More about vectors

Coplanar vectors

Q

A

B

Ca

b

c QA = λa

QB = μb

c = QA + QB

c = λa + μb a, b, and c are coplanar vectors

a = (1, 1)

b = (2, 1)

c = (3, 4)

λ = ?

μ = ?

5

-1

Coplanar vector in 3D

z

x

21

O

1 2 31

2

3

34

ac

b

c = λa + μb

Position vector A vector which has its initial point at the origin of

coordinates.

x

y

yQ

xQ

r

rx

ry

Q (xQ, yQ)

O

z

x

21

O

1 2 31

2

3

34

rP

2

3

P (2, 3, 3)

PrOP = (2, 3, 3)

r = (xQ, yQ)

Vector equations for curves & surfaces Circle

22 )()( byax

y

rC

rC (a, b)

O

P (x, y)

r - rC

r – rC = c

Proof: r – rC = (x, y) – (a, b) = (x-a, y-b)

r - rC

222 )()( cbyax

Vector equations for curves & surfaces Plane

z

x

O

A

C

B

AP = λAB + μAC x

rA rB

rC

rP

AB = rB - rA

AC = rC - rA

AP = r - rA

r – rA = λ (rB - rA)+ μ(rC – rA)

-- Parametric vector equation for a plane

Example-plane A (1,2,1), B(2,2,0), C(2,1,2)

(x, y, z) –(1, 2, 1) = λ{(2, 2, 0)-(1, 2, 1)} + μ{(2, 1, 2)-(1, 2, 1)}

r – rA = λ (rB - rA)+ μ(rC – rA)

= λ(1, 0, -1) + μ(1, -1, 1)

x = 1 + λ+ μ

y = 2 - μ

z = 1 – λ + μ x + 2 y + z = 6

Cartesian parametric equations for the plane

Cartesian (general) equation for a plane

(x, y, z) = (1+λ+μ, 2-μ, 1-λ+μ)

Vector parametric equation for a plane

1c’ d’

b’ d’

Plane cont’d

General form of a plane

A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3)

a

a

a

a’ d’x y z = + + cba

3

2

1

x

x

x

3

2

1

y

y

y

3

2

1

z

z

z

b

b

b

c

c

c

1

1

1

c

b

a

0d

General Cartesian form of a plane

A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3)

Plane cont’d

a

a

a

dx y z = + + cba

3

2

1

x

x

x

3

2

1

y

y

y

3

2

1

z

z

z

b

b

b

c

c

c

d

d

d

dC

dC

dC

c

b

a

3

2

1

General form of a plane

Special cases:

d=0: ax+by+cz=0a plane passing through the origin

a=0: by+cz=da plane parallel to the x axis

b=0: ax+cz=da plane parallel to the y axis

c=0: ax+by=da plane parallel to the z axis

Special planes

dx y z = + + cba

Vector equations for curves & surfaces Line

z

x

O

C

A

AP = λACx

rA

rC

rP

AC = rC - rA

AP = r - rA

r – rA = λ (rC – rA)

-- Parametric vector equation for a line

Example—Line

A(1,2,1), C(3, 0, -1)

2

1

2

2

2

1 zyx

z

x

O

C

A

x

rA

rC

rP

r – rA = λ (rC – rA)

(x, y, z)-(1, 2, 1)=λ{(3, 0, -1)-(1, 2,1)}

(x, y, z)=(1+2λ,2-2λ, 1-2λ)

21

22

21

z

y

x

AP = λ AC

Unit vector A vector of unit magnitude

e.g.

7

6,

7

3,

7

2a

17

6

7

3

7

2222

a

What about the vectors, b=(1,0,0), c=(0,1,0), d=(0,0,1)?

If a is a vector, then… the unit vector in the direction of a =(ax, ay, az) is:

a

aa ˆ

aa

z

x

y

θz

θx

θy

aaazyx aaa

,,

zyx cos,cos,cos

Basis vectors

i = (1, 0, 0) j = (0, 1, 0) k = (0, 0, 1) Any vector, a = (a1, a2, a3)

can be written as: a = a1 +a2 +a3

ki

j

x

y

z

1

1

1i j k

a = (a1, a2, a3) = (a1, 0, 0)+(0, a2, 0)+(0, 0, a3)

= a1(1, 0, 0)+a2 (0, 1, 0)+a3(0, 0, 1)

i j k = a1 +a2 +a3