Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Line (geometry) wikipedia , lookup
History of trigonometry wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Elementary mathematics wikipedia , lookup
Vector space wikipedia , lookup
Classical Hamiltonian quaternions wikipedia , lookup
Bra–ket notation wikipedia , lookup
The Law of Sines Law of Sines If A, B, and C are the measures of the angles of a triangle, and a, b, and c are the lengths of the sides opposite these angles, then sin A sin B sin C a b c The ratio of the length of the side of any triangle to the sine of the angle opposite that side is the same for all three sides of the triangle. Text Example • Solve triangle ABC if A 50º, C 33.5º, and b 76. Solution We begin by drawing a picture of triangle ABC and labeling it with the given information. The figure shows the triangle that we must solve. We begin by finding B. C A + B + C 180º b 76 a 83.5º + B 180º B 96.5º 50º A 50º + B + 33.5º 180º 33.5º c B The sum of the measurements of a triangle’s interior angles is 180º. A = 50º and C = 33.5º. Add. Subtract 83.5º from both sides. Text Example cont. Solve triangle ABC if A 50º, C 33.5º, and b 76. Solution Keep in mind that we must be given one of the three ratios to apply the Law of Sines. In this example, we are given that b 76 and we found that B 96.5º. Thus, we use the ratio b/sin B, or 76/sin96.5º, to find the other two sides. Use the Law of Sines to find a and c. C 33.5º b 76 a 50º A c B Find a: a b sin A sin B a 76 sin 50 sin 96.5 76sin 50 a 59 sin 96.5 Find c: This is the known ratio. The solution is B 96.5º, a 59, and c 42. c b sin C sin B c 76 sin 33.5 sin 96.5 76sin 33.5 c 42 sin 96.5 The Ambiguous Case (SSA) Consider a triangle in which a, b, and A are given. This information may result in: No Triangle a is less than h and not long enough to form a triangle. One Right Triangle a b h = b sin A A b a h = b sin A A Two Triangles a is greater than h and a is less than b. Two distinct triangles are formed. A a = h and is just the right length to form a right triangle. b a One Triangle a h = b sin A a is greater than h and a is greater than b. One triangle is formed. b A a Example • Solve the triangle shown with A=36º, B=88º and c=29 feet. b a c Solution: A=36º, B=88º so 180-88-36=56º C=56º sin 36 sin 88 sin 56 a b 29 .59 1 .83 a b 29 1 .83 so .83b 29 b 34.94 b 29 .59 .83 so .83a 17.11 a 20.61 a 29 Area of An Oblique Triangle • The area of a triangle equals one-half the product of the lengths of two sides times the sine of their included angle. In the following figure, this wording can be expressed by the formulas 1 1 1 Area bc sin A ab sin C ac sin B 2 2 2 C b A a h D B c Text Example • Find the area of a triangle having two sides of lengths 24 meters and 10 meters and an included angle of 62º. Solution The triangle is shown in the following figure. Its area is half the product of the lengths of the two sides times the sine of the included angle. Area 1/2 (24)(10)(sin 62º) 106 the area of the triangle is approximately 106 square meters. C b = 10 meters 62º A c = 24 meters B Example • Find the area of a triangle having two sides of lengths 12 ft. and 20 ft. and an included angle of 57º. Solution: 1 1 Area bc sin A (12)( 20) sin 57 2 2 120 * .84 100.8sq. ft. The Law of Cosines The Law of Cosines If A, B, mid C are the measures of the angles of a triangle, and a, b, and c are the lengths of the sides opposite these angles, then a2 b2 + c2 - 2bc cos A b2 a2 + c2 - 2ac cos B c2 a2 + b2 - 2ab cos C. The square of a side of a triangle equals the sum of the squares of the other two sides minus twice their product times the cosine of their included angle. Solving an SAS Triangle 1. Use the Law of Cosines to find the side opposite the given angle. 2. Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute. 3. Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180º. Text Example C b = 20 • Solve the triangle shown with A = 60º, b = 20, and c = 30. A a 60º c =30 B Solution We are given two sides and an included angle. Therefore, we apply the three-step procedure for solving a SAS triangle. Step l Use the Law of Cosines to find the side opposite the given angle. Thus, we will find a. a2 b2 + c2 - 2bc cos A a2 202 + 302 - 2(20)(30) cos 60º 400 + 900 - 1200(0.5) 700 a 700 26 Apply the Law of Cosines to find a. b = 20, c = 30, and A = 60°. Perform the indicated operations. Take the square root of both sides and solve for a. Text Example cont. C b = 20 a Solve the triangle shown with A = 60º, b = 20, and c = 30. A 60º c = 30 Solution Step 2 Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute. The shorter of the two given sides is b 20. Thus, we will find acute angle B. b a sin B sin A 20 700 sin B sin 60 700 sin B 20sin 60 sin B B B 41 41 20sin 60 0.6547 700 Apply the Law of Sines. We are given b = 20 and A = 60°. Use the exact value of a, 700, from step 1. Cross multiply. Divide by square root of 700 and solve for sin B. Find sin-10.6547 using a calculator. B Text Example cont. C Solve the triangle shown with A = 60º, b = 20, and c = 30. b = 20 A a 60º c = 30 Solution We are given two sides and an included angle. Therefore, we apply the three-step procedure for solving a SAS triangle. Step 3 Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180º. C 180º - A - B 180º - 60º - 41º 79º The solution is a 26, B 41º, and C 79º. B Solving an SSS Triangle 1. Use the Law of Cosines to find the angle opposite the longest side. 2. Use the Law of Sines to find either of the two remaining acute angles. 3. Find the third angle. Subtract the measures of the angles found in steps 1 and 2 from 180º. Text Example • Two airplanes leave an airport at the same time on different runways. One flies at a bearing of N66ºW at 325 miles per hour. The other airplane flies at a bearing of S26ºW at 300 miles per hour. How far apart will the airplanes be after two hours? Solution After two hours. the plane flying at 325 miles per hour travels 325 · 2 miles, or 650 miles. Similarly, the plane flying at 300 miles per hour travels 600 miles. The situation is illustrated in the figure. Let b the distance between the planes after two hours. We can use a north-south line to find angle B in triangle ABC. Thus, B 180º - 66º - 26º 88º. We now have a 650, c 600, and B 88º. Text Example cont. Two airplanes leave an airport at the same time on different runways. One flies at a bearing of N66ºW at 325 miles per hour. The other airplane flies at a bearing of S26ºW at 300 miles per hour. How far apart will the airplanes be after two hours? Solution We use the Law of Cosines to find b in this SAS situation. b2 a2 + c2 - 2ac cos B Apply the Law of Cosines. b2 6502 + 6002 - 2(650)(600) cos 88º Substitute: a= 650, c =600, and B= 88°. 755,278 Use a calculator. b 869 Take the square root and solve for b. After two hours, the planes are approximately 869 miles apart. Heron’s Formula The area of a triangle with sides a, b, and c is Area s ( s - a )( s - b)( s - c) 1 s (a + b + c) 2 Example • Use Heron’s formula to find the area of the given triangle: a=10m, b=8m, c=4m Solution: 1 s (a + b + c) 2 1 s (10 + 8 + 4) 2 1 s (22) 11 2 Area s( s - a)( s - b)( s - c) 11(11 - 10)(11 - 8)(11 - 4) 11(1)(3)(7) 231 sq.m. Polar Coordinates The Sign of r and a Point’s Location in Polar Coordinates. • The point P (r, ) is located |r| units from the pole. If r > 0, the point lies on the terminal side of . If r < 0 the point lies along the ray opposite the terminal side of . If r 0. the point lies at the pole, regardless of the value of . Text Example • Plot the points with the following polar coordinates: a. (2, 135°) Solution a. To plot the point (r, ) (2, 135°), begin with the 135° angle. Because 135° is a positive angle, draw 135° counterclockwise from the polar axis. Now consider r 2. Because r > 0, plot the point by going out two units on the terminal side of . Figure (a) shows the point. (2, 135°) (2, 135°) 135º Multiple Representation of Points If n is any integer, the point (r, ) can be represented as (r, ) (r, + 2n) or (r, ) (-r, + r + 2n ) Relations between Polar and Rectangular Coordinates x r cos y r sin x 2 + y 2 r2 y tan /x P = (r, ) = (x, y) Converting a Point from Rectangular to Polar Coordinates (r > 0 and 0 < < 2) 1. Plot the point (x, y). 2. Find r by computing the distance from the origin to (x, y). 3. Find using tan = y/x with lying in the same quadrant as (x, y). Example • Find the polar coordinates of a point whose rectangular coordinates are (2, 4) Solution: r x 2 + y 2 2 2 + 4 2 20 2 5 tan y 4 2 x 2 1.1 (2 5 , 1.1) Example • Convert 2x-y=1 to a polar equation. Solution: 2x - y 1 2r cos - r sin 1 r (2 cos - sin ) 1 1 r 2 cos - sin Graphs of Polar Equations Using Polar Grids to Graph Polar Equations Recall that a polar equation is an equation whose variables are r and è. The graph of a polar equation is the set of all points whose polar coordinates satisfy the equation. We use polar grids like the one shown to graph polar equations. The grid consists of circles with centers at the pole. This polar grid shows five such circles. A polar grid also shows lines passing through the pole, In this grid, each fine represents an angle for which we know the exact values of the trigonometric functions. 2 3 3 4 5 6 2 4 ˝ 6 7 6 5 4 4 3 3 2 3 2 4 0 11 6 7 5 4 3 Text Example • Graph the polar equation r 4 cos with in radians. Solution We construct a partial table of coordinates using multiples of 6. Then we plot the points and join them with a smooth curve, as shown. r 4 cos (r, ) 0 4 cos 0 = 4 • 1 = 4 (4, 0) /6 4 cos /6 = 4 •3/2=2 3=3.5 (3.5, /6) /3 4 cos /3 = 4 • 1/2 = 4 (2, /3) /2 4 cos /2 = 4 • 0 = 0 (0, /2) 2 /3 4 cos2 /3 = 4(- 1/2) = -2 (-2, 2 /3) 5/6 4 cos5 /6 = 4(- 3/2)=-2 3=-3.5 (-3.5,5 /6) ˝ 4 cos ˝ = 4(-1) = -4 (-4, ) (2, /3) (0, /2) 2 3 3 4 5 6 2 3 ˝ 4 6 (3.5, /6) 0 ˝ 2 4 (4, 0) or (-4, ) 7 11 6 6 5 7 (-2, 2 /3) 4 4 5 4 3 3 3 2 (-3.5, 5 /6) Circles in Polar Coordinates The graphs of r = a cos and r = a sin Are circles. r = a cos r = a sin /2 /2 a ˝ 0 ˝ 0 a 3 /2 3 /2 Text Example • Check for symmetry and then graph the polar equation: r 1 - cos . Solution We apply each of the tests for symmetry. Polar Axis: Replace by - in r 1 - cos : r 1 - cos (- ) Replace by - in r 1 - cos . r 1 - cos The cosine function is even: cos (- ) cos . Because the polar equation does not change when is replaced by - , the graph is symmetric with respect to the polar axis. Text Example cont. Solution The Line 2: Replace (r, ) by (-r, - ) in r 1 - cos : -r 1 - cos(-) Replace r by -r and by – in -r 1 - cos(- ). -r 1 – cos cos(- ) cos . r cos - 1 Multiply both sides by -1. Because the polar equation r 1 - cos changes to r cos - 1 when (r, ) is replaced by (-r, - ), the equation fails this symmetry test. The graph may of may not be symmetric with respect to the line 2. The Pole: Replace r by -r in r 1 - cos : -r 1 – cos Replace r by –r. r cos - 1 Multiply both sides by -1. Because the polar equation r 1 - cos changes to r cos - 1 when r is replaced by -r, the equation fails this symmetry test. The graph may or may not be symmetric with respect to the pole. Text Example cont. Solution Now we are ready to graph r 1 - cos . Because the period of the cosine function is 2r, we need not consider values of beyond 2. Recall that we discovered the graph of the equation r 1 - cos has symmetry with respect to the polar axis. Because the graph has symmetry, we may be able to obtain a complete graph without plotting points generated by values of from 0 to 2. Let's start by finding the values of r for values of from 0 to . r 0 0 /6 /3 2 2 /3 0.13 0.50 1.00 1.50 5 /6 1.87 2 The values for r and are in the table. Examine the graph. Keep in mind that the graph must be symmetric with respect to the polar axis. 2 3 3 4 ˝ 2 3 4 5 6 6 7 6 5 4 4 3 1 0 2 11 6 3 2 7 5 4 3 Text Example cont. Solution Thus, if we reflect the graph from the last slide about the polar axis, we will obtain a complete graph of r 1 - cos , shown below. 2 3 3 4 2 3 4 5 6 6 7 6 5 4 4 3 1 0 2 11 6 3 2 7 5 4 3 Limacons The graphs of r a + b sin , r a - b sin , r a + b cos , r a - b cos , a > 0, b > 0 are called limacons. The ratio ab determines a limacon's shape. Inner loop if ab < 1 Heart shaped if ab 1 Dimpled with no inner loop if 1< ab < 2 and called cardiods 2 3 2 2 0 3 2 loop if ab 2. 2 0 No dimple and no inner 2 0 3 2 0 3 2 Example • Graph the polar equation y= 2+3cos Example • Graph the polar equation y= 2+3cos Solution: Rose Curves The graphs of r a sin n and r a cos n, a does not equal 0, are called rose curves. If n is even, the rose has 2n petals. If n is odd, the rose has n petals. r a sin 2 r a cos 3 r a cos 4 r a sin 5 Rose curve with 4 petals Rose curve with 3 petals Rose curve with 8 petals Rose curve with 5 petals 2 2 2 2 n=4 a a n=3 0 0 a 0 0 a n=2 3 2 3 2 3 2 a 3 2 n=5 Example • Graph the polar equation y=3sin2 Example • Graph the polar equation y=3sin2 Solution: Lemniscates • The graphs of r2 = a2 sin 2 and r2 = a2 cos 2 are called lemniscates Lemniscate: r2 = a2 cos 2 Complex Numbers in Polar Form; DeMoivre’s Theorem The Complex Plane We know that a real number can be represented as a point on a number line. By contrast, a complex number z = a + bi is represented as a point (a, b) in a coordinate plane, shown below. The horizontal axis of the coordinate plane is called the real axis. The vertical axis is called the imaginary axis. The coordinate system is called the complex plane. Every complex number corresponds to a point in the complex plane and every point in the complex plane corresponds to a complex number. Imaginary axis b z = a + bi a Real axis Text Example Plot in the complex plane: a. z = 3 + 4i b. z = -1 – 2i Solution • • c. z = -3 We plot the complex number z = 3 + 4i the same way we plot (3, 4) in the rectangular coordinate system. We move three units to the right on the real axis and four units up parallel to the imaginary axis. The complex number z = -1 – 2i corresponds to the point (-1, -2) in the rectangular coordinate system. Plot the complex number by moving one unit to the left on the real axis and two units down parallel to the imaginary axis. d. z = -4i 5 4 3 2 1 -5 -4 -3 -2 z = -1 – 2i z = 3 + 4i 1 2 3 4 5 -3 -4 -5 Text Example cont. Plot in the complex plane: a. z = 3 + 4i b. z = -1 – 2i Solution • • c. z = -3 Because z = -3 = -3 + 0i, this complex number corresponds to the point (-3, 0). We plot –3 by moving three units to the left on the real axis. Because z = -4i = 0 – 4i, this complex number corresponds to the point (0, -4). We plot the complex number by moving three units down on the imaginary axis. d. z = -4i z = -3 5 4 3 2 1 -5 -4 -3 -2 z = -1 – 2i z = 3 + 4i 1 2 3 4 5 -3 -4 -5 z = -4i The Absolute Value of a Complex Number • The absolute value of the complex number a + bi is 2 2 z a + bi a + b Example • Determine the absolute value of z=2-4i Solution: z a + bi a + b 2 2 2 + (-4) 4 + 16 2 2 20 2 5 Polar Form of a Complex Number The complex number a + bi is written in polar form as z = r (cos + i sin ) where a = r cos , b = r sin , r a 2 + b 2 and tan =b/a The value of r is called the modulus (plural: moduli) of the complex number z, and the angle is called the argument of the complex number z, with 0 < < 2 Text Example Plot z = -2 – 2i in the complex plane. Then write z in polar form. Solution The complex number z = -2 – 2i, graphed below, is in rectangular form a + bi, with a = -2 and b = -2. By definition, the polar form of z is r(cos + i sin ). We need to determine the value for r and the value for , included in the figure below. Imaginary axis 2 è 2 -2 r -2 z = -2 – 2i Real axis Text Example cont. Solution 2 2 2 2 r a + b (-2) + (-2) 4 + 4 8 2 2 tan b -2 1 a -2 Since tan 4 = 1, we know that lies in quadrant III. Thus, 4 5 + + + 4 4 4 4 5 5 z r(cos + i sin ) 2 2(cos + i sin ) 4 4 Product of Two Complex Numbers in Polar Form Let z1 = r1 (cos 1+ i sin 1) and z2 = r2 (cos 2 + i sin 2) be two complex numbers in polar form. Their product, z1z2, is z1z2 = r1 r2 (cos ( 1 + 2) + i sin ( 1 + 2)) To multiply two complex numbers, multiply moduli and add arguments. Text Example Find the product of the complex numbers. Leave the answer in polar form. z1 = 4(cos 50º + i sin 50º) z2 = 7(cos 100º + i sin 100º) Solution z1 z 2 Form the product of the given = [4(cos 50º + i sin 50º)][7(cos 100º + i sin 100º)] numbers. = (4 · 7)[cos (50º + 100º) + i sin (50º + 100º)] Multiply moduli and add arguments. = 28(cos 150º + i sin 150º) Simplify. Quotient of Two Complex Numbers in Polar Form Let z1 = r1 (cos 1 + i sin 1) and z2 = r2 (cos 2 + i sin 2) be two complex numbers in polar form. Their quotient, z1/z2, is z1 r1 [cos(1 - 2 ) + i sin( 1 - 2 )] z 2 r2 To divide two complex numbers, divide moduli and subtract arguments. DeMoivre’s Theorem • Let z = r (cos + i sin ) be a complex numbers in polar form. If n is a positive integer, z to the nth power, zn, is z [r (cos + i sin )] n r n (cos n + i sin n ) n Text Example Find [2 (cos 10º + i sin 10º)]6. Write the answer in rectangular form a + bi. Solution By DeMoivre’s Theorem, [2 (cos 10º + i sin 10º)]6 = 26[cos (6 · 10º) + i sin (6 · 10º)] = 64(cos 60º + i sin 60º) 1 3 64 2 + i 2 32 + 32 3i Raise the modulus to the 6th power and multiply the argument by 6. Simplify. Write the answer in rectangular form. Multiply and express the answer in a + bi form. DeMoivre’s Theorem for Finding Complex Roots • Let =r(cos+isin) be a complex number in polar form. If 0, has n distinct complex nth roots given by the formula + 360k + 360k zk r cos + i sin n n where k 0,1,2,3,..., n - 1 n Example • Find all the complex fourth roots of 81(cos60º+isin60º) Solution: + 360k + 360k z k n r cos + i sin n n 60 + 360 * 0 60 + 360 * 0 4 81 cos + i sin 4 4 3(cos 15 + i sin 15 ) Example cont. • Find all the complex fourth roots of 81(cos60º+isin60º) Solution: 60 + 360 *1 60 + 360 *1 81 cos + i sin 4 4 4 3(cos 105 + i sin 105 ) 3(cos 195 + i sin 195 ) 3(cos 285 + i sin 285 ) Vectors Directed Line Segments and Geometric Vectors A line segment to which a direction has been assigned is called a directed line segment. The figure below shows a directed line segment form P to Q. We call P the initial point and Q the terminal point. We denote this directed line segment by PQ. Q Initial point Terminal point P The magnitude of the directed line segment PQ is its length. We denote this by || PQ ||. Thus, || PQ || is the distance from point P to point Q. Because distance is nonnegative, vectors do not have negative magnitudes. Geometrically, a vector is a directed line segment. Vectors are often denoted by a boldface letter, such as v. If a vector v has the same magnitude and the same direction as the directed line segment PQ, we write v = PQ. Vector Multiplication If k is a real number and v a vector, the vector kv is called a scalar multiple of the vector v. The magnitude and direction of kv are given as follows: The vector kv has a magnitude of |k| ||v||. We describe this as the absolute value of k times the magnitude of vector v. The vector kv has a direction that is: the same as the direction of v if k > 0, and opposite the direction of v if k < 0 The Geometric Method for Adding Two Vectors A geometric method for adding two vectors is shown below. The sum of u + v is called the resultant vector. Here is how we find this vector. 1. Position u and v so the terminal point of u extends from the initial point of v. 2. The resultant vector, u + v, extends from the initial point of u to the terminal point of v. Resultant vector u+v u Initial point of u v Terminal point of v The Geometric Method for the Difference of Two Vectors The difference of two vectors, v – u, is defined as v – u = v + (-u), where –u is the scalar multiplication of u and –1: -1u. The difference v – u is shown below geometrically. -u v–u -u v u The i and j Unit Vectors • Vector i is the unit vector whose direction is along the positive x-axis. Vector j is the unit vector whose direction is along the positive y-axis. y 1 j O i 1 x Representing Vectors in Rectangular Coordinates Vector v, from (0, 0) to (a, b), is represented as v = ai + bj. The real numbers a and b are called the scalar components of v. Note that a is the horizontal component of v, and b is the vertical component of v. The vector sum ai + bj is called a linear combination of the vectors i and j. The magnitude of v = ai + bj is given by 2 2 v a +b Text Example Sketch the vector v = -3i + 4j and find its magnitude. Solution For the given vector v = -3i + 4j, a = -3 and b = 4. The vector, shown below, has the origin, (0, 0), for its initial point and (a, b) = (-3, 4) for its terminal point. We sketch the vector by drawing an arrow from (0, 0) to (3, 4). We determine the magnitude of the vector by using the distance formula. Thus, the magnitude is v a +b 2 2 (-3) + 4 9 + 16 25 5 2 Terminal point 5 4 3 2 v = -3i + 4j 2 1 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 1 2 3 4 5 Initial point Representing Vectors in Rectangular Coordinates • Vector v with initial point P1 = (x1, y1) and terminal point P2 = (x2, y2) is equal to the position vector • v = (x2 – x1)i + (y2 – y1)j. Adding and Subtracting Vectors in Terms of i and j • If v = a1i + b1j and w = a2i + b2j, then • v + w = (a1 + a2)i + (b1 + b2)j • v – w = (a1 – a2)i + (b1 – b2)j Text Example If v = 5i + 4j and w = 6i – 9j, find: a. v + w b. v – w. Solution • v + w = (5i + 4j) + (6i – 9j) = (5 + 6)i + [4 + (-9)]j = 11i – 5j • v + w = (5i + 4j) – (6i – 9j) = (5 – 6)i + [4 – (-9)]j = -i + 13j These are the given vectors. Add the horizontal components. Add the vertical components. Simplify. These are the given vectors. Subtract the horizontal components. Subtract the vertical components. Simplify. Scalar Multiplication with a Vector in Terms of i and j • If v = ai + bj and k is a real number, then the scalar multiplication of the vector v and the scalar k is • kv = (ka)i + (kb)j. Example • If v=2i-3j, find 5v and -3v Solution: 5v (5 * 2)i + (5 * -3) j 10i - 15 j - 3v (-3 * 2)i + (-3 * -3) j -6i + 9 j The Zero Vector • The vector whose magnitude is 0 is called the zero vector, 0. The zero vector is assigned no direction. It can be expressed in terms of i and j using • 0 = 0i + 0j. Properties of Vector Addition and Scalar Multiplication If u, v, and w are vectors, then the following properties are true. Vector Addition Properties 1. u + v = v + u Commutative Property 2. (u + v) + w = v + (u + w)Associative Property 3. u + 0 = 0 + u = u Additive Identity 4. u + (-u) = (-u) + u = 0 Additive Inverse Properties of Vector Addition and Scalar Multiplication If u, v, and w are vectors, and c and d are scalars, then the following properties are true. Scalar Multiplication Properties 1. (cd)u = c(du) Associative Property 2. c(u + v) = cv + cu Distributive Property 3. (c + d)u = cu + du Distributive Property 4. 1u = u Multiplicative Identity 5. 0u = 0 Multiplication Property 6. ||cv|| = |c| ||v|| Finding the Unit Vector that Has the Same Direction as a Given Nonzero Vector v • For any nonzero vector v, the vector v v • is a unit vector that has the same direction as v. To find this vector, divide v by its magnitude. Example • Find a unit vector in the same direction as v=4i-7j Solution: v 4 + (-7) 2 2 16 + 49 65 v 4 7 ij v 65 65 The Dot Product Definition of a Dot Product If v=a1i+b1j and w = a2i+b2j are vectors, the dot product is defined as v w a1a2 + b1b2 The dot product of two vectors is the sum of the products of their horizontal and vertical components. Text Example If v = 5i – 2j and w = -3i + 4j, find: a. v · w b. w · v c. v · v. Solution To find each dot product, multiply the two horizontal components, and then multiply the two vertical components. Finally, add the two products. a. v · w = 5(-3) + (-2)(4) = -15 – 8 = -23 b. w · v = (-3)(5) + (4)(-2) = -15 – 8 = -23 c. v · v = (5)(5) + (-2)(-2) = 25 + 4 = 29 Properties of the Dot Product If u, v, and w, are vectors, and c is a scalar, then 1. u · v = v · u 2. u · (v + w) = u · v + u · w 3. 0 · v = 0 4. v · v = || v ||2 5. (cu) · v = c(u · v) = u · (cv) Alternative Formula for the Dot Product • If v and w are two nonzero vectors and is the smallest nonnegative angle between them, then • v · w = ||v|| ||w|| cos. Alternative Formula for the Dot Product • If v and w are two nonzero vectors and is the smallest nonnegative angle between them, then v w v w cos Formula for the Angle between Two Vectors • If v and w are two nonzero vectors and is the smallest nonnegative angle between v and w, then vw -1 v w cos and cos v w v w Example • Find the angle between v=2i-4j and w=3i+2j. -1 v w cos Solution: cos -1 cos -1 cos -1 v w 2 * 3 + -4 * 2 2 2 + (-4) 2 * 32 + 2 2 6-8 -2 cos -1 20 13 2 65 -1 97.1 65 The Dot Product and Orthogonal Vectors • Two nonzero vectors v and w are orthogonal if and only if v•w=o. Because 0•v=0, the zero vector is orthogonal to every vector v. Example • Are the vectors v=3i-2j and w=3i+2j orthogonal? Solution: v w 3 * 3 + 2 * -2 9-4 5 0 The vectors are not orthogonal. The Vector Projection of v Onto w • If v and w are two nonzero vectors, the vector projection of v onto w is proj wv vw w 2 w Example • If v=3i+4j and w=2i-5j, find the projection of v onto w Solution: projwv vw w 2 w 3 * 2 + 4 * -5 2 (2i - 5 j ) 2 [2 + (-5) ] - 14 - 28 70 (2i - 5 j ) i+ j 29 29 29 The Vector Components of v • Let v and w be two nonzero vectors. Vector v can be expressed as the sum of two orthogonal vectors v1 and v2, where v1 is parallel to w and v2 is orthogonal to w. v1 projwv vw w 2 w, v2 v - v1 Thus, v = v1 + v2. The vectors v1 and v2 are called the vector components of v. The process of expressing v as v1 and v2 is called the decomposition of v into v1 and v2. Example • Let v=3i+j and w=2i-3j. Decompose v into two vectors, where one is parallel to w and the other is orthogonal to w. v1 Solution: vw w 2 w, v2 v - v1 3 * 2 + 1* -3 (2i - 3 j ) 2 2 2 + (-3) 3 6 9 (2i - 3 j ) i j 13 13 13 6 9 v2 (3 - )i + (-3 + ) j 13 13 33 30 ij 13 13 v1 Definition of Work • The work W done by a force F in moving an object from A to B is • W = F · AB.