14 Vectors Operations - Handout.pdf

Vector Operations and its Applications

Mathematics 54 - Elementary Analysis 2

Vector Operations

Dot Product

Definition (Dot Product)

Let~A_{= 〈}a1,a2,a3〉and~B= 〈b1,b2,b3〉. We define their dot product as

~A_·~B₌a1b1+a2b2+a3b3

Note:~A_·~B_∈R

Example

Let~A= 〈2,−1, 4〉,~B= 〈−3, 2, 0〉and~C= 〈7, 4, 1〉

1 ~_A·~_{B=(2)(−3)+(−1)(2)+(4)(0)= −8} 2 ~C_·~B_{=(7)(−3)+(4)(2)+(1)(0)= −13}

Theorem (Properties of the dot product)

Let~A,~B and~C be vectors inR3and c_∈R.

1 ~A_·~B₌~B_·~A

2 ~_A·¡_c~_B¢_=c¡~_A·~_B¢

3 ~_A¡~_B±~_C¢₌¡~_A·~_B¢_±¡~_A·~_C¢

4 ~A_·~0₌0

5 ~A_·~A₌°°~A ° °

Angle Between Vectors

We define theangle between two vectors~Aand~Bto be the angle_θ∈[0,_π] formed by the two vectors in their position representation.

Theorem

Ifθis the angle between the two vectors~A and~B, then

~A·~B=°°~A ° ° ° °~B

° °cosθ.

Proof: Consider°°~A−~B ° °

2 .

° °~A−~B

° °

2

= (~A₋~B)_·(~A₋~B)

= ~A·(~A−~B)−~B·(~A−~B)

= ~A_·~A₋~A_·~B₋~B_·~A₊~B_·~B

= °°~A ° °

2

−2~A_·~B₊° °~B

° °

2

By cosine law:

~A

~B

~A−~B

° °~A−~B

° °

2 =°°~A

° °

2 +°°~B

° °

2 −2°°~A

° ° ° °~B

Examples

Example

1 _{Find the angle between}~_{A=3ˆı−2 ˆ}_+3ˆkand~_{B=ˆı−3 ˆ}_+5ˆk

Solution:

cosθ ₌ 〈3,−2, 3〉 · 〈1,−3, 5〉

k〈3,₋2, 3_{〉k k〈}1,₋3, 5_〉k

= 3p+6+15

22p35

= p24

770 θ = cos−1

µ

24

p

770

¶

Examples

Example

2 _{Find the angle between}~A_=3ˆı_{−2 ˆ}_+3ˆkand~B₌ˆı_{−3 ˆ}_−3ˆk

Solution:

cosθ = 〈3,−2, 3〉 · 〈1,−3,−3〉 k〈3,−2, 3〉k k〈1,−3,−3〉k

= p3+6−9

22p19

= 0

θ = π

2

Examples

Example

3 In Physics, Work (W) = Force (~F)_·Displacement (~d).

Find the work done by a man carrying a load of 100 N, while walking across a 10 meter, level surface.

Solution:

W ₌ ~F_·~d_{= k}F_kkd_kcosθ

= (100)(10) cos¡

Remarks

Two vectors~Aand~Bare parallel if and only if~A₌c~Bfor some nonzero scalarc.

~Aand~Bare perpendicular or orthogonal if and only if~A·~B=0.

Indeed, ifθ=π

2 is the angle between them, then

~A_·~B_{= k}A_{k k}B_kcos³π 2

´ =0

and conversely if~A·~B, then cosθ=0, soθ=π

2.

Note.

Projections

A nice application of the dot product is with projections.

Definition

Projections

Remark

1 Given two non-zero vectors~aand~b, the_projectionof~bonto the~ais

illustrated below

2 proj_~

Projection

The magnitude of proja~bis given byk~bk |cosθ|, whereθis the angle between~aand~b.

Moreover,

proj_~a~b ₌ ³_k~b_kcosθ´ ~a

k~a_k=

Ã

k~akk~bkcosθ

k~a_k2

!

~a

= Ã

~a_·~b

k~a_k2

!

~a

Hence, given two non-zero vectors~aand~b, thevector projectionof~bonto ~ais given by the following formula:

proj_~a~b₌~a·~b

Projection

Example

1 _{Determine the projection of}~_{A= 〈2, 1,−1〉onto}~_{B= 〈1, 0,−2〉.}

Solution.The projection is given by proj_~B~A=(~A_°·~B)~B °~B

° °

2

We need the dot product of~Aand~Band the magnitude of~B.

~A_·~B₌2₊0₊2₌4 ° °~B

° °

2

=1₊0₊4₌5

Then,

proj_~B~A = 4

5〈1, 0,−2〉 =

¿₄

5, 0,

−8 5

Projection

Example

2 _{Determine the projection of}~_{a= 〈1, 0,−2〉onto}~_{b= 〈2, 1,−1〉.}

Solution.Solving for~a_·~band the magnitude of~b.

~a·~b=2+0+2=4

° ° °~b

° ° °

2

=4+1+1=6

The projection is then,

projb~a = (~a°·~b)~b ° °~b

° ° °

2 = 4

6〈2, 1,−1〉

= ¿₄

3, 2 3,

−2 3

Projection

Example

3 _Express~_{A= 〈2, 3, 5〉as a sum of a vector parallel to}~_{B= 〈2,−1,−2〉and}

a vector perpendicular to~B.

Solution.Note that proj_~B~Ais parallel to~B. Thus,

proj_~B~A₌(~B·~A)~B

° °~B

° °

2 =

(4₋3₋10)_〈2,₋1,₋2_〉

4₊1₊4 = 〈−2, 1, 2〉.

Moreover, based from the construction of the vector projection, we can let ~C₌~A₋proj_~B~A, then one can verify that~Cand~Aare perpendicular.

Hence,

~C_{= 〈}2, 3, 5_{〉 − 〈−}2, 1, 2_{〉 = 〈}4, 2, 3_〉.

Vector Operations

Cross Product

Definition

Let~A_{= 〈}a1,a2,a3〉and~B= 〈b1,b2,b3〉

~A×~B= 〈a2b3−a3b2,a3b1−a1b3,a1b2−a2b1〉

Note:

~A_×~B is a vector.

Moreover, we can write~A_×~Bas

~A×~B = det





ˆ

ı ˆ kˆ a1 a2 a3

b1 b2 b3





= det

µ

a2 a3

b2 b3

¶

ˆ

ı−det

µ

a1 a3

b1 b3

¶

ˆ +det

µ

a1 a2

b1 b2

¶

ˆ

Determinant of a 2

×

2 matrix

Note that for the matrixA₌

µ

a1 a2

b1 b2

¶

,

detA=a1b2−a2b1.

Example

1 det

µ

2 0

−1 3

¶

=2(3)₋0(₋1)₌6

2 _det

µ −5 2

4 1

¶

Cross Product

Example

1 Let~A_{= 〈}1, 3, 4_〉and~B_{= 〈}2, 7,₋5_〉. Find~A_×~B.

Solution:

~A×~B = det





ˆ

ı ˆ kˆ

1 3 4 2 7 −5





= det

µ

3 4 7 −5

¶

ˆ

ı₋det

µ

1 4 2 −5

¶

ˆ +det

µ 1 3 2 7 ¶ ˆ k

= (−15−28)ˆı−(−5−8) ˆ+(7−6)ˆk

Theorem (Properties of the cross product)

Let~A,~B and~C be vectors inR3and c∈R.

1 ~A_×~B_{= −}~B_×~A

2 ~A_סc~B¢₌c¡~A_×~B¢

3 ~_Aס~_B±~_C¢₌¡~_A×~_B¢_±¡~_A×~_C¢

Examples

Example

1 _Suppose~_{A= 〈1, 0, 1〉and}~_{B= 〈2, 0,−1〉}. Find~_A×~_Band~_B×~_A.

~A×~B = det





ˆ

ı ˆ kˆ

1 0 1 2 0 −1





= det

µ

0 1 0 −1

¶

ˆ

ı−det

µ

1 1 2 −1

¶

ˆ +det

µ 1 0 2 0 ¶ ˆ k

= 3ˆ While,

~B_×~A ₌ det





ˆ

ı ˆ_ kˆ

2 0 −1 1 0 1





= det

µ

0 ₋1¶

ˆ

ı−det

µ

2 ₋1¶

ˆ +det

µ

2 0¶

ˆ

Examples

Example

2 Show that~A_×~A₌~0.

Let~A= 〈a,b,c〉, then

~A_×~A ₌ det





ˆ

ı _ˆ kˆ a b c a b c

  = det µ b c b c ¶ ˆ

ı₋det

µ

a c a c

¶

ˆ +det

µ a b a b ¶ ˆ k

= (bc−bc)ˆı−(ac−ac) ˆ+(ab−ab)ˆk

= 0ˆı₊0ˆ_+0kˆ

Theorem

Ifθis the angle between the two vectors~A and~B, then

° °~A×~B

° °=

° °~A

° ° ° °~B

° °sinθ.

Proof: Let~A_{= 〈}a1,a2,a3〉and~B= 〈b1,b2,b3〉.

° °~A×~B

° °

2

= 〈a2b3−a3b2,a3b1−a1b3,a1b2−a2b1〉

= (a2b3−a3b2)2+(a3b1−a1b3)2+(a1b2−a2b1)2 = (a2₁₊a2₂₊a2₃)(b2₁₊b2₂₊b2₃)₋(a1b1+a2b2+a3b3)2

= °°~A ° °

2° °~B

° °

2 −¡~

A·~B¢2

= °°~A ° °

2° °~B

° °

2 −°°~A

° °

2° °~B

° °

2 cos2θ

= °°~A ° °

2° °~B

° °

2

(1−cos2θ)= °°~A ° °

2° °~B

° °

2 sin2θ

Vector Cross Product

Remark

If~A×~B=~0, then the vectors are parallel.

Indeed, two vectors are parallel if and onlyθ=0, orθ=π.

In either case, sinθ=0.

Thus,°°~A×~B °

The last theorem can be geometrically interpreted as a formula for the area of a parallelogram, defined by two vectors.

Let~aand~bbe represented by directed line segments with same initial point, and angleθin between. They then determine a parallelogram with basek~akand altitude

° ° °~b

° °

°sinθ, thus the area of the parallelogram is given

by,

Area_{= k}~a_k°° °~b

° ° °sinθ=

° ° °~a×~b

Examples

Example

1 Find the area of the parallelogram defined by the vectors

~

G_{= −}3ˆi₊ˆj₋7ˆkand~H₌ˆj₊kˆ.

Solution:

~G×~H=det





ˆ_i ˆ_{j k}ˆ

−3 1 ₋7

0 1 1



=8ˆı+3 ˆ−3ˆk

∴Area =°°8ˆı+3 ˆ−3ˆk ° °=

p

Examples

Example

2 _{Find the area of the triangle with verticesA(1, 2, 4),B(−2, 1, 6) and}

C(1,−1, 3).

Solution:

−→

AB_{= −}3ˆı₋ˆ_+2ˆk, while−→AC_{= −}3 ˆ−kˆ.

So−→AB×−→AC=det





ˆ

ı ˆ kˆ

−3 −1 2 0 −3 −1



=7ˆı−3 ˆ+9ˆk.

The area of the parallelogram with adjacent sides−→ABand−→ACis

° ° °

−→

AB_×−→AC°° °=

°

°7ˆı−3 ˆ+9ˆk ° °=

p

49₊9₊81₌p139

The area of the triangleABCis half the area of this parallelogram, that is

p

Theorem

The vector~A_×~B is orthogonal to both~A and~B.

Proof:

Let~A= 〈a1,a2,a3〉and~B= 〈b1,b2,b3〉, then

¡~

A×~B¢

·~A = 〈a2b3−a3b2,a3b1−a1b3,a1b2−a2b1〉 · 〈a1,a2,a3〉 = (a1a2b3−a1a3b2)+(a2a3b1−a2a1b3)

+(a3a1b2−a3a2b1)

= a1a2b3−a2a1b3+a2a3b1−a3a2b1+a3a1b2−a1a3b2

= 0

A similar computation will show that¡~

A_×~B¢

Theorem

If~A,~B and~C are vectors inR3, then

~A_·¡~

B_×~C¢

=¡~A_×~B¢ ·~C

These equalities can be proven by letting ~A_{= 〈}a1,a2,a3〉,

~B_{= 〈}b1,b2,b3〉, and ~C_{= 〈}c1,c2,c3〉

and using the definition of the cross and dot products.

The geometric significance of the scalar triple product can be seen by considering the parallelepiped determined by the vectors~a,~b, and~c.

The area of the base parallelogram isA₌°° °~b×~c

° °

°. Ifθis the angle between

~aand~b_×~c, then the heighthof the parallelepiped ish_{= k}~ak |cosθ_|. Therefore, the volume of the parallelepiped is

V=Ah= ° ° °~b×~c

° °

°k~ak |cosθ| = ¯ ¯ ¯~a·

³

Examples

1 _{Find the volume of the parallelipiped defined by the vectors}

~u=3ˆı−ˆ+4ˆk,~v=2ˆı+ˆ−2ˆkand~w=5ˆı−kˆ.

Solution.

~u_·¡

~v_×~w¢

= det





3 ₋1 4 2 1 ₋2 5 0 −1





=3_·det

µ

1 −2 0 −1

¶

− (₋1)_·det

µ

2 −2 5 −1

¶ +4_·det

µ

2 1 5 0

¶

= −3₊8₋20

= −15

Therefore Volume =¯¯~u· ¡

Examples

2 Use the scalar triple product to show that the vectors

~A_{= 〈}1, 4,₋7_〉,~B_{= 〈}2,₋1, 4_〉and~C_{= 〈}0, 2,₋4_〉are coplanar.

Solution.

~A·¡~

B×~C¢

= det





1 4 ₋7 2 ₋1 4 0 2 ₋4





=1·det

µ −1 4

2 ₋4

¶

− 4·det

µ

2 4 0 ₋4

¶ −7·det

µ

2 ₋1 0 2

¶

= −4+32−28=0

Therefore, the volume of the parallelepiped determined by~A,~B, and ~

Exercises

1 _If~_{u= 〈2,−3, 1〉,}~_{v= 〈1, 0,−1〉and}~_{w= 〈−1, 3,−2〉, find:} 1 2~u₋~v

2 ~u₋~v₊~w 3 ~w₋¡0.5~u₊1.5~v¢

2 _{Determine whether the following vectors are parallel} 1 ~u= 〈3, 2,−1〉and~v= 〈−9,−6,−3〉

2 ~A= −4ˆı+ˆ−2ˆkand~B=

¿ 1 2,

−1 8 ,

1 4 À

3 _{Find a unit vector that points in the same direction as}~_{T=3ˆı−5 ˆ}_+kˆ_. 4 Find a vector of length 3 that is parallel to the vector from (2, 0,₋1) to

Exercises

5 Let~u₌ˆı₋2 ˆ₊kˆ,~v₌4ˆı₊ˆ₋3ˆkand~w₌2 ˆ₋kˆ. Find 1 ~u·~v

2 ~u×~v 3 ~w×~u 4 ~u·~v×~w 5 ~u_×~w_×~v

6 A unit vector that lies in thexy-plane that is orthogonal to~u.

6 _{The area of the parallelogram having 3 of its vertices to be the points}

(0, 1), (4, 3) and (1, 6).

7 _{Find a vector of length 5 that is orthogonal to both the vectors}

〈4,−1, 0〉and〈−3, 2,−1〉.

8 _{Find the volume of a parallelipiped defined by the vectors}