Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Vectors

Cartesian plane
A Cartesian plane is a two-dimensional space and it is also called R 2
...
Each point in this plane
has a x and a y coordinate and it is symbolized like (x,y)
...


y
3

A(2,3)

x’

x
O(0,0)

4

y’

Vector of R 2

There are some physical quantities that can be described completely by a
number
...
There are also
some physical quantities that need to be described by a number and also , a
direction
...
These physical quantities are called vectors
...
The vector below starts at A and
finishes at B
...


B
A

Vector on Cartesian plane

2

A(0,0)

B(3,2)

3

The vector start at (0,0) and ends at (2,3)
...
To find a vector we always deduct the end minus the
start
...
For
example AB  (3, 2), AB  32  22  9  4  13
...


Single vectors
There are vectors which their value is 1
...
The
vectors that are symbolized with ̂, ̂ are single vector
...
These vectors
are useful to express a vector’s x’x part and y’y part
...
Then
we draw a line from the start of the first vector to the end of the second vector as in
the picture above
...
For example:
𝐴 = 3𝑥 + 4𝑦
̂
̂
⃗𝐵 = 2𝑥 + 5𝑦
̂
̂
𝐴 + ⃗𝐵 = 3𝑥 + 4𝑦 + 2𝑥 + 5𝑦 = 5𝑥 + 9𝑦
̂
̂
̂
̂
̂
̂
This is the adding of two vectors
...

⃗
𝐴 ∙ ⃗𝐵 = |𝐴||𝐵|𝑐𝑜𝑠𝜃
Cosθ is the cosine of the angle between the two vectors
...
This happens either if the measure of one vector is 0 or both 0 or most
likely when their angle is 90o or


so the cosine of their angle is zero
...


Outer product of 2 vectors

The outer product of 2 vectors is a vector, not just a number
...

⃗
⃗
𝐴 𝑋𝐵 = |𝐴||𝐵|𝑠𝑖𝑛𝜃𝑛
̂
Sinθ is the sine of their angle and ̂ is the vector which is perpendicular to the
𝑛
level that the two vectors make
...


̂
𝑥
⃗
𝐴 𝑋𝐵 = | 𝑥 𝐴
𝑥𝐵

̂
𝑦
𝑦𝐴
𝑦𝐵

𝑧̂
𝑧 𝐴 | = ̂( 𝑦 𝐴 𝑧 𝐵 − 𝑧 𝐴 𝑦 𝐵 ) − ̂ ( 𝑥 𝐴 𝑧 𝐵 − 𝑧 𝐴 𝑥 𝐵 ) + 𝑧̂ ( 𝑥 𝐴 𝑦 𝐵 − 𝑦 𝐴 𝑥 𝐵 )
𝑥
𝑦
𝑧𝐵

The properties of the outer product are:

1) 𝐴 𝑋𝐴 = 0
⃗
⃗
2) 𝐴 𝑋𝐵 = −𝐵 𝑋𝐴
⃗
⃗
⃗
3) 𝜆(𝛢)𝛸𝛣 = (𝜆𝛢)𝛸𝛣 = 𝜆(𝛢 𝛸𝛣)
⃗
⃗
4) 𝛢 𝛸(𝛣 + 𝐶 ) = 𝐴 𝑋𝐵 + 𝐴 𝑋𝐶
⃗
⃗
5) 𝐴 ∙ (𝐵 𝑋𝐶 ) = (𝐴 𝑋𝐵 ) ∙ 𝐶
⃗
6) 𝐴 𝑋(𝐵 𝑋𝐶 ) = ⃗𝐵(𝐴 ∙ 𝐶 ) − 𝐶 (𝐴 ∙ ⃗𝐵)

Exercise 1
Find the absolute value of the following vectors:

A  (3,3), B  (1, 3), C  (0, 2), D  (5,0), E  (1, 6)
Solution
We will use the Pythagorean theorem in order to solve this problem
...
B  (2,6)
Solution

A  B  (1, 2)  (2,6)  1 (2)  2  6  2  12  10
Exercise 3
Given the vectors

a)
b)
c)

2 A  3B
1
2 A  B
2

2  A B

A  (2,5), B  (3, 4)

find:

Solution
a)

2 A  3B  2(2,5)  3(3, 4)  (4,10)  (9,12)  (4  9,10  12)  (5, 22)

1
1
 3 4
2 A  B  2(2,5)  (3, 4)  (4, 10)    ,  
2
2
 2 2
b)


  13
 3

  4     , 10  2     , 12 
 4


  4

2  A  B  2  (2,5)  (3, 4)  (4,10)  (3, 4) 
c)

 4  (3)  10  4  12  40  28
Exercise 4

Find the outer product of the following vectors:

A  (1, 2, 2), B  (3, 2, 4)
Solution

i
j k
2 2
1 2
1 2
A B  1 2 2  i
j
k

2 4
3 4
3 2
3 2 4
 i (2  4  2  (2))  j (1  4  2(3))  k (1(2)  2(3)) 
 12i  10 j 4 k  (12, 10, 4)
Exercise 5
Find the angle between the vectors

A  (1, 2), B  (1,5)

Solution
The cos of the two vectors will be given by the formula of the inner product of
two vectors:

A  B  A B cos   cos  

A B
A B

---