Lecture 4 (27)

Multiplication of 2 Vectors by:
BAC

 (read as C equals A cross B)
 CROSS PRODUCT
 DOT PRODUCT
Dot product of vectors A and B is written as:
BAC

. (read as C equals A dot B)
Cross product of vectors A and B is written as:

DOT PRODUCT
Dot product of vectors A and B is written as:
BAC

.
(read as C equals A dot B)

Dot Product
• Laws of Operation
1. Commutative law
A·B = B·A
2. Multiplication by a scalar
a(A·B) = (aA)·B = A·(aB) = (A·B)a
3. Distribution law
A·(B + D) = (A·B) + (A·D)

Angle between 2-Vectors
‘Dot Product of Vectors’
‘Scalar Product of Vectors’
• First define the magnitudes of A and B
• The angle between their tails is θ
A·B = |A||B|cosθ
where 0° ≤ θ ≤ 180°
• The result is a scalar

• Cartesian Vector Formulation
- Dot product of Cartesian unit vectors
Eg: i·i = (1)(1)cos 0° = 1 and
i·j = (1)(1)cos 90° = 0
- Similarly
i·i = 1 j·j = 1 k·k = 1
i·j = 0 i·k = 0 j·k = 0

- Dot product of 2 vectors A and B
A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk)
= AxBx(i·i) + AxBy(i·j) + AxBz(i·k)
+ AyBx(j·i) + AyBy(j·j) + AyBz(j·k)
+ AzBx(k·i) + AzBy(k·j) + AzBz(k·k)
= AxBx + AyBy + AzBz
Note: since the result is a scalar, be careful of not
including any unit vectors in your result.

The angle formed between:
two vectors or intersecting lines are:
θ = cos-1 [(A·B)/(|A||B|)] 0°≤ θ ≤180°
Note 1: if A·B = 0, cos-1 0 = 90°,
so A is perpendicular to B
Note 2: if A·B = 1, cos-1 1 = 0°,
so A is parallel to B

U
V
Example:
The components of two vectors U and V are:
U= 6i – 5j – 3k and V= 4i + 2j + 2k.
(a) What is the value of U.V
(b) What is the angle between U and V when
they placed tail to tail.

x x y y z z
2 2 2 2 2 2
-1
U.V U V U V U V
(6)(4) ( 5)(2) ( 3)(2) 8
U.V
U.V U V cos so cos =
U V
8
cos 0.1952
(6) ( 5) ( 3 ). (4) (2) (2 )
=cos 0.1952 78.7436
  
    
  
  
     
   
r r
r r
r r

Multiplication of 2 Vectors by
Cross Product
BAC


read as C equals A cross B

Direction: Right Hand Rule
λC
Magnitude: C ABsin 

Cross Product
• Laws of Operation
1. Commutative law not applicable
A X B = B X A
2. Multiplication by a scalar
a(AxB) = (aA)xB = Ax(aB) = (AxB)a
3. Distribution law
Ax(B + D) = (AxB) + (AxD)

Cross Product of 2 Vectors
Not Commutative.
ABBA


)AB(BA



- Cross product of Cartesian unit vectors
Eg: i x i = (1)(1) sin 0° = 0 and
i x j = (1)(1) sin 90° = 1
Hence;
i x i = 0 j x j = 0 k x k = 0
i x j = k i x k = -j j x k = i
j x i = -k k x i = j k x j = -i
i
k
j
+

Unit Vectors Cross Product
o
90 sin 1
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi i 0 i j k i k j
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆj i k j j 0 j k i
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆk i j k j i k k 0
    
      
      
      

Consider cross product of vector A and B
A X B = (Ax i + Ay j + Az k) X (Bx i + By j + Bz k)
= AxBx (i X i) + AxBy (i X j) + AxBz (i X k)
+ AyBx (j X i) + AyBy (j X j) + AyBz (j X k)
+ AzBx (k X i) + AzBy (k X j) + AzBz (k X k)
= + (AyBz – AzBy) i – (AxBz - AzBx) j + (AxBy – AyBx) k

zyx
zyx
BBB
AAA
kji
BA


X
In determinant form,
Sarrus’ Rule

Cross Product
Of even more utility, the cross product can be written as
Each component can be determined using 2  2 determinants.

zyx
zyx
zyx
zyx
BBB
AAA
kji
BBB
AAA
kji
BA


X
In determinant form:
+ + +- - -
= + (AyBz – AzBy) i + (AzBx -AxBz) j + (AxBy – AyBx) k
Sarrus’ Rule

U
V
Example:
The components of two vectors U and V are:
U= 6i – 5j – 3k and V= 4i + 2j + 2k.
(a) Determine the parallelface area formed by
these vectors? i.e find [U x V]?
(b) determine the smallest angle formed
between these vectors?

224
356
kji



VXU
Determinant form:
(a)
= (-10+6) i + (-12-12) j + (12+20) k = - 4 i -24 j + 32 k
= 1616 = 40.200 𝑢𝑛𝑖𝑡2
U = 70 V = 24
θ = sin-1 [(A X B)/(|A||B|)]
θ = sin-1 [( 1616)/( 70 24)]=78.7448°
(b)

Example:
A=(4,0,-1), B=(1,1,2) and C=(-2,1,2) are the
corners of the tringle, determine its area?
z
y
x
A
C
B

AB = (1-4) i + (1-0) j + (2-(-1)) k = -3i+1j+3k
A=(4,0,-1), B=(1,1,2), C=(-2,1,2)
AC = (-2-4) i + (1-0) j + (2-(-1)) k = -6i+1j+3k
AB X AC = area of the rectangle
So (AB X AC) / 2 area of the triangle
316
313
kji



ABXAC
= (3-3) i + (9-18) j + (-3+6) k = -9 j + 3 k = 90 = 9.487 𝑢𝑛𝑖𝑡2
Area of the triangle = 9.487/2= 4.744 𝑢𝑛𝑖𝑡2

Example:
Calculate the volume of the parallel surface
bounded by lines OP, OQ and OT, where O= (0,0,0),
P= (4,0,-1), Q= (1,1,2) and T= (-2,1,2)?
Note: To determine the volume; use the mixed triple
vector product approach [(OP X OQ ) . OT].
O P
Q
T

OP = 4 i -1 k
O=(0,0,0) P=(4,0,-1), Q=(1,1,2), T=(-2,1,2)
OT = -2 i + 1 j + 2 k
OP X OQ = area of the rectangle
211
104
kji


OPXOQ
= (0-(-1) i + (-8-1) j + (4-0) k = 1i -9 j + 4 k 𝑢𝑛𝑖𝑡2
OQ = 1 i +1j + 2 k
So (OP X OQ ) . OT=
(1i -9 j + 4 k) . (-2i +1j+2k)= -2-9 +8 =3 𝑢𝑛𝑖𝑡3

Chapter Summary
• Dot product between two vectors A and B
(vectors expressed as Cartesian form)
A · B = AxBx + AyBy + AzBz
• Angle between the tails of two vectors
θ = cos-1 [(A·B)/(IAI IBI)]
• The projected component of A onto an axis defined
by its unit vector λ
A = A cos θ = A·λ
Dot Product
• Dot product between two vectors A and B
A·B = AB cos θ

Lecture 4 (27)

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