2. Chapter 10: Applications of Trigonometry and Vectors
10.1 The Law of Sines
10.2 The Law of Cosines and Area Formulas
10.3 Vectors and Their Applications
10.4 Trigonometric (Polar) Form of Complex Numbers
10.5 Powers and Roots of Complex Numbers
10.6 Polar Equations and Graphs
10.7 More Parametric Equations

3. 10.3 Vectors and Their Applications
Basic Terminology
A scalar is a magnitude
E.g. 45 pounds
A vector quantity is a
magnitude with direction
E.g. 50 mph east
Vectors
Represented in boldface type or with an arrow over the letters
E.g. OP, and OP represent the vector OP

4. 10.3 Basic Terminology Vector OP First letter represents the
initial point
Second letter represents
the terminal point
Vector OP and vector PO
are not the same vectors.
They have the same magnitude,
but in opposite directions.

5. 10.3 Basic Terminology
Two vectors are equal if and only if they both have the same magnitude and direction.
The sum of two vectors A and B
Place the initial point of B at the terminal point of A. The vector with the same initial point as A and the same terminal point as B is the sum A + B.
The parallelogram rule: place the vectors so that their initial points coincide. Then complete the parallelogram as shown in the figure.
Figure 22 pg 10-47

6. 10.3 Basic Terminology
Vectors are commutative: A + B = B + A, and the sum A + B is called the resultant of A and B.
For every vector v there is a vector –v such that
v + (–v) = 0, the zero vector.
The scalar product of a real number (scalar) k and a vector u is the vector k·u, which has magnitude |k| times the magnitude of u. If k < 0, then k·u is in the opposite direction as u.

7. 10.3 Algebraic Interpretation of Vectors
A vector with its initial point at the origin is called a position vector.
A position vector u with endpoint (a,b) is written as u = a, b, where a is called the horizontal component and b is called the vertical component of u.

8. 10.3 Algebraic Interpretation of Vectors
The magnitude (length) of vector u = a, b is given by
The direction angle  satisfies tan  = where a  0.
Magnitude and Direction Angle of a Vector a, b

9. 10.3 Finding the Magnitude and Direction Angle
Example Find the magnitude and direction angle for
u = 3, –2.
Analytic Solution
Vector u has a positive x-component and a negative y-
component, placing u in quadrant IV. Adding 360°
yields the direction of

10. 10.3 Finding Horizontal and Vertical Components
The horizontal and vertical components, respectively, of a vector u having magnitude |u| and direction angle  are given by
a = |u| cos  and b = |u| sin .
That is, u = a, b =  |u| cos  , |u| sin  .

11. 10.3 Finding Horizontal and Vertical Components
Example From the figure,
the horizontal component is
a = 25.0 cos 41.7°  18.7.
The vertical component is
b = 25.0 sin 41.7°  16.6.

12. 10.3 Finding the Magnitude of a Resultant
Example Two forces of 15 and 22 newtons act on a point in
the plane. If the angle between the forces is 100°, find the
magnitude of the resultant force.
Solution From the figure, the angles
of the parallelogram adjacent to angle P
each measure 80º, since they are
supplementary to angle P. The resultant
force divides the parallelogram into two
triangles. Use the law of cosines on
either triangle.
|v|2 = –2(15)(22) cos 80º  |v|  24 newtons

13. 10.3 Some Properties of a Parallelogram
Properties of Parallelograms
A parallelogram is a quadrilateral whose opposite sides are parallel.
The opposite sides and opposite angles of a parallelogram are equal, and adjacent angles are supplementary.
The diagonals of a parallelogram bisect each other but do not necessarily bisect the angles.

14. 10.3 Operations with Vectors

15. 10.3 Operations with Vectors
Vector Operations
For any real numbers a, b, c, d, and k,

16. 10.3 Performing Vector Operations
Example Let u = –2, 1 and v = 4, 3. Find each of
the following: (a) u + v, (b) –2u, (c) 4u – 3v.
Solution
u + v = –2, 1 + 4, 3 = –2 + 4, 1 + 3 = 2, 4
–2u = –2 · –2, 1 = –2(–2), –2(1) = 4, –2
4u – 3v = 4 · –2, 1 – 3 · 4, 3
= 4(–2) – 3(4), 4(1) – 3(3)
= –8 – 12, 4 –9
= –20, –5

17. A unit vector is a vector that has magnitude 1.
10.3 The Unit Vector
A unit vector is a vector that has magnitude 1.
Two very useful unit vectors
are defined as
i = 1, 0 and j = 0, 1.
i, j Forms for Unit Vectors
If v = a, b, then v = ai + bj.

18. 10.3 Dot Product Dot Product
The dot product of two vectors u = a, b and
v = c, d  is denoted u · v, read “u dot v,” and given by
u · v = ac + bd.
Example Find the dot product 2, 3 · 4, –1.
Solution 2, 3 · 4, –1 = 2(4) + 3(–1)
= 8 – 3
= 5

19. Properties of the Dot Product
For all vectors u, v, and w and real numbers k,
u · v = v · u
u · (v + w) = u · v + u · w
(u + v) · w = u · w + v · w
(ku) · v = k(u · v) = u · (kv)
0 · u = 0
u · u = |u|2.

20. Geometric Interpretation of the Dot Product
If  is the angle between the two nonzero vectors
u and v, where 0º    180º, then

21. 10.3 Finding the Angle Between Two Vectors
Example Find the angle between the two vectors
u = 3, 4 and v = 2, 1.
Solution
Therefore,   26.57º.
NOTE If a · b = 0, then cos  = 0 and  = 90º. Thus a and b
are perpendicular or orthogonal vectors.

22. 10.3 Applying Vectors to a Navigation Problem
Example A plane with an airspeed of 192 mph is headed on a
bearing of 121º. A north wind is blowing (from north
to south) at 15.9 mph. Find the groundspeed and the
actual bearing of the plane.
Solution Let |x| be groundspeed. We
must find angle . Angle AOC = 121º.
Find |x| using the law of cosines .

23. 10.3 Finding a Required Force
Example Find the force required to pull a wagon weighing
50 lbs up a ramp inclined at 20º to the horizontal.
(Assume no friction.)
Solution The vertical 50 lb force BA
represents the force of gravity. BA is the
sum of the vectors BC and –AC. Vector
BC represents the force with which the
weight pushes against the ramp. Vector
BF represents the force required to pull
the weight up the ramp. Since BF and AC are equal, | AC | gives
the magnitude of the required force.