1. The document discusses vectors in two and three dimensions, including the dot product, properties of the dot product, calculating the dot product, and the dot product theorem. 2. It also covers 2D vectors including unit vectors, addition and subtraction of vectors, and multiplication of a vector by a scalar. 3. Examples are provided to demonstrate calculating magnitudes of vectors, algebraic operations on vectors, and expressing vectors in terms of i and j.

1. Vectors in Two and Three
Dimensions-‖
Business Mathematics І
Ms. Inomi Anjala Gayashani
Department of Computer Science
Faculty of Computer Science and Engineering

2. The Dot Product
Definition of the Dot Product
If u = 𝑎1, 𝑎2 𝑎𝑛𝑑 𝐯 = 𝑏1, 𝑏2 are vectors, then their dot
product, denoted by u .v, is defined by
u .v= 𝑎1 𝑏1+ 𝑎2 𝑏2
The dot product is not a vector; it is a real number, or scalar.

3. Calculating Dot Product
Example 1
If u = 3, −2 𝑎𝑛𝑑 v = 4,5 𝑡ℎ𝑒𝑛
u .v = (3)(4)+(-2)(5) = 2
Example 2
If u = 2i+j and v= 5i-6j, then
u .v = (2)(5)+(1)(-6) = 4

4. Properties of the Dot Product
1. u.v = v.u
2. (cu).v = c(u.v)=u.(cv)
3. (u+v).w= u.w + v.w
4. |𝒖|𝟐
= 𝒖. 𝒖

5. The Dot Product Theorem
If 𝜃 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑛𝑜𝑛𝑧𝑒𝑟𝑜 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
u and v , then
u.v= 𝒖 𝒗 cos 𝜃

6. Angle between two vectors.
If 𝜃 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑛𝑜𝑛𝑧𝑒𝑟𝑜 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝒖 𝑎𝑛𝑑 𝒗, 𝑡ℎ𝑒𝑛

7. Vectors in Two Dimensions(2-D Vectors)
Unit Vectors
A vector of length 1.
i= 1,0 J= 0,1

8. Vectors in Two Dimensions(2-D Vectors)
Addition of Vectors
𝐴𝐵+ 𝐵𝐶 = 𝐴𝐶

9. Vectors in Two Dimensions(2-D Vectors)
Subtraction of Vectors
𝑢 − 𝑣 = 𝑢 − 𝑣

10. Vectors in Two Dimensions(2-D Vectors)
Multiplication of a vector by a scalar
If c is a real number and v is a vector,
If c>0,Then c𝑣 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑠 𝒗.
If c<0,Then c𝑣 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝒗.
If c=0,Then c𝑣 = 0.

11. Vectors in the coordinate planes
v = 𝑎1, 𝑎2
𝑎1 - horizontal component of v
𝑎2 - vertical component of v

12. Component form of a vector
If a vector v is represented in the plane with
initial point P(𝑥1, 𝑦1 ) and terminal point
Q(𝑥2, 𝑦2), then
v = 𝑥2 − 𝑥1, 𝑦2 − 𝑦1

13. Exercise 1

14. Exercise 2
(a)Find the component form of the vector u with initial point (-2,5)and
terminal point (3,7)
The desired vector is
u = 3 − (−2), 7 − 5 = 5,2

15. Exercise 2
(b) If the vector v = 𝟑, 𝟕 is sketched with initial point (2,4)what is its
terminal point?
Let the terminal point of v be (x, y). Then
v = (𝑥 − 2), (𝑦 − 4)
𝟑, 𝟕 = (𝑥 − 2), (𝑦 − 4)
𝑥, 𝑦 = (3 + 2), (7 + 4)
𝑥, 𝑦 = (5), (11)

16. Exercise 2
(c) Sketch representations of the vector w= 2,3 with initial points
at (0,0) , (2,2) , (-2,-1) and (1,4)

17. Exercise 3

18. Magnitude of a vector
The magnitude or the length of a vector v = 𝑎1, 𝑎2 𝑖𝑠
(𝑎1
2
+ 𝑎2
2
)

19. Exercise 4
(a)Find the magnitude of each vector.
u = 2, −3
v = 5,0
w =
3
5
,
4
5
𝑢 = (22 + (−3)2) = 13
𝑣 = (52 + (0)2) = 25 = 5
𝑤 = ((
3
5
)2 + ((
4
5
)2) =
9
25
+
16
25
= 1

20. Algebraic Operations on Vectors
If u = 𝑎1, 𝑎2 𝑎𝑛𝑑 𝒗 = 𝑏1, 𝑏2 , 𝑡ℎ𝑒𝑛
𝒖 + 𝒗 = 𝑎1 + 𝑏1, 𝑎2 + 𝑏2
𝒖 − 𝒗 = 𝑎1 − 𝑏1, 𝑎2 − 𝑏2
𝑐𝒖= 𝑐𝑎1, 𝑐𝑎2 , c € R

21. Exercise 5
If u= 2, −3 𝑎𝑛𝑑 𝒗 = −1,2 , 𝑓𝑖𝑛𝑑 𝒖 + 𝒗, 𝒖 − 𝒗, 𝟐𝒖 − 𝟑𝒗 𝑎𝑛𝑑 𝟐𝒖 + 𝟑𝒗
By the definitions of vector operations we have,
𝒖 + 𝒗= 2, −3 + −1,2 = 1, −1
𝒖 − 𝒗= 2, −3 - −1,2 = 3, −5
𝟐𝒖 − 𝟑𝒗=2( 2, −3 )-3( −1,2 ) = 7, −12
𝟐𝒖 + 𝟑𝒗=2( 2, −3 )+3( −1,2 ) = 1,0

22. Properties of Vectors
Vector addition
𝒖 + 𝒗 = 𝒗 + 𝒖
𝒖 + 𝒗 + 𝒘 = 𝒖 + 𝒗 + 𝒘
u+0 = u
u+(-u) = 0

23. Properties of Vectors
Length of a Vector
𝒄𝒖 = 𝒄 𝒖

24. Properties of Vectors
Multiplication by a scalar
c(𝒖 + 𝒗) = c𝒖 + 𝑐𝒗
𝑐 + 𝑑 𝒖 = 𝑐𝒖 + 𝑑𝒖
(cd)u = c(du) = d(cu)
1u=u
0u=0
c0=0

25. Vectors in terms of i and j
The vector v = a1, a2 can be expressed in terms of 𝐢 and 𝐣
v= a1, a2 = a1𝐢 + a2𝐣

26. Exercise 6

27. Exercise 7
(a)Write the vector 𝒖 = 5, −8 in terms of i and j.
(b)If u = 3i + 2j and v = -i+6j , write 2u+5v in terms of i and j
(a)𝒖= 5𝒊 + −8 𝒋 = 5𝒊 − 8𝒋
(b) 2u+5v = 2(3i + 2j )+5(-i+6j )
= 6i + 4j - 5i+30j
= i + 34j

28. Horizontal and vertical components of a vector
Let v be a vector with magnitude 𝑣 𝑎𝑛𝑑 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 θ.
𝑇ℎ𝑒𝑛 𝑣= a1, a2 = a1𝐢 + a2𝐣 where
a1 = 𝑣 cos 𝜃 𝑎𝑛𝑑 a2 = 𝑣 sin 𝜃
Thus we can express v as
v= 𝑣 cos 𝜃 𝐢 + 𝑣 sin 𝜃 𝐣

29. Exercise 8
(a) A vector v has length 8 and direction π/3. Find the horizontal and
vertical
components, and write v in terms of i and j.
(b) Find the direction of the vector u = - 3 i + j.
(a) We have v= 𝑎, 𝑏 , where the components are given by
a= 8 cos
𝜋
3
= 4 and b = 8 sin
𝜋
3
= 4 3
Thus v=(4, 4 3) = 4 i + 4 3 j

30. Exercise 8
(b) Find the direction of the vector u = - 3 i + j.
(b) tan 𝜃 =
1
− 3
=
− 3
3
Thus the reference angle for θ is π/6.
Since the terminal point of the vector u
is in Quadrant II,it follows that θ =5π/6.

31. Exercise 9

32. Exercise 10

34. Using Vectors to model Velocity and Force
An airplane heads due north at 300 mi/h. It experiences a 40 mi/h crosswind flowing in the direction N
300
E, as shown below.
(1). Express the velocity v of the airplane relative to the air and the velocity u of the wind, in component form.
(2) Find the true velocity of the airplane as a vector.
(3) Find the true speed and direction of the airplane.
The True speed and Direction of an airplane

35. Using Vectors to model Velocity and Force
(1)The velocity of the airplane relative to the air is v = 0 i+300 j= 300j. By the formulas for
the components of a vector we find that the velocity of the wind is
u= (40 cos 60)i+(40 sin 60)j
= 20i+20 𝟑𝒋
≈ 𝟐𝟎𝒊 + 𝟑𝟒. 𝟔𝟒𝒋
(2) The true velocity of the airplane is given by the vector w=u+v
w=u+v=(20i+20 𝟑𝒋) + 𝟑𝟎𝟎𝒋
=20i+(20 𝟑 + 𝟑𝟎𝟎)𝒋
≈ 𝟐𝟎𝒊 + 𝟑𝟑𝟒. 𝟔𝟒𝒋
• The true velocity is the velocity relative to the ground.
The True speed and Direction of an airplane

36. Using Vectors to model Velocity and Force
The true speed of the airplane is given by the magnitude of w:
𝒘 ≈ ((𝟐𝟎)𝟐 + 𝟑𝟑𝟒. 𝟔𝟒 𝟐) ) ≈ 𝟑𝟑𝟓. 𝟐𝒎𝒊/𝒉
The direction of the airplane is the direction 𝜃 of the vector w.
The angle 𝜃 has the property that tan 𝜃 ≈334.64/20 =16.732, so 𝜃 ≈ 86.6
Thus the airplane is heading in the direction N 3.40
E.
The True speed and Direction of an
airplane

37. Using Vectors to model Velocity and Force
A woman launches a boat from one shore of a straight river and wants to land
at the point directly on the opposite shore. If the speed of the boat (relative to
the water) is 10 mi/h and the river is flowing east at the rate of 5 mi/h, in what
direction should she head the boat in order to arrive at the desired landing
point?
Calculating a Heading

38. Using Vectors to model Velocity and Force
We choose a coordinate system with the origin at the initial position of the boat as shown below. Let u and v
represent the velocities of the river and the boat, respectively. Clearly, u =5 i, and since the speed of the boat
is 10 mi/h, we have 𝒗 =10, so
v=(10cos 𝜃)𝑖 + (10 sin 𝜃)𝑗
The true course of the boat is given by the vector w = u + v.
We have
w = u + v=5i+ (10cos 𝜃)𝑖 + (10 sin 𝜃)𝑗
Since the woman wants to land at a point directly across the river, her direction
should have horizontal component 0.
In other words, she should choose u in such a way that
5+ 10 cos 𝜃 = 0
cos 𝜃 =
−1
2
Calculating a Heading