Vector Algebra - Calculus II - Lecture Slides | MATH 114, Study notes of Mathematics

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Math 114-004, Fall 2009

Tong Zhu

Department of Mathematics University of Pennsylvania

September 24, 2009

Vector Algebra

I (^) Vector Addition

I (^) Scalar Multiplication

Vector Algebra

I (^) Vector Addition

I (^) Scalar Multiplication

I (^) How about the product of two vectors?

Vector Algebra

I (^) Vector Addition

I (^) Scalar Multiplication

I (^) How about the product of two vectors?

1. Dot Product ~v · ~u
2. Cross Product ~v × ~u

Before the geometry,

Determinant

Before the geometry,

Determinant

I (^) Determinant of a 2 × 2 matrix,

∣

∣

∣

∣

a b

c d

∣

∣

∣

∣ =^ ad^ −^ bc

I (^) Determinant of a 3 × 3 matrix,

∣ ∣ ∣ ∣ ∣ ∣

a 1 a 2 a 3

b 1 b 2 b 3

c 1 c 2 c 3

∣ ∣ ∣ ∣ ∣ ∣

= a 1

∣

∣

∣

∣

b 2 b 3

c 2 c 3

∣

∣

∣

∣

− a 2

∣

∣

∣

∣

b 1 b 3

c 1 c 3

∣

∣

∣

∣

• a 3

∣

∣

∣

∣

b 1 b 2

c 1 c 2

∣

∣

∣

∣

=a 1 (b 2 c 3 − b 3 c 2 ) − a 2 (b 1 c 3 − b 3 c 1 ) + a 3 (b 1 c 2 − b 2 c 1 )

Example 1

1. ∣

∣

∣

∣

− 1 2

∣

∣

∣

∣

Example 1

1. ∣

∣

∣

∣

− 1 2

∣

∣

∣

∣

= − 5 − 6 = − 11

Example 1

1. ∣

∣

∣

∣

− 1 2

∣

∣

∣

∣

= − 5 − 6 = − 11

∣ ∣ ∣ ∣ ∣ ∣

− 1 3 5

∣ ∣ ∣ ∣ ∣ ∣

= 2

∣

∣

∣

∣

∣

∣

∣

∣

− 4

∣

∣

∣

∣

− 1 5

∣

∣

∣

∣

∣

∣

∣

∣

− 1 3

∣

∣

∣

∣

=2(18 − 10) − 4(− 6 − 35) + 6(− 2 − 21) = 42

Example 1

1. ∣

∣

∣

∣

− 1 2

∣

∣

∣

∣

= − 5 − 6 = − 11

∣ ∣ ∣ ∣ ∣ ∣

− 1 3 5

∣ ∣ ∣ ∣ ∣ ∣

= 2

∣

∣

∣

∣

∣

∣

∣

∣

− 4

∣

∣

∣

∣

− 1 5

∣

∣

∣

∣

∣

∣

∣

∣

− 1 3

∣

∣

∣

∣

=2(18 − 10) − 4(− 6 − 35) + 6(− 2 − 21) = 42

∣ ∣ ∣ ∣ ∣ ∣

− 1 3 5

− 1 3 5

∣ ∣ ∣ ∣ ∣ ∣

= 2

∣

∣

∣

∣

∣

∣

∣

∣

− 4

∣

∣

∣

∣

− 1 5

− 1 5

∣

∣

∣

∣

∣

∣

∣

∣

− 1 3

− 1 3

∣

∣

∣

∣

= 0

Cross Product

In 3-D,

Definition If two vectors ~a =< a 1 , a 2 , a 3 > and ~b =< b 1 , b 2 , b 3 >,

then the cross product of ~a and ~b is defined as

~a × ~b =< a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 >

Cross Product

In 3-D,

Definition If two vectors ~a =< a 1 , a 2 , a 3 > and ~b =< b 1 , b 2 , b 3 >,

then the cross product of ~a and ~b is defined as

~a × ~b =< a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 >

Remark: Cross product gives a vector.

Cross Product

In 3-D,

Definition If two vectors ~a =< a 1 , a 2 , a 3 > and ~b =< b 1 , b 2 , b 3 >,

then the cross product of ~a and ~b is defined as

~a × ~b =< a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 >

Remark: Cross product gives a vector.

But the components of this vector is hard to memorize. There

must be a neat formula!

By the notation of determinant, if ~a =< a 1 , a 2 , a 3 > and

~b =< b 1 ,^ b 2 ,^ b 3 >,

~a × ~b =

∣ ∣ ∣ ∣ ∣ ∣

~i ~j ~k

a 1 a 2 a 3

b 1 b 2 b 3

∣ ∣ ∣ ∣ ∣ ∣

Example ~a =< 2 , 3 , 4 >, ~b =< 1 , 5 , − 6 >

By the notation of determinant, if ~a =< a 1 , a 2 , a 3 > and

~b =< b 1 ,^ b 2 ,^ b 3 >,

~a × ~b =

∣ ∣ ∣ ∣ ∣ ∣

~i ~j ~k

a 1 a 2 a 3

b 1 b 2 b 3

∣ ∣ ∣ ∣ ∣ ∣

Example ~a =< 2 , 3 , 4 >, ~b =< 1 , 5 , − 6 >

~a × ~b =

∣ ∣ ∣ ∣ ∣ ∣

~i ~j ~k

1 5 − 6

∣ ∣ ∣ ∣ ∣ ∣ =

∣

∣

∣

∣

5 − 6

∣

∣

∣

∣

~i −

∣

∣

∣

∣

1 − 6

∣

∣

∣

∣

~j +

∣

∣

∣

∣

∣

∣

∣

∣

~k

= − 38 ~i + 16~j + 7~k

Some properties of cross product:

I (^) ~a × ~a = ~ 0

Some properties of cross product:

I (^) ~a × ~a = ~ 0

~a × ~a =

∣ ∣ ∣ ∣ ∣ ∣

~i ~j ~k

a 1 a 2 a 3

a 1 a 2 a 3

∣ ∣ ∣ ∣ ∣ ∣

= ~ 0

Some properties of cross product:

I (^) ~a × ~a = ~ 0

~a × ~a =

∣ ∣ ∣ ∣ ∣ ∣

~i ~j ~k

a 1 a 2 a 3

a 1 a 2 a 3

∣ ∣ ∣ ∣ ∣ ∣

= ~ 0

I (^) ~a × ~b = −~b × ~a

Some properties of cross product:

I (^) ~a × ~a = ~ 0

~a × ~a =

∣ ∣ ∣ ∣ ∣ ∣

~i ~j ~k

a 1 a 2 a 3

a 1 a 2 a 3

∣ ∣ ∣ ∣ ∣ ∣

= ~ 0

I (^) ~a × ~b = −~b × ~a

~a × ~b =

∣ ∣ ∣ ∣ ∣ ∣

~i ~j ~k

a 1 a 2 a 3

b 1 b 2 b 3

∣ ∣ ∣ ∣ ∣ ∣

=

∣

∣

∣

∣

a 2 a 3

b 2 b 3

∣

∣

∣

∣

~i −

∣

∣

∣

∣

a 1 a 3

b 1 b 3

∣

∣

∣

∣

~j +

∣

∣

∣

∣

a 1 a 2

b 1 b 2

∣

∣

∣

∣

~k

~b × ~a =

∣ ∣ ∣ ∣ ∣ ∣

~i ~j ~k

b 1 b 2 b 3

a 1 a 2 a 3

∣ ∣ ∣ ∣ ∣ ∣

=

∣

∣

∣

∣

b 2 b 3

a 2 a 3

∣

∣

∣

∣

~i −

∣

∣

∣

∣

b 1 b 3

a 1 a 3

∣

∣

∣

∣

~j +

∣

∣

∣

∣

b 1 b 2

a 1 a 2

∣

∣

∣

∣

~k

I (^) (c~a) × ~b = c(~a × ~b) = ~a × (c~b)

I (^) (c~a) × ~b = c(~a × ~b) = ~a × (c~b)

I (^) ~a × (~b + ~c) = ~a × ~b + ~a × ~c

I (^) (c~a) × ~b = c(~a × ~b) = ~a × (c~b)

I (^) ~a × (~b + ~c) = ~a × ~b + ~a × ~c

I (^) (~a + ~b) × ~c = ~a × ~c + ~b × ~c

I (^) (c~a) × ~b = c(~a × ~b) = ~a × (c~b)

I (^) ~a × (~b + ~c) = ~a × ~b + ~a × ~c

I (^) (~a + ~b) × ~c = ~a × ~c + ~b × ~c

I (^) ~a · (~b × ~c) = (~a × ~b) · ~c

I (^) (c~a) × ~b = c(~a × ~b) = ~a × (c~b)

I (^) ~a × (~b + ~c) = ~a × ~b + ~a × ~c

I (^) (~a + ~b) × ~c = ~a × ~c + ~b × ~c

I (^) ~a · (~b × ~c) = (~a × ~b) · ~c

Proof : Method 1: algebraic way

I (^) (c~a) × ~b = c(~a × ~b) = ~a × (c~b)

I (^) ~a × (~b + ~c) = ~a × ~b + ~a × ~c

I (^) (~a + ~b) × ~c = ~a × ~c + ~b × ~c

I (^) ~a · (~b × ~c) = (~a × ~b) · ~c

Proof : Method 1: algebraic way

Method 2:

~a · (~b × ~c) =

∣ ∣ ∣ ∣ ∣ ∣

a 1 a 2 a 3

b 1 b 2 b 3

c 1 c 2 c 3

∣ ∣ ∣ ∣ ∣ ∣