Line in R2 The line through ~p with direction vector ~d is {~p + t ~d | t ∈ R}, which has
vector equation
~x = ~p + t ~d, t ∈ R.
This line is parallel to the line through the origin with equation
~x = t ~d,
x2 t ∈ R.
d + p line x = td + p
p d
x1 We say the line
has been translated by~p.
2
0
2
Example: ~x = t
is parallel to ~x =
+t
1
1
1 Parametric Equations and Scalar Form
Parametric Equations
Sometimes the components of a vector equation are written separately:
(
x1 = p1 + td1 ,
~x = ~p + t ~d becomes
t ∈ R.
x2 = p2 + td2 ,
This is the parametric equation of the line.
(
x1 = 2t,
0
2
Example: ~x =
+t
=⇒
1
1
x2 = 1 + t, t ∈ R. If we eliminate t we get(the scalar form:
x1 = 2t,
0
2
~x =
=⇒
t ∈ R. =⇒ x2 = 1 + 12 x1
+t
1
1
x2 = 1 + t,
Exercise for you: Scalar Form The familiar scalar form of the equation of the line is obtained by
eliminating the parameter t.
If d1 6= 0, d2 6= 0, we have
x1 − p1
x2 − p2
=t=
d1
d2 or x2 = p2 + d2
(x1 − p1 ).
d1 Line in Higher Dimensions Definition (Line in Rn ) Let ~p , ~d ∈ Rn , with ~d 6= ~0. Then the set with vector equation
~x = ~p + t ~d, t ∈ R, is called a line in Rn that passes through ~p .
To find the parametric equations we carry out the scalar multiplication and addition. x1 = p1 + td1 ,
..
~x = ~p + t ~d becomes
t ∈ R.
. x = p + td ,
n
n
n
Remark: There are infinite ways to write a vector equation for a line and
similarly parametric equations. Use different P on the line or a non-zero
scalar multiple of ~d. Example Example: Find the line through the points A(1, 1, −1) and B(4, 0, −3).
Present both vector equation and parametric equations.
x3
Solution:
x2
~d x1
B A 4
1
3
~ OA
~ = 0 − 1 = −1
~d = AB
~ = OB−
−3
−1
−2
~ + t ~d
~x = OA 1
3
=⇒ Vector equation: ~x = 1 + t −1 , t ∈ R
−1
−2 x1 = 1 + 3t,
parametric equations: x2 = 1 − t,
t ∈ R. Plane in R3 and Higher Dimensions
Definition (Plane in Rn )
Let ~v1 , ~v2 , ~p ∈ Rn , with ~v1 , ~v2 non-zero and not parallel to each other.
Then the set with vector equation
~x = ~p + t1~v1 + t2~v2 , t1 , t2 ∈ R, is called a plane in Rn that passes through ~p .
x3
v~2 P
~p x1 v~1
x2 Again note that the representation is not unique.
Depends on choice of P, ~v1 and ~v2 .
Exercise: Show if ~v1 and ~v2 are parallel, then we get a line instead of plane. x3
C Example
Find an equation of the plane
containing points
A(1, 1, 1), B(1, 2, 3) and C(−1, 1, 2). B
x2 A x1
x1
1
0
−2
Solution: x2 = 1 + r 1 + s 0 , r , s ∈ R
x3
1
2
1
If we sets = 0 then we get a line:
x1
1
0
x2 = 1 + r 1 , r ∈ R
x3
1
2
If we set r = 0 then we getanother
x1
1
−2
line: x2 = 1 + s 0 , s ∈ R
x3
1
1
Both lines are on the plane. x3
C B
A x1 x2 To find the parametric equations we carry out the scalar multiplication and addition. x1 = p1 + rd1 + sv1 ,
..
~x = ~p + r ~d + s~v becomes
r , s ∈ R.
. x = p + rd + sv ,
n
n
n
n
Example: x1
1
0
−2
x2 = 1 + r 1 + s 0 , r , s ∈ R
x3
1
2
1 x1 = 1 − 2s
=⇒ x2 = 1 + r x3 = 1 + 2r + s
Again not unique! r , s ∈ R. Example
Find the vector equation
theplane, containing the point (1, −1, −2) of 1
1
and the line ~x = 3 + t 1 , t ∈ R.
−1
4
B
~v ~d A 1
1
0
~ = −1 − 3 = −4
Solution: ~v = AB
−2
−1
−1 x1
1
1
0
=⇒ x2 = 3 + t 1 + s −4
x3
−1
4
−1 Exercise: Write parametric equations and then eliminate t and s to find a
scalar equation of the plane. Answer: 5x1 + x2 − 4x3 = 22 Scalar equation of a plane in R3
R3 In
we can also determine a plane
if we have one point on the plane
and the direction of a line orthogonal
to it. ~n
P0 Normal Vector to a plane
A vector ~n 6= 0 is a normal vector for a plane, if ~n is orthogonal to every
vector in the plane.
It is not unique; every non-zero scalar multiple of ~n is also normal vector
to the plane. Finding the Normal to a Plane
The normal to the plane with vector equation ~x = ~p + s~u + t~v , where ~u
and ~v are non-zero and not parallel is ~n = ~u × ~v . Scalar equation of a plane in R3 Suppose P0 (p1 , p2 , p3 ) is a known point on the plane with normal vector
n1
~n = n2 . If Q(x1 , x2 , x3 ) is any other point on the plane:
n3
~ lies in the plane
P and Q lie on the plane ⇐⇒ PQ
~ =0
⇐⇒ ~n · PQ
~ − OP)
~ =0
⇐⇒ ~n · (OQ
~ = ~n · OP
~
⇐⇒ ~n · OQ n1
x1
n1
p1
n2 · x2 = n2 · p2 =⇒ n1 x1 + n2 x2 + n3 x3 = n1 p1 + n2 p2 + n3 p3
n3
x3
n3
p3
If we can calculate RHS:
d ≡ n1 p1 + n2 p2 + n3 p3 =⇒ n1 x1 + n2 x2 + n3 x3 = d Example
Find a scalar equation of the plane containing the points
A(3, 1, 2), B(1, 2, 3) and C(−2, 1, 3).
Also check if ~n is the normal vector for the plane then
~ = ~n · OC
~ = ~n · OA.
~
~n · OB
Remark: the scalar equation for a plane in R3 is not unique. Why? Parallel planes in R3
Two planes in R3 are parallel, if their normal vectors are parallel. Finding the Line of Intersection of Two Planes
The direction vector of the line of intersection of two planes is the
cross-product of the normals to those planes. With this direction vector
and one point on the line, one can find the line of intersection. Projections ~ as a sum of a
We want to write w
vector parallel to a non-zero ~v and a
vector perpendicular to ~v :
~ = proj~v w
~ + perp~v w
~
w
that means:
~ = c~v ,
proj~v w ~ ) · ~v = 0
(perp~v w ~ =w
~ − c~v =⇒ (~
perp~v w
w − c~v ) · ~v = 0
~v 6=0 ~ · ~v − c ~v · ~v = 0 =⇒ c =
=⇒ w
~ =
=⇒ proj~v w ~ · ~v
w
~v ,
k~v k2 ~ · ~v
~ · ~v
w
w
=
~v · ~v
k~v k2 ~ =w
~ − proj~v w
~
perp~v w Projections
Definition (Projection)
~ ∈ Rn , with ~v 6= ~0, the projection of w
~ onto ~v is
For vectors ~v , w
~ =
proj~v w ~ · ~v
w
~v .
k~v k2 Definition (Perpendicular of a Projection)
~ ∈ Rn , with ~v 6= 0, the projection of w
~ perpendicular to ~v
For vectors ~v , w
is
~ =w
~ − proj~v w
~.
perp~v w Projections
proj~v ~u = ~u ~u
θ
proj~v ~
u ~v
π
0≤θ<
2
~u · ~v > 0 proj~v ~u and ~v in same
direction. ~u · ~v
~v
k~v k2 θ
proj~v ~
u ~v π
<θ≤π
2
~u · ~v < 0
proj~v ~u and ~v in
opposite direction. ~u θ
~v
θ= π
2 ~u · ~v = 0
proj~v ~u = ~0 Example 1
−1
Let ~u = 2 and ~d = 1 .
3
2
~
~
Find proj~d u , perp~d u , proj~u ~d, (perp~d ~u ) · ~d, proj~d (proj~d ~u ).
Solution:
~u · ~d = −1 + 2 + 6 = 7, k~dk2 = 1 + 1 + 4 = 6, k~u k2 = 1 + 4 + 9 = 14 7
−
−1
~
~u · d ~
7 76 ~
1
proj~d u =
d=
= 6
6
7
k~dk2
2
3 7 13 −6
1
6
13 5 4
perp~d ~u = ~u −proj~d ~u = 2− 76 = 56 , (perp~d ~u )·~d = − + + = 0
6 6 3
7
2
3
3
3 1 7
−
1
−1
~d · ~u
2
7 7 76 ~
~u =
2 = 1 , proj~d (proj~d ~u ) =
1 = 6
proj~u d =
14
6
k~u k2
7
3
3
2
3
2 Some Properties of Projections
Projection property
proj~x (proj~x ~y ) = proj~x ~y for all ~x , ~y ∈ Rn . Proof:
proj~x (proj~...

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