Dr. Neal, WKU MATH 117 Vectors and Polar Form Any point (x, y) in the xy plane forms a directed line segment from the origin (0, 0) to the point (x, y). Such a segment is called a vector. When we want to consider the vector and not just the point, then we generally label it as v = (x, y). v = (7, 4)! Length and Direction A vector v = (x, y) has a length (or norm) denoted by v which is simply the distance to the origin given by the hypotenuse. The direction ! is the standard angle determined by (x, y) as measured from the positive x axis. Thus we have v = x2+ y2 and tan!=yx To find the direction !, compute tan!1(y / x) and then adjust the angle to the proper quadrant. A vector written in terms of its length and direction is then in polar form. Example 1. Let u = (–6, 8) and v = (–2, –10). Find the length and direction of each vector and write the vectors in polar form. Solution. Vector u has length u = 62+ 82 = 10. Here, tan!1( y / x) is tan!1(8 / !6) ≈ –53.13º. So in Quad. II, ! = !53.13º +180º ≈ 126.87!. So u = (10, 126.87!) in polar form. uv Vector v has length v = 22+ 102 = 104. Its angle in Quadrant III is given by ! = tan!1(10 / 2) + 180º ≈ 258.69!. Then v = (104, 258.69!) in polar form.Dr. Neal, WKU Converting Polar Form Back to Rectangular Form If a vector is given in polar form (v, !), then we recover the x and y coordinates by x = v cos! and y = v sin! Here, v is taking the place of the radius r, where x = r cos! and y = r sin!. Example 2. Find the rectangular form of the vectors u = (30, 120º) and v = (20, 330º). Solution. For u = (30, 120º), we have x = 30 cos120º = 30 ! "12# $ % & ' ( = "15 and y =30 sin120º = 30 !32" # $ $ % & ' ' = 15 3; so u = (–15, 15 3). For v, x = 20 cos330º = 20 !32" # $ $ % & ' ' = 10 3 and y = 20 sin330º = 20 ! "12# $ % & ' ( = "10; so then v = (10 3, –10). Adding Vectors Given two vectors u = (x1, y1) and v = (x2, y2), both in rectangular form, we obtain the sum of vectors by adding component wise: u + v = (x1+ x2, y1+ y2). If we make a parallelogram out of the vectors u and v, then vector u + v is the diagonal that starts at the origin. Example 3. Let u = (3, 6) and v = (–8, –2). Graph u, v, and u + v. What is the length and direction of u + v? Solution. First, u + v = (3 + (–8), 6 + (–2)) = (–5, 4), which we see to be the diagonal of the parallelogram determined by u and v. (See graphs on next page.) Then u + v = 52+ 42= 41 ! 6. 4, and the direction of u + v is given by != tan"1("4 / 5) + 180º # 141. 34º.Dr. Neal, WKU u= (3, 6)v = (–8, –2)u + v uuvvu + v Adding Forces Often, a force is given in terms of its magnitude and direction. In order to add two forces, we convert each to rectangular form, add the x and y components to get the sum, then convert the result back to polar form. The sum of two forces is called the resultant force. Example 4. Let F1 be a force of 50 Newtons in the direction 30º East of South, and let F2 be a force of 80 Newtons in the direction 10º South of East. Find the magnitude and direction of the resultant F1+ F2. Solution. First, the angle for F1 is 30º + 270º = 300º, and the angle for F2 is 360º – 10º = 350º. Next, the x and y components for each force and the resultant are: F1 = (50 cos300º, 50sin 300º) F2 = (80cos350º, 80sin350º) SEF110º30ºF2 So the resultant force is F1+ F2 = (50 cos300º + 80cos350º, 50sin 300º + 80sin350º) = (103.78462, –57.19312) So F1+ F2 has magnitude F1+ F2= 103.784622+ 57.193122! 118.5 Newtons. Its direction is in the 4th Quadrant is tan!1(!57.19312 / 103.78462) + 360º ≈ 331.142º, or about 28.858º South of East.Dr. Neal, WKU Subtraction and Distance Between Vectors Given two vectors u = (x1, y1) and v = (x2, y2), both in rectangular form, we obtain the vector from u to v by the difference v ! u which is obtained by subtracting component wise: v ! u = (x2! x1, y2! y1). If we make a parallelogram out of the vectors u and v, then vector v ! u is equivalent to the diagonal from the end of u to the end of v after it is picked up and moved to the origin. The length of v ! u (or u ! v) gives the distance between the endpoints of u and v, which we call the distance between the vectors. This length of v ! u is equivalent to the common “distance formula” in the xy plane: u = (x1, y1) v = (x2, y2) v ! u = (x2! x1, y2! y1) distance = (x2! x1)2+ (y2! y1)2= v ! u Example 5. Let u = (3, 6) and v = (–8, –2). Find the vector from u to v and the distance between u and v. Graph all the vectors. Solution. By subtracting component wise, we get v ! u = (–11, –8). Now we can pick up the vector (–11, –8) and place it between the original u and v to see that v ! u is the diagonal from u to v. u= (3, 6)v = (–8, –2)v ! u u= (3, 6)!u = (–3, –6)vv ! u(!11, !8)v = (!8, !2) The distance between u and v is v ! u = 112+ 82= 185 " 13 . 6.Dr. Neal, WKU Dot Product and Angle Between Vectors Given vectors u = (x1, y1) and v = (x2, y2), the dot product is given by u ! v = x1x2+ y1y2 The dot product can be used to find the angle ! between the vectors u and v. Because v ! u is the length of the third side of a triangle and is opposite angle !, the Law of Cosines gives v ! u2 = u2+ v2! 2 u v cos" We now can solve for cos! and simplify the expression: uvv ! u" cos!=u2+ v2" v " u22 u v=x12+ y12( )+ x22+ y22( )" (x2" x1)2+ ( y2" y1)2( )2 u v=" "2 x1x2" 2 y1y2( )2 u v=x1x2+ y1y2u v=u # vu v. Thus, cos!=u " vu v and != cos"1u # vu v$ % & ' …