More Vectors

Linear Combination of Vectors

Recall that two non-zero vectors and are if, and onlcollinear , for somy if e u v u kv k R

3, 4u

9,12v 1

3u v

or

3v u

These two vectors are on the same line (collinear)

Linear Combination

3,4,Find and such that and are collinea7 ( , 7, ) r.a b u v a b

Since u and v are collinear: 3,4, 7 , 7,k a b

Therefore:

3 1

4 7 2

7 3

ka

k

kb

From (2): 4

7k

Thus: 21 49

,4 4

a b

Linear Combination

Definition: any two non-collinear vectors form a basis for the plane in which they lie, and any other vector in that plane can be written as a linear combination of these basis vectors

(-1,3) (2,4Show that and form a basis for the plane) .u v

We must demonstrate that these two vectors are not collinear 1,3 2,4

u kv

k

Assume

From the 1st element:1

2k

From the 2nd element:3

4k

Since these are different, u and v are not collinear and hence form a basis for the plane.

Linear CombinationWrite as a linear combina(5,8) (-1,3)tion of and (2,4).w u v

w uk lv

5,8 1,3 2,4k l

2 ,3 4k l k l

5 2

8

(1)

(2)3 4

k l

k l

2 (1) 2 2 5

2

5

k

k

23

10l

2 23

5 10w u v

Therefore:

Linear CombinationThe vectors , , and are not coplanar.

Write

(1, 1, 1) (1, -1, 1) (1, 1, -1)

(1 as a linear combination of , , and ,3,1a) )

u v w

x u v w

k lx v mu w

1,3,1 1,1,1 1, 1,1 1,1 1k l m

1,3,1 , ,k l m k l m k l m

(1)1

3 )

(31

(2

)

k l m

k l m

k l m

(1) (3) (2 42 )2k l

(2) (3) 5)2 (k

1

0

l

m

2 0

2

x u v w

x u v

Note: x is coplanar with u and vTherefore:

Thus we have

Linear Combination

For vectors , , and . if and only if, , ( ) 0 , and are coplanar u v w u v w u v w

Theorem:

Are u=(2, -1, -2), v=(1, 1, 1) and w=(1, -5, -7) coplanar?

2, 1, 2 1,1,1 1, 5, 7u v w

1, 4,3 1, 5, 7u v w

1 1 4 5 3 7u v w

0u v w

Therefore the vectors are coplanar

Equations of Lines in the Plane

In order to determine a straight line it is enough to specify either of the following sets of information:

a) Two points on the line, orb) One point on the line and its direction

For a line a fixed vector is called a direction vector for the line if it is parallel to

ll

d

Note: every line has an infinite number of direction vectors that can be represented as where is one direction vector for the line and t is a non-zero real number.

d

td

Equations of Lines in a Plane

Find a direction vector for each linea) The line l1 through points A(4,-5) and B(3, -7)b) The line l2 with slope 4/5

3 4, 7 5 1, 2AThe vector B ��������������

Therefore a direction vector for l1 can be given by vector (-1, -2)

Any scalar multiple of (-1,-2) could also be used as a direction vector of l1a)

b) A line with slope 4/5 that passes through the origin would pass through the point (5,4). Thus we can use direction vector (5,4) for l2

Equations of Lines in the Plane 0Develop a vector equation for the line through the point with direction vector P 3,5 1,2d

Pick any point P(x,y) on the line.

Because P is on the line, the vector P0P (from PO to P, can be written as a scalar multiple of the direction vector d=(1,2): that is

3, 5 1, for any real number2OP P

x t t

d

y

t

��������������

0 0 1 2

through with direction vector is for any real num

A Vec

ber

tor Equation of the Li

, ,

e

,

n

O OP d P P td t

xor y x y t d d

��������������

Equations of Lines (2D)

For each real value of the scalar t in the vector equations corresponds to a point on the line. This scalar is called the parameter for the equation of the line.

0, 0 1 2

0 1 0 2

through with direction vector

Parametric Equ

P ,

ations of the Lin

re

e

aO x y d d d

x x td y y td

d1 and d2 are called direction numbers of the line

Understanding

Find vector and parametric equations of the line through points A(1,7) and B(4,0)

A direction vector for this line is: 4 1,0 7 3, 7d AB ��������������

Thus, a vector equation of this line is: 1,7 3, 7OP t ��������������

From the vector equation we can obtain the parametric equation

1 3

7 7

x t

y t

Equations of Lines (2D)

Yet another form of equation of a line evolves from solving the parametric equations for the parameter.

00 1

1

x xx x td t

d

0

0 22

y yy y td t

d

Therefore

0 0 1 2

0 0

1 2

through with direction

A Symmetric Equation of the Line

vector , , isx y d d

x x y y

d d

Understanding

For each pair of equations, determine whether or not they describe the same line

1, 6 3, 2

4, 4 6,4

) x y r

x s

a

y

2 5

35

)

1

x t y tb

yx

5, 3 2,1

5 3

)

2 1

x y s

x y

c

For each pair of equations, determine whether or not they describe the same line

1, 6 3, 2

4, 4 6,4

) x y r

x s

a

y

Understanding

Step 1: compare the direction vector in both lines 2 3, 2 6,4

The direction vectors are parallel

Step 2: see if a point on one line is also on the other line

Pick the point (4,4) and check

1, 6 3, 2

14 , 6 3, 2

3, 2 3,

4

2

x r

r

r

y

When r=1, we have a match Therefore these equations are for the same line

Understanding

For each pair of equations, determine whether or not they describe the same line

2 5

35

)

1

x t y tb

yx

Step 1: compare the direction vector in both lines

Step 2: see if a point on one line is also on the other line

1 2 11,5 , 1, 5d d d

The direction vectors are parallel

Pick the point (2,0) and check

13

51

30

51

2

15

yx

Since the left-side does not match the right side, the lines are different

Understanding

For each pair of equations, determine whether or not they describe the same line

Step 1: compare the direction vector in both lines 1 22,1 2,1d d

The direction vectors are not parallel, therefore these lines cannot be identical

5, 3 2,1

5 3

)

2 1

x y s

x y

c

Equations of Lines in Space

Vector Equation:

0

A vector Equation through with direction

vector is for any real number

P

O

d OP P td t ��������������

Parametric Equations: 0

0 1

0 2

0

0 0 1 2

3

3through with direction vector , , d= , , areO

x x td

y y td

z z t

d

d

P x y z d d

Symmetric Equations: 0

0

0 0 1 2

0

3

0

1 2 3

through with direction vector are, , d= , ,O

x x

P x y z d d d

y y z z

d d d

Find the vector, parametric, and symmetric equations for the line through the points A(1, 7, -3) and B(4, 0, 2).

First determine the direction vector: 4 1,0 7,2 3 3, 7,5d AB ��������������

A vector equation is: 1,7, 3 3, 7,5OP t ��������������

A parametric equation is: 1 3

7 7

3 5

x t

y t

z t

A symmetric equation is: 1 7 3

3 7 5

x y z

Understanding

Direction NumbersOne alternative technique for describing the direction of a line focusses on the direction angles of the line.

The direction angles of a line in the plane are the angles,

and , , , between a direction vector in the upper

half-plane (where ) and the positive and axi

0

0 s.

l

l

y x y

Direction Numbers in a SpaceThe direction angles of a line in space are the angles, and

, , between a direction vector in the upper

half-space (where ) and the positive and and axe

, ,

0 , ,

0 s.

l

l

z x y z

Direction Cosines

In the Plane: 1 2 2For a line with directi , 0on vector , l d d d d

1cosd

d 2cos

d

d

1 2 3 3For a line with direction vector , , , 0l d d d d d

1cosd

d 2cos

d

d 3cos

d

d

In Space:

Note: 1 2 3

1cos ,cos ,cos , ,

dd d d

d d

The direction cosines of a line are the components of a unit vector in the direction of the line

Understanding

The line l has direction vector (1,3,5). Find its direction cosines and thus its direction angles

1,3,5Let . Thend

2 2 21 3 5

35

d

So, 1cos

1

35

d

d

80

2cos

3

35

d

d

60

3cos

5

35

d

d

32

Understanding

Determine the angle, to the nearest degree, that (1, 2, -3) makes with the positive x-axis.

2 22

cos2 3

1

14

1

1

74.49

75

1cosd

d

UnderstandingFind the cosines for the line: 5

1 2

3

x t

y t

z t

The direction vector is: 1,2, 1d

This vector is not in the upper half-space (last coordinate is not positive)

So we choose: 1, 2,1d

This direction vector (parallel to the first) is in the upper half-space 2 2 2

1 2 1 6d

1cos

6

2cos

6

1cos

6

Understanding

1 3 20,3,2

1A li

4 14ne through the point has direction cosines , , and .

Find parametric equations of

the

l1

e4

in .

l

For the line we could use the direction vector cos ,cos ,cos

1 3 2, ,

14 14 14

d

However, we can obtain “nicer numbers” is we use

1 3 214 , ,

14 14 14

1, 3,2

d

Then a vector equation is: 0,3,2 1, 3,2OP t ��������������

Parametric equations are: 3 3

2 2

x t

y t

z t