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

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

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.

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

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.

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.

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

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

10.3 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  .

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.

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 = 152 + 222 –2(15)(22) cos 80º  |v|  24 newtons

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.

10.3 Operations with Vectors

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

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

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.

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

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.

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

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.

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 .

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.