Physics 1250
Vector, Matrix & Complex Analysis
Physical Quantities that have directions and magnitudes are commonly called as vectors. Examples of
vectors are velocity, acceleration, force, torque, momentum etc.
Vectors can be added, subtracted multiplied but cannot be “divided”.
⃗ and ⃗𝒃 be two vectors. To get the sum of the two vectors, place the tail of ⃗𝒃 onto the head of 𝒂
⃗
Let 𝒂
⃗
⃗ to the head of 𝒃. The distance between the tail of 𝒂
⃗ and head of
and then draw an arrow from the tail of 𝒂
⃗𝒃 is |𝒂
⃗
⃗
⃗ + 𝑏 | and the vector sum is 𝒂
⃗ + 𝑏.
⃗ + 𝑏⃗
𝒂
O
O
⃗𝑨 = |𝑨𝒙 |𝒙
̂ + |𝑨𝒚 |𝒚
̂ + |𝑨𝒛 |𝒛̂
⃗𝑨 = [|𝑨𝒙 | − 𝟎] 𝒙
̂ + [|𝑨𝒚 | − 𝟎] 𝒚
̂ + [|𝑨𝒛 | − 𝟎] 𝒛̂
(Ax , Ay ,Az)
(0, 0, 0)
⃗|
|𝒃
⃗𝒃
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
|𝒂
+ 𝒃|
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝒂+𝒃
⃗
𝒂
|𝒂
⃗|
Given the vectors:
⃗ = −𝟑𝒙
̂ − 𝒚
̂ + 𝟐𝒛̂
𝑨
⃗⃗ = −𝟐𝒙
̂ + 𝟑𝒚
̂ − 𝒛̂
𝑩
⃗ = 𝑨
⃗ + 𝑩
⃗⃗ = −𝟓𝒙
̂ + 𝟐𝒚
̂ + 𝒛̂
𝑪
⃗ | = √(−𝟓)𝟐 + (𝟐)𝟐 + 𝟏𝟐 = √𝟑𝟎 𝒖𝒏𝒊𝒕𝒔
|𝑪
So the vector sum ⃗𝑪 has coordinates (-5, 2, 1) for its head; and the tail is at the origin.
Tail coordinates (1, 2, 1) and Head coordinates (2, 0, 2), the expression for the vector is:
⃗⃗ = (𝟐 − 𝟏)𝒙
̂ + (𝟎 − 𝟐)𝒚
̂ + (𝟐 − 𝟏)𝒛̂
𝑽
⃗⃗ | = √𝟏𝟐 + (−𝟐)𝟐 + 𝟏𝟐 = √𝟔 𝒖𝒏𝒊𝒕𝒔
|𝑽
Example
⃗𝑨
⃗ = −𝒙
̂ +𝟑𝒚
̂ + 𝟐𝒛̂
| ⃗𝑨| = √𝟏 + 𝟗 + 𝟒 = √𝟏𝟒 𝒖𝒏𝒊𝒕𝒔
What is the direction of the vector?
From the definition of a vector,
⃗𝑨
⃗ = | ⃗𝑨
⃗ |𝒂
̂
The direction is therefore,
̂=
𝒂
̂ =
𝒂
̂ +𝟑𝒚
̂ + 𝟐𝒛̂
−𝒙
√𝟏𝟒
⃗
𝑨
⃗⃗ |
|𝑨
|𝒂
̂| = ?
In matrix form, the previous vectors are
−𝟑
−𝟐
⃗𝑨
⃗ = (−𝟏) , ⃗𝑩
⃗ = (𝟑)
𝟐
−𝟏
𝒔𝒐
−𝟓
⃗𝑨 + ⃗𝑩
⃗ = [𝑨] + [𝑩] = ( 𝟐 )
𝟏
⃗
𝐴 − 𝐵
⃗
𝐵
⃗
𝐴+ 𝐵
⃗
𝐵
𝐴
IS NOT THE SAME
𝐴
𝐴
⃗
−𝐵
⃗
𝐴 + −𝐵
⃗|
|𝒃
⃗𝒃
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
|𝒂
+ 𝒃|
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝒂+𝒃
⃗
𝒂
|𝒂
⃗|
Multiplication of a vector by a positive scalar k multiplies the magnitude but leaves the direction
⃗ just doubles the magnitude only but the direction
unchanged. If k = 2 then the magnitude of 𝒂
remains the same.
Multiplication of Vectors
Multiplication of a vector by a positive scalar k multiplies
the magnitude but leaves the direction unchanged. If k = 2,
⃗ just doubles but the direction
then the magnitude of vector 𝒂
remains the same. See figure on the right:
⃗
2𝒂
⃗
𝒂
Dot product of two vectors - is the product of a vector to the projection of the other vector
It is just the vector product of two parallel or antiparallel vectors. If the vectors are neither
parallel or antiparallel, then the component (|𝑎| cos 𝜃) is the one to be multiplied to the other
component ⃗𝒃.
⃗ cos 𝜃
𝒂
⃗
𝒂
θ
⃗𝒃
⃗
𝑭
⃗|
|𝑭
⃗
𝒔
θ
θ
|𝒔
⃗|
⃗ |𝒄𝒐𝒔 𝜽
|𝑭
⃗ ∙ ⃗𝒃 is called the dot product of the two vectors, defined by
The dot product 𝒂
⃗ | 𝑖𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
|𝒂
⃗ ||𝒃
⃗ | cos 𝜃 = {−|𝒂
⃗ ∙ ⃗𝒃 = |𝒂
⃗ ||𝒃
𝒂
⃗ | 𝑖𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 𝑎𝑛𝑡𝑖𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
⃗ ||𝒃
0 𝑖𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
⃗⃗ 𝑎𝑛𝑑 𝑩
⃗⃗ 𝑏𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
Example: Let 𝑨
⃗𝑩
⃗ = 3𝑥̂
⃗𝑨 = 2𝑥̂
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
⃗⃗ = 3𝑥̂
𝑩
⃗ = 2𝑦̂
𝑨
⃗⃗ ∙ 𝑩
⃗⃗ = (2)(3) cos 90 = 0
𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 = 𝑨
⃗𝑩
⃗ = 3𝑥̂
⃗𝑨 = −2𝑥̂
𝑎𝑛𝑡𝑖𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
⃗𝑨 ∙ ⃗𝑩
⃗ = (2)(3) cos 0 = 6 𝑢𝑛𝑖𝑡𝑠
⃗𝑨 ∙ ⃗𝑩
⃗ = (2)(3) cos 180 = −6 𝑢𝑛𝑖𝑡𝑠
⃗⃗ ∙ 𝑩
⃗⃗ = 2(−𝑥̂) ∙ 3𝑥̂ = (2)(3) [(−𝑥̂) ∙ (𝑥̂)] = 6 [(1)(1) cos 180] = −6 𝑢𝑛𝑖𝑡𝑠
𝑨
⃗ 𝑎𝑛𝑑 𝑩
⃗⃗ 𝑏𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
Example: Let 𝑨
⃗𝑨 = 3𝑥̂ + 2𝑦̂ 𝑎𝑛𝑑 ⃗𝑩
⃗ = 𝑧̂
⃗ ∙ 𝑩
⃗⃗ = (3𝑥̂ + 2𝑦̂) ∙ 𝑧̂ = 0 , 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
𝑨
Example: Let
⃗𝑨 = 2𝑥̂ − 𝑦̂ + 2𝑧̂ 𝑎𝑛𝑑 ⃗𝑩
⃗ = −2𝑥̂ + 2 𝑦̂ − 𝑧̂
⃗⃗ ∙ 𝑩
⃗⃗ = (2𝑥̂ − 𝑦̂ + 2𝑧̂ ) ∙ (−2𝑥̂ + 2 𝑦̂ − 𝑧̂ )
𝑨
= (2𝑥̂) ∙ (−2𝑥̂ + 2 𝑦̂ − 𝑧̂ ) + (− 𝑦̂) ∙ (−2𝑥̂ + 2 𝑦̂ − 𝑧̂ )
+ (2𝑧̂ ) ∙ (−2𝑥̂ + 2 𝑦̂ − 𝑧̂ )
= [−4] + [−2] + [−2] = −8
Example: Let 𝐴 = 2𝑥̂ − 𝑦̂ + 2𝑧̂
⃗ = −2𝑥̂ + 2 𝑦̂ − 𝑧̂ find the angle between them
𝑎𝑛𝑑 𝐵
Using the definition:
⃗ = |𝐴||𝐵| cos 𝜃
𝐴∙𝐵
cos 𝜃 =
−8
8
= −
(3)(3)
9
→
⃗
𝐴∙𝐵
= cos 𝜃
|𝐴||𝐵|
→
8
𝜃 = arccos [− ]
9
⃗ = |𝐴||𝐵| sin 𝜃 𝜇̂ = [|𝐴||𝐵| sin 𝜃] 𝜇̂ ,
Cross Product of any two vectors is defined by 𝐴 𝑥 𝐵
̂ is a unit vector (vector of length 1) pointing
where 𝜇̂ = 𝒏
⃗
𝑪
⃗ . But as there are
⃗ 𝑎𝑛𝑑 𝒃
perpendicular to the plane of 𝒂
two directions perpendicular to any plane, the ambiguity
is resolved by the right hand rule: let your fingers point
in the direction of the first vector and curl around
(via the smaller angle) towards the second; then your
thumb indicates the direction of the unit vector.
𝑛̂
Plane defined the vectors 𝑎 𝑎𝑛𝑑 𝑏⃗
which area is |𝑎||𝑏⃗ | sin 𝜃
⃗
𝒃
θ
⃗
𝒂
Unit vector is a vector whose magnitude is 1 and
points to a particular direction. Without loss of
generality, we can assume 𝑥̂ = 𝑖, 𝑦̂ = 𝑗, 𝑧̂ = 𝑘
to be three distinct unit vectors along the x, y, and
z-axis relatively.
y-axis
j
i
k
z-axis
Applying the Dot Product definition,
x-axis
𝑥̂ ∙ 𝑥̂ = 1
𝑦
{ ̂ ∙ 𝑥̂ = 0
𝑧̂ ∙ 𝑥̂ = 0
,
,
,
𝑥̂ ∙ 𝑦̂ = 1
𝑦̂ ∙ 𝑦̂ = 1
𝑧̂ ∙ 𝑦̂ = 0
,
,
,
𝑥̂ ∙ 𝑧̂ = 1
𝑦̂ ∙ 𝑧̂ = 1
𝑧̂ ∙ 𝑧̂ = 1
Likewise applying the Cross Product definition,
𝑥̂ × 𝑥̂ = 0 ,
{𝑦̂ × 𝑥̂ = −𝑧̂ ,
𝑧̂ × 𝑥̂ = 𝑦̂ ,
𝑥̂ × 𝑦̂ = 𝑧̂ ,
𝑦̂ × 𝑦̂ = 0 ,
𝑧̂ × 𝑦̂ = −𝑥̂ ,
𝑥̂ × 𝑧̂ = −𝑦̂
𝑦̂ × 𝑧̂ = 𝑥̂
𝑧̂ × 𝑧̂ = 0
A simplified guide to this definition is achieved by using the
cyclic permutation as shown where counterclockwise
permutation is considered positive and clockwise permutation
is negative. Neither clockwise or counterclockwise gives the
zero cross product.
𝑥̂
𝑦̂
𝑧̂
⃗ = 𝐴𝜌 𝜌̂ + 𝐴𝜑 𝜑̂ + 𝐴𝑧 𝑧̂ , where 𝜌̂, 𝜑̂, 𝑧̂ are the unit
In cylindrical coordinate systems, a vector 𝑨
vectors of the coordinate system.
From rectangular to cylindrical
𝜌 = √𝑥 2 + 𝑦 2
𝑦
𝜑 = 𝑡𝑎𝑛−1
𝑥
𝑧 = 𝑧
From cylindrical to rectangular
𝑥 = 𝜌 cos 𝜑
𝑦 = 𝜌 cos 𝜑
𝑧 = 𝑧
𝒛 − 𝒂𝒙𝒊𝒔
𝒚 − 𝒂𝒙𝒊𝒔
𝒙 − 𝒂𝒙𝒊𝒔
𝝋
𝜌
̂ 𝒔𝒊𝒏 𝝋
−𝝋
𝒚 − 𝒂𝒙𝒊𝒔
̂
𝝆
̂ 𝒄𝒐𝒔 𝝋
𝝆
𝝋
𝝆
̂
−𝝋
𝝋
̂
𝒙
̂
𝝆
𝒙 − 𝒂𝒙𝒊𝒔
𝝋
Unit Vectors: Cylindrical to rectangular
̂ = 𝒄𝒐𝒔 𝝋 𝝆
̂ − 𝒔𝒊𝒏 𝝋 𝝋
̂
𝒙
̂ = 𝒔𝒊𝒏 𝝋 𝝆
̂ + 𝒄𝒐𝒔 𝝋 𝝋
̂
𝒚
𝒛̂ = 𝒛̂
In matrix form:
𝒄𝒐𝒔 𝝋
̂
𝒙
(𝒚
)
=
(
̂
𝒔𝒊𝒏 𝝋
𝟎
𝒛̂
̂
− 𝒔𝒊𝒏 𝝋 𝟎 𝝆
)
(
𝒄𝒐𝒔 𝝋 𝟎 𝝋
̂)
𝟎
𝟏
𝒛̂
⃗ = 𝐴𝑟 𝑟̂ + 𝐴𝜃 𝜃̂ + 𝐴𝜑 𝜑̂, where 𝑟̂ , 𝜃̂, 𝜑̂ are the unit
In spherical coordinate system, a vector 𝑨
vectors of the coordinate system.
Examples of Cross Product
𝐴 = 2𝑥̂ − 𝑦̂ + 2𝑧̂
⃗ = −2𝑥̂ + 2 𝑦̂ − 𝑧̂
𝑎𝑛𝑑 𝐵
⃗ = 2𝑥̂ 𝑥 (−2𝑥̂ + 2 𝑦̂ − 𝑧̂ ) + (−𝑦̂) 𝑥 (−2𝑥̂ + 2 𝑦̂ − 𝑧̂ ) + 2𝑧̂ 𝑥 (−2𝑥̂ + 2 𝑦̂ − 𝑧̂ )
𝐴 𝑥 𝐵
⃗ = 2𝑥̂ 𝑥 (2 𝑦̂ − 𝑧̂ ) + (−𝑦̂) 𝑥 (−2𝑥̂ − 𝑧̂ ) + 2𝑧̂ 𝑥 (−2𝑥̂ + 2 𝑦̂)
𝐴 𝑥 𝐵
⃗ = (4𝑧̂ + 2𝑦̂) + (−2𝑧̂ + 𝑥̂) + (−4𝑦̂ − 4𝑥̂) = −𝟑𝒙
̂ − 𝟐𝒚
̂ + 𝟐𝒛̂
𝐴 𝑥 𝐵
𝑥̂
𝑦̂
⃗
𝐴 𝑥 𝐵 = | 2 −1
−2 2
𝑧̂
𝑥̂
𝑦̂
𝑧̂
𝑥̂
𝑦̂
2 | → | 2 −1 2 | 2 −1 = (𝑥̂ − 4𝑦̂ + 4𝑧̂ ) − (4𝑥̂ − 2𝑦̂ + 2𝑧̂ )
−1
−2 2 −1 −2 2
𝑥̂
𝑦̂
⃗
𝐴 𝑥 𝐵 = | 2 −1
−2 2
𝑧̂
̂ − 𝟐𝒚
̂ + 𝟐𝒛̂
2 | = (𝑥̂ − 4𝑦̂ + 4𝑧̂ ) − (4𝑥̂ − 2𝑦̂ + 2𝑧̂ ) = −𝟑𝒙
−1
Vector Practice
1. Consider three vectors:
𝐴 = 3𝑥̂ + 0𝑦̂
⃗ = 2√3𝑥̂ + 2𝑦̂
{𝐵
𝐶 = −5𝑥̂ + 5√3𝑦̂
a. With three vectors, find the angles
between vectors, and find the vector
sum.
b. What is the length or magnitude of A , B and C ?
Length of the vector A  A 
Ax2  Ay2 = ?
Length of the vector B  B  Bx2  B y2 = ?
Length of the vector C  C  C x2  C y2 = ?
c. What is the angle between A and C , A and B , B and C ?
We know that A  C = A C cos  ,


AC 
So,   cos 1 
A C
AC


Now, 𝐴 ∙ 𝐶 = 𝐴𝑥 𝑥̂ + 𝐴𝑦 𝑦̂ ∙ 𝐶𝑥 𝑥̂ + 𝐶𝑦 𝑦̂
= Ax C x  Ay C y
So,
cos  
So, cos  
AC
AC
AC
   either
? or
?.
But notice that the y-component of C is positive and the x-component negative. So it is in the 2nd
quadrant and the A vector is in the first quadrant. So, the angle between them cannot be more than
?.
180 . So the angle between A and C is
Similarly, the angle between A and B is
A B
,
cos  
AB
cos  
A B
=
?    either ? or
?. But   330 , as none of the components
AB
of A and B are negative. So,    ?
Similarly, the angle between B and C is
B C
cos  
,
BC
cos  
B C
=
?    either
? or
?. But   270 , as none of the
BC
components of B and C are negative. So,   
?
2. Consider three vectors:
𝐴 = 4𝑥̂ + 6𝑦̂ − 2𝑧̂
⃗ = 2𝑥̂ + 7𝑦̂ − 𝑧̂
{𝐵
𝐶 = 0𝑥̂ + 3𝑦̂ + 5𝑧̂
a. What is the length or magnitude of A , also written as A ?
Length of the vector A 
Ax2  Ay2  Az2 = 4 2  6 2  (2) 2 =
b. Write the expression for 2 A .
2 A = 2( A ) = 2(4𝑥̂ + 6𝑦̂ − 2𝑧̂ ) = ?
c. What is A  B ?
A  B = 𝐴𝑥 𝑥̂ + 𝐴𝑦 𝑦̂ + 𝐴𝑧 𝑧̂ + 𝐵𝑥 𝑥̂ + 𝐵𝑦 𝑦̂ + 𝐵𝑧 𝑧̂
?
= (𝐴𝑥 + 𝐵𝑥 )𝑥̂ + 𝐴𝑦 + 𝐵𝑦 𝑦̂ + (𝐴𝑧 + 𝐵𝑧 )𝑧̂
d. What is C  A ?
C  A = 𝐶𝑥 𝑥̂ + 𝐶𝑦 𝑦̂ + 𝐶𝑧 𝑧̂ − 𝐴𝑥 𝑥̂ + 𝐴𝑦 𝑦̂ + 𝐴𝑧 𝑧̂
= (𝐶𝑥 − 𝐴𝑥 )𝑥̂ + 𝐶𝑦 − 𝐴𝑦 𝑦̂ + (𝐶𝑧 − 𝐴𝑧 )𝑧̂
e. What is C  A ?
With
𝐴 = 4𝑥̂ + 6𝑦̂ − 2𝑧̂
⃗ = 2𝑥̂ + 7𝑦̂ − 𝑧̂
{𝐵
𝐶 = 0𝑥̂ + 3𝑦̂ + 5𝑧̂
C  A = 𝐶𝑥 𝑥̂ + 𝐶𝑦 𝑦̂ + 𝐶𝑧 𝑧̂ × 𝐴𝑥 𝑥̂ + 𝐴𝑦 𝑦̂ + 𝐴𝑧 𝑧̂
𝑥̂
𝐶 × 𝐴 = | 𝐶𝑥
𝐴𝑥
𝐶×𝐴 =
𝑦̂
𝐶𝑦
𝐴𝑦
𝑧̂
𝐶
𝐶𝑧 | = | 𝑦
𝐴𝑦
𝐴𝑧
𝐶𝑧
𝐶
| 𝑥̂ − | 𝑥
𝐴𝑧
𝐴𝑥
𝐶𝑥
𝐶𝑧
| 𝑦̂ + |
𝐴𝑧
𝐴𝑥
𝐶𝑦
| 𝑧̂
𝐴𝑦
𝐶𝑦 𝐴𝑧 − 𝐶𝑧 𝐴𝑦 𝑥̂ − (𝐶𝑥 𝐴𝑧 − 𝐶𝑧 𝐴𝑥 )𝑦̂ + 𝐶𝑥 𝐴𝑦 − 𝐶𝑦 𝐴𝑥 𝑧̂
𝐶 × 𝐴 = [3(−2) − 5(6)]𝑥̂ − [0 − 5(4)]𝑦̂ + [0 − 3(4)]𝑧̂
𝐶 × 𝐴 = [3(−2) − 5(6)]𝑥̂ − [0 − 5(4)]𝑦̂ + [0 − 3(4)]𝑧̂
𝐶 × 𝐴 = −36𝑥̂ + 20𝑦̂ − 12𝑧̂
f. What is the magnitude of C  A ?
Magnitude of C  A = C  A
Let us name C  A = M
Then magnitude of M = M  M x2  M y2  M z2 = 4 115
g. What is B  C ?
B  C = Bx C x  B y C y  Bz C z
= 2  0  7  3  (1)  5
=16
Note:
B  C is the dot product of the two vectors.
B  C = B C cos  , where  is the angle formed by B and C when they are place tail-to-tail.
h. What is the angle between A and C ?
 
We know that A  C = A C cos  ,
So,
cos  
AC
AC


AC 

So,   cos
A C




 2 
8
So   cos 1 
  cos 1 

 2 14  34 
 119 
1
i. Does B  C equal C  B ?
Yes. We can see that B  C is a scalar (hence the alternative name scalar product). So the dot
product is commutative. That is B  C = C  B
j. How is C  A and A C related?
Using the formula for the cross product,
𝐶 × 𝐴 = 𝐶𝑥 𝑥̂ + 𝐶𝑦 𝑦̂ + 𝐶𝑧 𝑧̂ × 𝐴𝑥 𝑥̂ + 𝐴𝑦 𝑦̂ + 𝐴𝑧 𝑧̂
𝑥̂
𝐶 × 𝐴 = | 𝐶𝑥
𝐴𝑥
𝐶×𝐴 =
𝑦̂
𝐶𝑦
𝐴𝑦
𝑧̂
𝐶
𝐶𝑧 | = | 𝑦
𝐴𝑦
𝐴𝑧
𝐶𝑧
𝐶
| 𝑥̂ − | 𝑥
𝐴𝑧
𝐴𝑥
𝐶𝑥
𝐶𝑧
| 𝑦̂ + |
𝐴𝑧
𝐴𝑥
𝐶𝑦
| 𝑧̂
𝐴𝑦
𝐶𝑦 𝐴𝑧 − 𝐶𝑧 𝐴𝑦 𝑥̂ − (𝐶𝑥 𝐴𝑧 − 𝐶𝑧 𝐴𝑥 )𝑦̂ + 𝐶𝑥 𝐴𝑦 − 𝐶𝑦 𝐴𝑥 𝑧̂
And
𝐴 × 𝐶 = 𝐴𝑥 𝑥̂ + 𝐴𝑦 𝑦̂ + 𝐴𝑧 𝑧̂ × 𝐶𝑥 𝑥̂ + 𝐶𝑦 𝑦̂ + 𝐶𝑧 𝑧̂
𝑥̂
𝐴
𝐴×𝐶 = | 𝑥
𝐶𝑥
𝐴×𝐶 =
𝑦̂
𝐴𝑦
𝐶𝑦
𝑧̂
𝐴
𝐴𝑧 | = | 𝑦
𝐶𝑦
𝐶𝑧
𝐴𝑧
𝐴
| 𝑥̂ − | 𝑥
𝐶𝑧
𝐶𝑥
𝐴𝑥
𝐴𝑧
| 𝑦̂ + |
𝐶𝑧
𝐶𝑥
𝐴𝑦
| 𝑧̂
𝐶𝑦
𝐴𝑦 𝐶𝑧 − 𝐴𝑧 𝐶𝑦 𝑥̂ − (𝐴𝑥 𝐶𝑧 − 𝐴𝑧 𝐶𝑥 )𝑦̂ + 𝐴𝑥 𝐶𝑦 − 𝐴𝑦 𝐶𝑥 𝑧̂
𝐴 × 𝐶 = − 𝐶𝑦 𝐴𝑧 − 𝐶𝑧 𝐴𝑦 𝑥̂ + (𝐶𝑥 𝐴𝑧 − 𝐶𝑧 𝐴𝑥 )𝑦̂ − 𝐶𝑥 𝐴𝑦 − 𝐶𝑦 𝐴𝑥 𝑧̂
Equating 1) and 2), we can clearly see that
( C  A ) = ― ( A C )
k. Give an example of the use of dot product in Physics and explain.
Mechanical work is a dot product between the force and the displacement.
W  F r
It means that the work is amount of displacement times the projection of force along the
displacement.
l. Give an example of the use of cross product in Physics and explain.
Magnetic force is a good example of a cross product of velocity of a charged particle times the
magnetic field.

 
F  q(v  B)
It means that the force is perpendicular to both v and B , and that the magnitude of the force is equal
to the area of the parallelogram they span.
m. Imagine that the vector A is a force whose units are given in Newtons. Imagine vector B is a radius
vector through which the force acts in meters. What is the value of the torque (  r  F ) , in this case?
Torque is the cross product between the radius (lever arm) vector and the force vector, (  r  F ) .
⃗ 𝑎𝑛𝑑 𝐹 = 𝐴, so that, in this case   B  A . Using the given expressions of vectors,
Let 𝑟 = 𝐵
𝐴 = 4𝑥̂ + 6𝑦̂ − 2𝑧̂
⃗ = 2𝑥̂ + 7𝑦̂ − 𝑧̂
{𝐵
𝐶 = 0𝑥̂ + 3𝑦̂ + 5𝑧̂
⃗ × 𝐴 = 𝐵𝑥 𝑥̂ + 𝐵𝑦 𝑦̂ + 𝐵𝑧 𝑧̂ × 𝐴𝑥 𝑥̂ + 𝐴𝑦 𝑦̂ + 𝐴𝑧 𝑧̂
𝐵
𝑥̂
⃗ × 𝐴 = | 𝐵𝑥
𝐵
𝐴𝑥
⃗ ×𝐴 =
𝐵
𝑦̂
𝐵𝑦
𝐴𝑦
𝑧̂
𝐵
𝐵𝑧 | = | 𝑦
𝐴𝑦
𝐴𝑧
𝐵𝑧
𝐵
| 𝑥̂ − | 𝑥
𝐴𝑧
𝐴𝑥
𝐵𝑥
𝐵𝑧
| 𝑦̂ + |
𝐴𝑧
𝐴𝑥
𝐵𝑦
| 𝑧̂
𝐴𝑦
𝐵𝑦 𝐴𝑧 − 𝐵𝑧 𝐴𝑦 𝑥̂ − (𝐵𝑥 𝐴𝑧 − 𝐵𝑧 𝐴𝑥 )𝑦̂ + 𝐵𝑥 𝐴𝑦 − 𝐵𝑦 𝐴𝑥 𝑧̂
𝜏 = [7(−2) − (−1)(6)]𝑥̂ − [2(−2) − (−1)(4)]𝑦̂ + [2(6) − 7(4)]𝑧̂
𝜏 = 8𝑥̂ − 0𝑦̂ − 16𝑧̂ = 8𝑥̂ − 16𝑧̂ 𝑁. 𝑚
n. Now imagine that A continues to be a force vector and C is a displacement vector whose units are
meters. What is the work done in applying force A through a displacement C ?
Work done in applying force A through a displacement C is
W  AC
= ( Ax iˆ  Ay ˆj  Az kˆ)  (C x iˆ  C y ˆj  C z kˆ)
W = Ax C x  Ay C y  Az C z
= 4  0  6  3  (2)  5
= 8 joules


o. What is the vector sum of a vector D given by 40 m, 30 degrees and a vector E given by
12 m, 225 degrees? Use the method of resolving vectors into their components and then
adding the components.
Let us start by resolving the vectors into their x-components and y-components.
It can be assumed that the angles given in the question are w.r.t the x-axis.

For D :


The magnitude of D = D is 40 and the angle is 30o


So, as the projection of D along the x-axis = D cos 
3
Dx = 40 cos 30 = 40 
 20 3
2


The projection of D along the y-axis = D sin 
1
D y  40 sin 30  40   20
2

Similarly for E :
 1 
E x  12 cos 225  12  
  6 2
 2
 1 
E y  12 sin 225  12  
  6 2
 2
 
So, D + E = ( Dx xˆ  D y yˆ )  ( E x xˆ  E y yˆ ) = ( Dx  E x ) xˆ  ( D y  E y ) yˆ
= (20 3  6 2 ) xˆ  (20  6 2 ) yˆ
3. Consider three vectors:

A  3iˆ  3 ˆj  2kˆ

B  2iˆ  4 ˆj  2kˆ

C  2iˆ  3 ˆj  1kˆ
  
a. Find A  ( B  C ) .
First we shall find
 
B  C = ( Bx iˆ  B y ˆj  Bz kˆ)  (C x iˆ  C y ˆj  C z kˆ) = ( Bx  C x )iˆ  ( B y  C y ) ˆj  ( Bz  C z )kˆ
= ((2)  2)iˆ  ((4)  3) ˆj  (2  1)kˆ
= 0iˆ  1 ˆj  3kˆ
  
Let us name B  C  E
    
Now, A  ( B  C ) = A  E  ( Ax iˆ  Ay ˆj  Az kˆ)  ( E x iˆ  E y ˆj  E z kˆ)
= Ax E x  Ay E y  Az E z
= (3)  0  3  (1)  2  3
=3
  
b. Find A  ( B  C ) .
First we shall find
 
B  C = ( Bx iˆ  B y ˆj  Bz kˆ)  (C x iˆ  C y ˆj  C z kˆ)
=
iˆ
Bx
Cx
ˆj
By
Cy
kˆ
Bz
Cz
= ( B y C z  Bz C y )iˆ  ( Bx C z  Bz C x ) ˆj  ( Bx C y  B y C x )kˆ
= ((4)  1  2  3)iˆ  ((2)  1  2  2) ˆj  ((2)  3  (4)  2)kˆ
=  10iˆ  6 ˆj  2kˆ
  
Let us name B  C  D
  
 
Now, A  ( B  C ) = A  D  ( Ax iˆ  Ay ˆj  Az kˆ)  ( Dx iˆ  D y ˆj  Dz kˆ)
= Ax Dx  Ay D y  Az Dz
= (3)  (10)  3  6  2  2
= 52

 
c. Find A  ( B  C ) .
First we shall find
 
B  C = ( Bx iˆ  B y ˆj  Bz kˆ)  (C x iˆ  C y ˆj  C z kˆ) = ( Bx  C x )iˆ  ( B y  C y ) ˆj  ( Bz  C z )kˆ
= ((2)  2)iˆ  ((4)  3) ˆj  (2  1)kˆ
= 0iˆ  1 ˆj  3kˆ
  
Let us name B  C = F

 
 
Now A  ( B  C ) = A  F
 
A  F = ( Ax iˆ  Ay ˆj  Az kˆ)  ( Fx iˆ  Fy ˆj  Fz kˆ)
=
iˆ
ˆj
kˆ
Ax
Fx
Ay
Fy
Az
Fz
= ( Ay Fz  Az Fy )iˆ  ( Ax Fz  Az Fx ) ˆj  ( Ax Fy  Ay Fx )kˆ
= (3  3  2  (1))iˆ  ((3)  3  2  0) ˆj  ((3)  (1)  3  0)kˆ
= 11iˆ  9 ˆj  3kˆ
Prove the following:
⃗𝑨 𝒙 ⃗𝑩
⃗ 𝒙 ⃗𝑪 =
⃗𝑨 ∙ ⃗𝑪 ⃗𝑩
⃗ −
⃗𝑨 ∙ ⃗𝑩
⃗ ⃗𝑪
⃗ = 𝑩
⃗ − 𝑪
⃗ 𝑨
⃗ 𝒙 𝑩
⃗⃗ 𝒙 𝑪
⃗⃗ 𝑨
⃗ ∙𝑪
⃗ ∙𝑩
⃗⃗
𝑨
Reciprocal of a Vector
Reciprocal of a Vector in One-Dimensional Basis
For a single 3-D vector, say
⃗⃗ = |𝐴|𝒂
̂
𝑨
(1)
The reciprocal vector is one that is directed in the same direction as the given vector A, but has
magnitude equal to
1
⃗ −1 =
𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑨
|𝐴|
So, the reciprocal vector
⃗ −1 =
𝑨
1
̂
𝒂
|𝐴|
From equation (1)
⃗
⃗
𝟏
𝑨
𝑨
⃗ −1 = ( )[ ] =
𝑨
|𝑨| |𝑨|
|𝑨|𝟐
Therefore for the given single vector
⃗⃗⃗
̂ − 2𝒚
̂ + 5𝒛̂
𝑨 = 3𝒙
2
⃗⃗⃗ | = (3)2 + (−2)2 + (5)2
|𝑨
The |𝐴|2 = 38 , so the reciprocal vector of the given single vector ⃗𝑨 is
1
̂ − 2𝒚
̂ + 5𝒛̂)
(3𝒙
38
⃗𝑨− 1 =
Given:
𝐴 = 3𝑥̂ − 2𝑦̂ + 2𝑧̂
Find:
⃗ = 𝑡ℎ𝑒 𝑟𝑒𝑐𝑖𝑝𝑟𝑜𝑐𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑣𝑒𝑐𝑡𝑜𝑟
𝐵
⃗ =1
𝐴 ∙ 𝐵
→
Looking for
⃗ =
𝐵
𝐴
|𝐴|
2
=
3𝑥̂ − 2𝑦̂ + 2𝑧̂
3𝑥̂ − 2𝑦̂ + 2𝑧̂
=
2
2
+ (−2) + (2)
17
(3)2
To check
⃗ ?̿ 1
𝐴 ∙ 𝐵
(3𝑥̂ − 2𝑦̂ + 2𝑧̂ ) ∙
Take for example,
Look for the Reciprocal of the vector of,
3𝑥̂ − 2𝑦̂ + 2𝑧̂
17
=
= 1
17
17
⃗ = −2𝑥̂ − 3𝑦̂ + 2𝑧̂
𝑨
⃗ −1 =
𝑨
−2𝑥̂ − 3𝑦̂ + 2𝑧̂
17
Reciprocal of Vectors in Two-Dimensional Basis
⃗ ,𝑩
⃗⃗ , it is not easy to do this but such difficulty can be by-passed
For 2 given 3-D vectors 𝑨
̂ . For this, the cyclic permutation
by considering an additional vector which is a unit vector 𝒏
⃗
[𝐴 𝐵 𝐶 ] can be helpful.
⃗ = 𝐶, 𝐵
⃗ ×𝐶 = 𝐴, 𝐶×𝐴 = 𝐵
⃗ . This applies to [𝑨
⃗ 𝑩
⃗⃗ 𝒏
̂ ], where 𝒏
̂ is the unit
That is, 𝐴 × 𝐵
̂. In equation form,
vector normal to the plane formed by the other 2 vectors and 𝐶 = |𝐶 |𝒏
⃗ = |𝑨
⃗ 𝑥𝑩
⃗⃗ |𝑛̂ = |𝐶 |𝒏
̂
𝐴×𝐵
,
̂=
𝒏
⃗ ×𝐶 = 𝐴
𝐵
⃗ 𝑥𝑩
⃗⃗
𝑨
⃗ 𝑥𝑩
⃗⃗ |
|𝑨
,
⃗
𝐶×𝐴=𝐵
Note that for
⃗ ×𝐶 = 𝐴
𝐵
⃗ |𝒏
⃗𝑩
⃗ × |𝑪
̂ = ⃗𝑨
⃗𝑨 = ⃗𝑩
⃗ × |𝑨
⃗ 𝑥 ⃗𝑩
⃗ |𝒏
⃗ 𝑥 ⃗𝑩
⃗ | { ⃗𝑩
⃗ ×𝒏
̂ = |𝑨
̂}
→
⃗ , is therefore
The reciprocal of vector 𝑨
⃗⃗ 𝑥 𝒏
𝑩
̂
⃗𝑨− 1 =
⃗ 𝑥𝑩
⃗⃗ |
|𝑨
⃗𝑨− 1 =
⃗𝑩
⃗ −1 =
⃗⃗ 𝑥 (𝑨
⃗ 𝑥𝑩
⃗⃗ )
𝑩
⃗⃗ 𝑥 ⃗𝑩
⃗|
|𝑨
⃗
𝒏
̂𝑥𝑨
⃗ 𝑥𝑩
⃗⃗ |
|𝑨
⃗𝑩
⃗ −1 =
2
⃗⃗ 𝑥 𝑩
⃗⃗ ) 𝑥 𝑨
⃗
(𝑨
⃗⃗ 𝑥 ⃗𝑩
⃗|
|𝑨
2
Cite your own examples
⃗⃗ = 𝟑𝒙
̂ − 𝟐𝒚
̂ + 𝟑𝒛̂
𝑨
𝑥̂
⃗⃗ 𝑥 𝑩
⃗⃗ = |3
𝑨
1
𝑦̂ 𝑧̂
−2 3| = [(−2)2 − 3(4)]𝑥̂ − [3(2) − 3(1)]𝑦̂ + [3(4) − (−2)]𝑧̂
4 2
2
⃗𝑨
⃗ 𝑥 ⃗𝑩
⃗ = −16𝑥̂ − 3𝑦̂ + 14𝑧̂
Note that,
⃗⃗ = 𝒙
̂ + 𝟒𝒚
̂ + 𝟐𝒛̂
𝑩
⃗ 𝑥 ⃗𝑩
⃗ | = 256 + 9 + 196 = 461
|𝑨
𝑥̂
𝑦̂
𝑧̂
⃗⃗ × 𝑨
⃗ × 𝑩
⃗⃗ = | 1
𝑩
4
2 | = 62𝑥̂ − 46𝑦̂ + 61𝑧̂
−16 −3 14
⃗ is
The reciprocal of vector ⃗𝑨
⃗𝑨− 1 =
⃗𝑩
⃗ 𝑥 (𝑨
⃗⃗ 𝑥 ⃗𝑩
⃗)
⃗ 𝑥𝑩
⃗⃗ |
|𝑨
2
=
62𝑥̂ − 46𝑦̂ + 61𝑧̂
461
To check,
⃗⃗ ∙ ⃗𝑨− 1 = 1 ?
𝑨
⃗𝑨
⃗ = 𝟑𝒙
̂ − 𝟐𝒚
̂ + 𝟑𝒛̂
⃗⃗ ∙ ⃗𝑨− 1 =
𝑨
⃗𝑨
⃗ −1 =
62𝑥̂ − 46𝑦̂ + 61𝑧̂
461
3(62) + 2(46) + 3(61)
= 1
461
Find
⃗𝑩
⃗ −1 =
⃗ 𝑥 ⃗𝑩
⃗ ) 𝑥 ⃗𝑨
(𝑨
⃗⃗ 𝑥 𝑩
⃗⃗ |
|𝑨
2
=?
Reciprocal of Vectors in Three-Dimensional Basis: Reciprocal Lattice
⃗ and ⃗𝑪 , we consider the cyclic permutation [𝐴𝐵
⃗ 𝐶 ] which is
For 3 given 3-D vectors ⃗𝑨, ⃗𝑩
actually expressed commonly as,
⃗ 𝐶] = 𝐴 ∙ 𝐵
⃗ ×𝐶 =𝑉
[𝐴𝐵
⃗ ] = 𝐶∙𝐴×𝐵
⃗ =𝑉
[𝐶 𝐴𝐵
⃗ 𝐶 𝐴] = 𝐵
⃗ ∙𝐶×𝐴 =𝑉
[𝐵
In writing the first scalar triple product,
⃗ ×𝐶
𝐴 ∙ 𝐵
=1
𝑉
→
𝐴 ∙ {
⃗ ×𝐶
𝐵
} =1
𝑉
It is apparent that
⃗ ×𝐶
⃗ ×𝐶
𝐵
𝐵
=
𝑉
⃗ ×𝐶
𝐴 ∙ 𝐵
𝐴−1 =
Writing the other scalar triple product,
⃗ ∙ 𝐶×𝐴
𝐵
=1
𝑉
It is apparent that
⃗ −1 =
𝐵
𝐶×𝐴
𝐶×𝐴
=
𝑉
⃗ ×𝐶
𝐴 ∙ 𝐵
Lastly for the scalar triple product,
⃗
𝐶 ∙ 𝐴×𝐵
=1
𝑉
𝐶 −1 =
Cite your own examples
⃗
⃗
𝐴×𝐵
𝐴×𝐵
=
𝑉
⃗ ×𝐶
𝐴 ∙ 𝐵
Inverse of Matrix
𝑨𝟏𝟏
[𝑨] = [𝑨𝟐𝟏
𝑨𝟑𝟏
𝑨𝟏𝟐
𝑨𝟐𝟐
𝑨𝟑𝟐
𝑨𝟏𝟑
𝟑
𝑨𝟐𝟑 ] = [𝟐
𝑨𝟑𝟑
𝟎
𝟎 𝟐
𝟎 −𝟐]
𝟏 𝟏
Minor of 𝑨𝟏𝟏 :
Is the determinant of the submatrix after eliminating (deleting) the row and column
where A11 is located.
𝑨
𝑨𝟐𝟑
𝒅𝒆𝒕 [ 𝟐𝟐
] = [𝑨𝟐𝟐 𝑨𝟑𝟑 − 𝑨𝟐𝟑 𝑨𝟑𝟐 ] = 𝟐
𝑨𝟑𝟐 𝑨𝟑𝟑
Minor of 𝑨𝟏𝟐 :
Is the determinant of the submatrix after eliminating (deleting) the row and column
where A12 is located.
𝑨
𝑨𝟐𝟑
𝒅𝒆𝒕 [ 𝟐𝟏
] = [𝑨𝟐𝟏 𝑨𝟑𝟑 − 𝑨𝟐𝟑 𝑨𝟑𝟏 ] = 𝟐
𝑨𝟑𝟏 𝑨𝟑𝟑
Minor of 𝑨𝟑𝟐 :
Is the determinant of the submatrix after eliminating (deleting) the row and column
where A32 is located.
𝒅𝒆𝒕 [
𝑨𝟏𝟏
𝑨𝟐𝟏
𝑨𝟏𝟑
] = [𝑨𝟏𝟏 𝑨𝟐𝟑 − 𝑨𝟏𝟑 𝑨𝟐𝟏 ] = −𝟏𝟎
𝑨𝟐𝟑
The Matrix of Minors
[𝑨]𝒎𝒊𝒏𝒐𝒓
𝟐
= [−𝟐
𝟎
𝟐
𝟐
𝟑
𝟑]
−𝟏𝟎 𝟎
The Matrix of Cofactors
Change the sign of the elements found in the odd locations in the matrix
𝑨𝟏𝟏
[𝑨] = [𝑨𝟐𝟏
𝑨𝟑𝟏
𝟐
[𝑨]𝒄𝒐𝒇 = [𝟐
𝟎
𝑨𝟏𝟐
𝑨𝟐𝟐
𝑨𝟑𝟐
𝑨𝟏𝟑
𝑨𝟐𝟑 ]
𝑨𝟑𝟑
−𝟐 𝟐
𝟑 −𝟑]
𝟏𝟎 𝟎
Adjoint of a Matrix
Is found by getting the transpose of a matrix: exchange the row number with the column
number
𝟐
𝟐
[𝑨]𝒂𝒅𝒋 = [−𝟐 𝟑
𝟐 −𝟑
𝟎
𝟏𝟎]
𝟎
Determinant of Matrix [A]
𝟑
|𝑨| = |𝟐
𝟎
𝟎
𝟎
𝟏
𝟐
−𝟐| = 𝟑(𝟐) + 𝟐(𝟐) = 𝟏𝟎
𝟏
Inverse of Matrix [A] :
[𝑨−𝟏 ] =
[𝑨]𝒂𝒅𝒋
|𝑨|
𝟏
𝟏
𝟓
𝟓
−𝟏 𝟑
=
𝟓 𝟏𝟎
𝟏 −𝟑
[ 𝟓 𝟏𝟎
𝟎
𝟏
𝟎]
To check if the calculated matrix is the inverse of the given matrix:
[𝑨][𝑨−𝟏 ]
𝟑
[𝑨] = [𝟐
𝟎
𝟎
𝟎
𝟏
𝟐
−𝟐]
𝟏
𝟏
= 𝚰 = [𝟎
𝟎
𝒂𝒏𝒅
[𝑨][𝑨−𝟏 ]
𝟎
𝟏
𝟎
𝟎
𝟎]
𝟏
𝟏
𝟓
−𝟏
[𝑨−𝟏 ] =
𝟓
𝟏
[𝟓
𝟏 𝟎
= [𝟎 𝟏
𝟎 𝟎
𝟎
𝟎]
𝟏
𝟏
𝟎
𝟓
𝟑
𝟏
𝟏𝟎
−𝟑
𝟎]
𝟏𝟎
Vector Differentiation
For this part of the discussion, it is advantageous for the reader to have recalled a bit of
calculus. Some examples of ordinary differentiation can easily be found,
𝑑
𝑖𝑠 𝑟𝑒𝑎𝑑 𝑎𝑠 "𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 … 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥"
𝑑𝑥
Definition of a Derivative:
𝒅
𝒇(𝒙 + ∆𝒙) − 𝒇(𝒙)
∆𝒚
𝒇(𝒙) = 𝐥𝐢𝐦
= 𝐥𝐢𝐦
∆𝒙 →𝟎
∆𝒙 →𝟎 ∆𝒙
𝒅𝒙
∆𝒙
Y-axis
Y-axis
ΔY
dY
ΔX
dX
X-axis
X-axis
𝑛
𝑑
𝛼𝑥 𝑛 = 𝛼𝑛𝑥 𝑛−1 𝑓𝑜𝑟 { 𝑓(𝑥) = 𝑥
,
𝑑𝑥
𝛼 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑑 𝑥
𝑎 = 𝑎 𝑥 ln 𝑎 𝑓𝑜𝑟 𝑓(𝑥) = 𝑎 𝑥
𝑑𝑥
Derivative of alpha x to the n w/r to x
𝑑 𝑎𝑢
𝑒 = 𝑎𝑒 𝑎𝑢 𝑓𝑜𝑟 𝑓(𝑢) = 𝑒 𝑎𝑢 ,
𝑑𝑢
Derivative of e to the au w/r to u
Derivative of a to the x w/r to x
𝑑
sin 𝜃 = cos 𝜃 𝑓𝑜𝑟 𝑓(𝜃) = sin 𝜃
𝑑𝜃
Derivative of sine x w/r to theta
Assignment:
1.
2.
Simple examples of
𝑑
cos 𝜃 = ?
𝑑𝜃
𝑑
sec 𝜃 = ?
𝑑𝜃
𝑑2 𝑥
𝑎 =?
𝑑𝑥 2
𝑑 𝑛
𝑥 = 𝑛𝑥 𝑛−1
𝑑𝑥
1.
𝑑 3
𝑥 = 3𝑥 2 ,
𝑑𝑥
𝑑 −2
𝑥 = −2𝑥 −3
𝑑𝑥
,
𝑑
1
(2𝑥 2 + 𝑥 3 ) = 4𝑥 + 𝑥 2
𝑑𝑥
3
Note that,
𝑑
1
𝑑
𝑑 1 3
(2𝑥 2 ) +
(2𝑥 2 + 𝑥 3 ) =
( 𝑥 ) = 4𝑥 + 𝑥 2
𝑑𝑥
3
𝑑𝑥
𝑑𝑥 3
Example
𝑑
𝑑 ′
𝑑
(𝑥′ + 𝑥) =
(𝑥 ) +
(𝑥) = 1 + 0 = 1
𝑑𝑥
𝑑𝑥
𝑑𝑥
Example
Example
𝑑
𝑑
𝑑 ′
(𝑥 − 𝑥′) =
(𝑥) −
(𝑥 ) = 1 − 0 = 1
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑
𝑑
𝑑
(𝑥 − 𝑥′) =
(𝑥) −
(𝑥 ′ ) = 0 − 1 = −1
𝑑𝑥′
𝑑𝑥′
𝑑𝑥 ′
𝑑
1
𝑑
(𝑓(𝑥))−1 = − 𝑓(𝑥)
(
)=
𝑑𝑥 𝑓(𝑥)
𝑑𝑥
−2
𝑑
𝑓(𝑥)
𝑑𝑥
𝑑 1
𝑑 −1
𝑑
1
( ) =
𝑥 = −𝑥 −2
𝑥 = − 2
𝑑𝑥 𝑥
𝑑𝑥
𝑑𝑥
𝑥
The derivative of x w/r to x is one,
𝑑
𝑥 = 1,
𝑑𝑥
𝑑
𝑦 = 1,
𝑑𝑦
𝑑
𝑑
𝑥=
𝑥 𝑦 0 = 𝑥(0)𝑦 −1 = 0
𝑑𝑦
𝑑𝑦
Rewriting the general rule,
𝑑
1
𝑑
(𝑓(𝑥 ′ ))−1 = − 𝑓(𝑥 ′ )
(
)
=
𝑑𝑥 ′ 𝑓(𝑥 ′ )
𝑑𝑥 ′
−2
𝑑
𝑓(𝑥 ′ )
𝑑𝑥 ′
For 𝑓(𝑥 ′ ) = 𝑥 − 𝑥 ′
𝑑
1
𝑑
𝑑
(𝑥 − 𝑥 ′ )−1 = −(𝑥 − 𝑥 ′ )−2 ′ (𝑥 − 𝑥 ′ ) = (𝑥 − 𝑥 ′ )−2
(
) =
′
′
′
𝑑𝑥 𝑥 − 𝑥
𝑑𝑥
𝑑𝑥
𝑑
1
1
(
) =
′
′
(𝑥 − 𝑥 ′ )2
𝑑𝑥 𝑥 − 𝑥
Use this to remind that “the derivative of a one-dimensional potential is the negative of
the electric field.
Simple examples of
2.
𝑑 𝑎𝑥
𝑒 = 𝑎𝑒 𝑎𝑥
𝑑𝑥
𝑑 𝑎𝑥
𝑒 = 𝑎𝑒 𝑎𝑥 ,
𝑑𝑥
Simple examples of
𝑑 −2𝑥
𝑒
= −2𝑒 −2𝑥
𝑑𝑥
𝑑
𝑑
[𝑓(𝑥)]𝑛 = 𝑛[𝑓(𝑥)]𝑛−1
𝑓(𝑥)
𝑑𝑥
𝑑𝑥
3.
1
𝑑
3
1
3 −2
3
√2𝑥 2 −
= (2𝑥 2 − ) (4𝑥 + 2 )
𝑑𝑥
𝑥
2
𝑥
𝑥
Simple examples of
4.
𝑑
𝑑
𝑑
[𝑓(𝑥)𝑔(𝑥)] = 𝑔(𝑥) [𝑓(𝑥)] + 𝑓(𝑥) [𝑔(𝑥)]
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑 2 −𝛼𝑥
𝑓(𝑥) = 𝑥 2
[𝑥 𝑒
] 𝑤ℎ𝑒𝑟𝑒 {
𝑔(𝑥) = 𝑒 −𝛼𝑥
𝑑𝑥
𝑑 2 −𝛼𝑥
𝑑 2
𝑑 −𝛼𝑥
[𝑥 𝑒
] = 𝑒 −𝛼𝑥
[𝑥 ] + 𝑥 2
[𝑒
]
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑 2 −𝛼𝑥
[𝑥 𝑒
] = 𝑒 −𝛼𝑥 (2𝑥) + 𝑥 2 [−𝛼(𝑒 −𝛼𝑥 )]
𝑑𝑥
𝑑 2 −𝛼𝑥
[𝑥 𝑒
] = −𝛼𝑥 2 𝑒 −𝛼𝑥 + 2𝑥𝑒 −𝛼𝑥
𝑑𝑥
And some examples on partial differentiation,
𝜕 𝑥
𝑦 = 𝑥𝑦 𝑥−1
𝜕𝑦
𝜕 𝑟𝑠𝑖𝑛 𝜃
𝑒
= 𝑟𝑐𝑜𝑠 𝜃 𝑒 𝑟𝑠𝑖𝑛 𝜃
𝜕𝜃
𝜕 𝑥
𝑦 = 𝑦 𝑥 ln 𝑦
𝜕𝑥
𝜕 2 −𝛼𝑥 2
2
𝑒
= 𝑥 4 𝑒 −𝛼𝑥
2
𝜕𝛼
𝜕2 −
𝑒
𝜕𝑥 2
𝜕2 −
𝑒
𝜕𝑥 2
𝑥2+ 𝑦2
𝑥2+ 𝑦2
𝜕 𝜕 −
[ 𝑒
𝜕𝑥 𝜕𝑥
=
= [
𝑥 2 + 𝑦2
𝜕
(−2𝑥)] 𝑒 −
𝜕𝑥
]=
𝑥 2 + 𝑦2
𝜕
[−2𝑥 𝑒 −
𝜕𝑥
+ (−2𝑥) [
𝑥2+ 𝑦2
𝜕 −
𝑒
𝜕𝑥
]
𝑥2+ 𝑦2
]
Vector Differential Operator
Introduce the operator nabla or del
⃗ = 𝒙
̂
𝛁
𝝏
𝝏𝒙
̂
+ 𝒚
𝝏
𝝏𝒚
+ 𝒛̂
𝝏
𝝏𝒛
Introduce the Laplacian operator
⃗𝛁 𝟐 =
𝝏𝟐
𝝏𝒙𝟐
𝝏𝟐
+
𝝏𝒚𝟐
𝝏𝟐
+
𝝏𝒛𝟐
Example
⃗ 𝑓(𝑥) 𝑤ℎ𝑒𝑟𝑒 𝑓(𝑥) = −
∇
1
𝑥
⃗ 𝑓(𝑥) = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓(𝑥)
∇
⃗ 𝑓(𝑥) = (𝑥̂
∇
𝜕
𝜕𝑥
⃗∇ 𝑓(𝑥) = 𝑥̂
+ 𝑦̂
𝜕
𝜕𝑦
+ 𝑧̂
𝜕
1
) (− )
𝜕𝑧
𝑥
𝜕
1
1
(− ) = 𝑥̂ ( 2 )
𝜕𝑥
𝑥
𝑥
Gradient
The gradient of a scalar function is a vector function
Example
⃗𝛁 𝑓(𝑟) 𝑤ℎ𝑒𝑟𝑒 𝑓(𝑟) =
⃗𝛁 𝑓(𝑟) = (𝑥̂
𝜕
𝜕𝑥
1
1
=
𝑎𝑛𝑑 𝑟 = √𝑥 2 + 𝑦 2 + 𝑧 2
𝑟 √𝑥 2 + 𝑦 2 + 𝑧 2
+ 𝑦̂
𝜕
𝜕𝑦
+ 𝑧̂
𝜕
1
)[
]
𝜕𝑧 √𝑥 2 + 𝑦 2 + 𝑧 2
1
1
1
𝜕[
]
𝜕[
]
𝜕[
]
√𝑥 2 + 𝑦 2 + 𝑧 2
√𝑥 2 + 𝑦 2 + 𝑧 2
√𝑥 2 + 𝑦 2 + 𝑧 2
𝑥̂
+ 𝑦̂
+ 𝑧̂
𝜕𝑥
𝜕𝑦
𝜕𝑧
⃗𝛁
⃗ 𝑓(𝑟) =
(
)
Note that:
For the first term
1
𝜕[
]
2
1
3
𝜕
1
√𝑥 + 𝑦 2 + 𝑧 2
𝑥̂
= 𝑥̂ (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 = − (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 (2𝑥)𝑥̂
𝜕𝑥
𝜕𝑥
2
1
𝜕[
]
√𝑥 2 + 𝑦 2 + 𝑧 2
𝑥̂
= −
𝜕𝑥
𝑥𝑥̂
3
= −
(𝑥 2 + 𝑦 2 + 𝑧 2 )2
𝑥𝑥̂
𝑟3
For the second term
1
𝜕[
]
2
1
3
𝜕 2
1
√𝑥 + 𝑦 2 + 𝑧 2
(𝑥 + 𝑦 2 + 𝑧 2 )−2 = − (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 (2𝑦)𝑦̂
𝑦̂
= 𝑦̂
𝜕𝑦
𝜕𝑦
2
𝜕[
𝑦̂
√𝑥 2
1
]
+ 𝑦2 + 𝑧2
= −
𝜕𝑦
𝑦𝑦̂
(𝑥 2 + 𝑦 2 +
3
𝑧 2 )2
= −
𝑦𝑦̂
𝑟3
For the first term
1
𝜕[
]
2
1
3
𝜕
1
√𝑥 + 𝑦 2 + 𝑧 2
𝑧̂
= 𝑧̂ (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 = − (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 (2𝑧)𝑧̂
𝜕𝑧
𝜕𝑧
2
1
𝜕[
]
√𝑥 2 + 𝑦 2 + 𝑧 2
𝑧̂
= −
𝜕𝑧
𝑧𝑧̂
(𝑥 2 + 𝑦 2 +
The final expression is obtained by adding all 3 terms
3
𝑧 2 )2
= −
𝑧𝑧̂
𝑟3
⃗ 𝑓(𝑟) = −
∇
𝑥𝑥̂ + 𝑦𝑦̂ + 𝑧𝑧̂
𝑟
1
= − 2 = − 2 𝑟̂
3
𝑟
𝑟𝑟
𝑟
Finally,
1
1
⃗𝛁 ( ) = − 𝒓̂
𝑟
𝑟2
Divergence
The divergence of a vector function is a scalar function
⃗⃗ ∙ 𝑭
⃗ = (𝒙
̂
𝛁
𝜕
𝜕𝑥
̂
+ 𝒚
𝜕
𝜕𝑦
⃗ ∙ 𝑭
⃗ =
𝛁
+ 𝒛̂
𝜕
)∙
𝜕𝑧
𝜕𝐹𝑦
𝜕𝐹𝑥
+
𝜕𝑥
𝜕𝑦
Example
⃗ ∙ 𝑭
⃗ =
𝛁
̂
𝐹𝑥 𝒙
+
̂
+ 𝐹𝑦 𝒚
𝜕𝐹𝑧
𝜕𝑧
+ 𝐹𝑧 𝒛̂