Vectors

VocabularyMagnitude

Direction

Location

Scalar

Vector

Perpendicular

Unit

Graphical

Geometric

Component

Size, amount

Place

Measurement with only magnitude

Measurement with magnitude and direction

Graphical form

Unit vectors (a) have a magnitude of 1

Vectors are equal if they have:

– Same magnitude– Same direction– Not necessarily the same position

Magnitude Head

TailAB

^

A

B

aVector: =a =aa

Multiplying (by a scalar)

Product is a:

Vector

a

v

Adding Vectors

Triangle Law

av

u

u+v

Subtracting Vectors

u – v

= u + (-v)

v – ua

v

u

u-v

Ex. 26.21. Given that vector of magnitude 2, state

a) A vector of magnitude 4 in the direction of vb) A vector of magnitude 0.5 in the direction of vc) A vector of magnitude 6 in the opposite direction of

vd) A unit vector in the direction of ve) A unit vector in the opposite direction of v

S.E. 26.32. Which of the following are true and which are false?

a) a + b is the same as a + b

b) b – a is the same as a – b

c) a + b + c is the same as c + a + b

Explain the term ‘resultant as applied to two vectors

Ex 26.3• Using the diagram below, find:

a) AB+BC b) AC+CD+ DE c) AC – EC

d) DE + EA + AC e) AB-CB f) BC-AC

g) EC+CB+BA

B

D

EA

C

1 2 3 4 5 6 7 8

9

8

7

6

5

4

3

2

1

P

AOi

j

Component formOP =

i =

j =

OA =

OP =

OP =

OA + APUnit vector (x)Unit vector (y)6i8j

6i + 8j

S.E. 26.41. State a unit vector:

(a)In the x direction (b) in the y direction

1. State a unit vector: (a) In the negative x direction (b) in the negative y direction

2. What is the angle between

3. a) i and j b) i and i c) i and –i

4. State an expression for the magnitude of xi + yj

5. State a vector of magnitude 6 in the x direction

6. State a vector of magnitude 2 in the negative y direction

Addinga = a

1i + a

2 j

b = b1

i + b2

j

a + b = (a1

+ b

1)i + (a

2+ b

2)j

eg:

a = 7i + 3j

b = i + 2j

a + b = (7+1)i + (3+2)j

= 8i + 5j

1 2 3 4 5 6 7 8

9

8

7

6

5

4

3

2

1i

j

Multiplying (by a scalar)v = v

1i + v

2 j

av =a(v1

i + v2

j)

av =av1

i + av2

j

eg:

v = 2i + 3j

a = 3

av = 3(2i + 3j)

= 6i + 9j

1 2 3 4 5 6 7 8

9

8

7

6

5

4

3

2

1i

j

Ex 26.41. Calculate the magnitude of:

a) i + 3j b) 3i + 4j c) –i d)1/2(i + j)

2. Calculate the vector from

a) (0,0) to (3,7) b) 3,7) to (5,2) c) (5,2) to (-1,-2)

1. Given a = 3i – 2j, b= - i – 3j, c = 2 i + 5j, find

a) 4a b) 3b + 2c c) 2a – 3b – 4cd) IaI e) I4aI f) I3b + 2cI

Dot (scalar) product• A scalar product of two vectors

• The product shows the magnitude of the influence that one vector has on another.

Eg

• Running a race

• Pushing a piano

• Mario kart zooms

Component forma = a

1i + a

2 j

b = b1

i + b2

j

a.b = a1

b1

+ a2

b2

For two equal vectors:

a.a = a1

a1

+ a2

= a1

2 + a2

2

= IaI2 ij

IaI2

a12

a22

Cosine Rule:

c2 = a2 +b2 – 2ab.cosC

c =

c.c = (a-b).(a-b)

= a.a + b.b – 2a.b

IcI2 = IaI2 + IbI2 – 2a.b

a.b = IaIIbIcosC

Geometric form

a – b

Comparing the component and geometric forms

Show:

• i.i=1

• i.j=0

a.b

• i.i = I1II1IcosC = 1

• i.j = I1II1IcosC = 0

= (a1

i + a2

j).(b1

i+ b2

j)

=a1

b1

i.i + a2

b2

j.j + a2

b1

i.j + a1

b2

i.j

= a1

b1

+ a2

b2

…as shown before

Exercise 26.51. Find the dot product of the following pairs of vectors:

(a) 2i-j, 3i+2j (b) –i+3j, 4i-j c) i-j, -2i-j

2. Find the angles between the following pairs of vectors:

(a) 3i+4j, i+2j (b) i+j, 2i-3j c) i-j, i-j (d) j,-j

3. Show that the vectors a=2i-j and b=-i-2j are perpendicular

4. Points A, B and C have coordinates (-6,2), (3,8) and (-3,-2) respectively.

(a) Find AB and BC

(b) By using the dot product, calculate the angle between AB and BC

Cross (vector) product• A vector product produced by two vectors

• It also measures the interaction between the 2 vectors

• The vector should be the same irrespective of the vector set’s orientation

• Therefore, we need a vector which remains constant irrespective of orientation

• Which direction should the vector point in order to satisfy this requirement?

Right hand rulea x b

a. 1st term, 1st finger

b. 2nd term, 2nd finger

x product. Thumb

j

i

If k = i x j,which direction is k?

k

-k

ExerciseFind:

• i x j =

• i x k =

• k x i =

Notice:

u x v = -v x u

u x u = -u x u

u = -u =

i x i = j x j = k x k = 0

-j x i

-k x j

-i x k

kij

===

0

Component formu = u

1i + u

2j + u

3k

v = v1

i + v2

j + v3

k

u x v =

= (u1

i + u2

j + u3

k)(v1

i + v2

j + v3

k)

= u1

iv1

i + u1

i v2

j + u1

iv3

k + u2

jv1

i + u2

jv2

j + u2

j v3

k + u3

kv1

i + u3

kv2

j +

u3

k v3

k

= u1

i v2

j + u1

iv3

k + u2

jv1

i + u2

j v3

k + u3

kv1

i + u3

kv2

j

u x v = (u2

v3

– u3

v2

)i + (u3

v1

– u

1v3

)j + (u1

v2

– u2

v1

)k

ixj = -jxi = kjxk = -kxj = ikxi = -ixk =j

Magnitude of cross productu x v = (u

2v3

– u3

v2

) + (u3

v1

– u

1v3

) + (u1

v2

– u2

v1

)

From Pythagoras’ Theorem:

IuxvI2 = (u2

v3

– u3

v2

)2 + (u3

v1

– u

1v3

)2 + (u1

v2

– u2

v1

)2

= u2

2v3

2 – 2u

2u

3v2

v3

+ u3

2v2

2

+ u3

2v1

2 – 2u

1u

3v1

v3

+ u1

2v3

2

+ u1

2v2

2 – 2u

1u

2v1

v2

+ u2

2v1

2

= u1

2(v2

2+v3

2) + u2

2(v1

2+v3

2) + u3

2(v1

2+v2

2)

- 2(u2

u3

v2

v3

+ u1

u3

v1

v3

+ u1

u2

v1

v2

) 1

IuIIvIcosC = u.v = u1

v1

+ u2

v2

+ u3

v3

IuI2IvI2cos2C = (u.v)2 = (u1

v1

+ u2

v2

+ u3

v3

)2

= u1

2v1

2 u1

u2

v1

v2

u1

u3

v1

v3

u2

u3

v2

v3

u2

2v2

2 + u1

u2

v1

v2

+ u1

u3

v1

v3

+ u2

u3

v2

v3

u3

2v3

2

= u1

2v1

2 + u2

2v2

2 + u3

2v3

2

+ 2(u1

u2

v1

v2

+ u1

u3

v1

v3

+ u2

u3

v2

v3

) 2

+

= IuxvI + IuI2IvI2cos2C

= u1

2(v2

2+v3

2) + u2

2(v1

2+v3

2) + u3

2(v1

2+v2

2)

- 2(u2

u3

v2

v3

+ u1

u3

v1

v3

+ u1

u2

v1

v2

)

u1

2v1

2 + u2

2v2

2 + u3

2v3

2

+ 2(u1

u2

v1

v2

+ u1

u3

v1

v3

+ u2

u3

v2

v3

)

= u1

2(v2

2+v3

2) + u2

2(v1

2+v3

2) + u3

2(v1

2+v2

2)

+ u1

2v1

2 + u2

2v2

2 + u3

2v3

2

= u1

2(v1

2+v2

2+v3

2) + u2

2(v1

2+v2

2+v3

2) + u3

2(v1

2+v2

2+v3

2)

= (u1

2 + u2

2 + u3

2 )(v1

2+v2

2+v3

2)

1 2

By pythagoras’ Theorem:

(u1

2 + u2

2 + u3

2 )(v1

2+v2

2+v3

2)

= IuI2IvI2

In summary:

+ = Iu x vI2 + IuI2IvI2cos2C = IuI2IvI2

Iu x vI2 = IuI2IvI2 - IuI2IvI2cos2C

= IuI2IvI2 (1- cos2C)

= IuI2IvI2 (sin2C)

Iuxvi2 = IuI2IvI2 sin2C

1 2

Magnitude as a parallelogram