# Section V.2: Magnitudes, Directions, and Components of Vectors

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1 Section V.: Magnitudes, Directions, and Components of Vectors Vectors in the plane If we graph a vector in the coordinate plane instead of just a grid, there are a few things to note. Firstl, directions will most likel be specified as angles in standard position (positive angles rotate counterclockwise, starting from the positive -ais). Secondl, the vector does not need to begin or end at the origin. For eample, all of the vectors in the diagram below are equal, because each arrow has the same magnitude and direction. 3 3 A vector ma be moved all around the coordinate plane, and provided that the length and orientation of the arrow does not change, the vector itself will be considered unchanged. Calculating magnitudes of vectors To calculate the magnitude (length) of a vector, we ma use some plane geometr. First, sketch our vector on the coordinate plane. Because the angle between the - and -aes is 90º, it should be straightforward to draw a right triangle with the arrow as hpotenuse and sides parallel to the aes. Then the length of the hpotenuse ma be calculated from the lengths of the sides using the Pthagorean theorem, a + b = c, as in the eample below. What is the magnitude of vector A drawn in the diagram below? 3 3

2 Page Math 85 Vectors Answer: If ou zoom in on the arrow, the blown-up picture looks like: unit 4 units where the vector arrow is the hpotenuse of a right triangle that has a horizontal side with length of 4 units and vertical side of length unit. Plugging this into the Pthagorean theorem: c = a + b V = 4 + V = 7 V= 7 Since the magnitude of a vector is like a length measurement, the magnitude is alwas positive. If ou wish, ou could also plug the coordinates into the distance formula, ( ) ( ) d = +. As this formula is derived b putting generic coordinate points into the Pthagorean theorem, the calculation is essentiall identical to the geometric method used in the eample. In the eample above, if ou plug the points, 5,, =, into the distance formula, ou will calculate the ( ) = ( ) and ( ) ( ) same result, V= 7. Calculating directions of vectors When using a coordinate plane, directions should be specified using angles in standard position (positive angles rotate counterclockwise, starting from the positive -ais). Remember that adding an integral multiple of 360º gives an angle that s coterminal to the first. Therefore, the direction given b the angle 0º (0º below the positive -ais) could be equivalentl specified b the angle 340º. To determine the direction of a vector, graph it in the coordinate plane and consider once again the right triangle formed with the arrow as hpotenuse. Use right-triangle trig to determine the reference angle between the arrow and the -ais, and then calculate the size of the angle in standard position.

3 Math 85 Vectors Page 3 Calculate the direction of the vectors A, B, and C in the diagram below. A 3 C 3 B Answer: When ou zoom in on vector A, ou see the following. unit 4 units The angle ma be calculated from the arctangent function: tan = tan = 4 = tan 4 = 4 You ll notice from the diagram that vector B will have a right triangle that s congruent to A s triangle, and so will have the same reference angle (positive angle with respect to the -ais). However, B is pointed in the opposite direction, so to get the angle in standard position, we consider the sketch below

4 Page 4 Math 85 Vectors and add 80º to B s reference angle, to get the angle of 94º in standard position. If ou prefer, ou can consider the - and -coordinates of B s triangle to be negative, given the arrow s orientation, but then ou ll just have / 4 inside the tan function, which reduces to ¼, and ou ll get the same result. Remember that the tan - () function onl gives answers in the first and fourth quadrant, so if our vector has an angle in the second or third quadrant, ou ll have to do a further calculation. Vector C will also have a reference angle of 4º, and b consulting the sketch below, ou can see that to get the angle in standard position, ou ll have to subtract 4º from 80º to get a direction of 66º. If, instead, ou used 4 as our -coordinate, ou ll calculate an angle of 4º, which is in quadrant IV. B adding 80º to it to get an angle in QII, ou ll arrive at the same 66º as found earlier, which is reassuring. Component Form of a Vector Looking closel at our vector A from previous eamples, we have determined that its magnitude is 7 and its direction is 4º in standard position. However, this is a bit unsatisfactor in that the angle is an approimation onl (though theoreticall we could write it out with as man decimal places as we wanted). However, we do know the dimensions of the vector in the - and -directions eactl, as show in the diagram below. unit 4 units So rather than giving an approimation for this vector, it would be nice if there were a wa to state that this vector was 4 units long in the +-direction and unit long in the +direction. This description is the component form of a vector. There are a number of different notations for writing a vector in component form (also known as resolving a vector into components ). For the purposes of this course, we will use the same notation as our phsics and statics courses. Using this notation, the - component of the vector A is called A and the -component A. Our eample above could then be described as the vector with A = 4 and A =. Note that although the magnitude of vector A is alwas positive, the individual components ma be negative.

5 Math 85 Vectors Page 5 Give the component form of vectors B and C in the diagram below. A 3 C 3 B Answer: B looking at the diagram, we can see that B = 4 and B =, while C = 4 and C =. If a vector is given as a magnitude and direction, right-triangle trigonometr ma be used to resolve the vector into components. Give the component form of the vector D = 53 m/s at 35º. Answer: A quick sketch of the vector gives D = 35º D = 53 m/s D Note that an unbolded D is used for the hpotenuse, since we are referring onl to the vector s magnitude. From right-triangle trig,

6 Page 6 Math 85 Vectors D cos = D D = Dcos = 5cos and, similarl, D = D sin = 5 sin 35º As D points downward, though, the -component should be negative, so ou would therefore write as our final answer that D = 4 m/s and D = 9 m/s, giving the components the same units as the original vector magnitude. There is one other notation for component form that is commonl used in science and engineering tetbooks. But in order to understand how it works, we must first look at the ideas of scalar multiplication and unit vectors. Multiplication of a vector b a scalar If a vector A is multiplied b a positive real number b, then the product ba is a vector with in the same direction as the original vector A but with a magnitude that is b times as long as the original vector. If, however, a vector A is multiplied b a negative real number b, then the product ba is a vector with in the opposite direction to the original vector A but with a magnitude that is b times as long as the original vector. If a vector A is multiplied b zero, then the result is the zero vector, with 0 magnitude and no direction. For the vectors A and B sketched below, sketch 3A and ½ B. A B Answer: A B 3A ½ B

7 Math 85 Vectors Page 7 C = at 45º. Find the vectors 5C and /4 C. Answer Unit vectors For 5C, the vector is being multiplied b a positive number, so its direction remains unchanged, and the magnitude will be 5 =0: 5C = 0 at 45º For /4 C, the vector will change direction and be in QIII, so will be at an angle of (80º + 45º) = 5º. It will also onl be ¼ of the original length: /4 C = ½ at 5º A unit vector, like the name sounds, is simpl a vector that is one unit long. Such vectors are ver useful for indicating directions. The unit vector in the -direction has a special smbol associated with it: i (which unfortunatel can be easil confused with the smbol used in the comple number sstem). Similarl, the unit vector in the -direction is denoted b j. Combining our ideas of unit vectors and scalar multiplication, the product 4i is a vector 4 units long pointing in the +-direction. Let s look at an eample to see wh this is useful. Let us consider our old friend vector A, as shown in the diagram below. A 3 C 3 B Vector A can be considered as the sum of a vector 4 units long in the -direction plus a vector unit long in the -direction. Writing these two components as 4i and j, then we can write A as a vector sum: or, more generall, A = 4i + j

8 Page 8 Math 85 Vectors A = A i +A j with A and A being the vector components we discussed before. Resolve the following vector into components. Write our answer using the ij notation. Give eact answers. F = 35 at 5º Answer F cos = F F = Fcos = 35cos 5 35 = and, similarl, F = F sin = F sin 35º = 35. Therefore, F = 35 i 35 j. Calculate the magnitude and direction of the following vector. Give eact answers. D = 4 3 i 4 j Answer The -component is positive, 4 3, while the -component, 4, is negative, so the vector will look something like the following. D = 4 3, D D = 4

9 Math 85 Vectors Page 9 We can find D, the length of the vector (and the hpotenuse of the triangle) b The angle ma be found b c = a + b D = D + D ( ) = = = 64 D= 8 D tan = D 4 tan = 4 3 = tan = 30 3 So D = 8 units at 30º. Alternativel, ou could notice that this is our old friend, the triangle with sides in the ratio : 3 :, and make the angle negative because it s below the -ais. Resolve the following vector into components. Write our answer using the ij notation. Round answers to the nearest decimal place. F = 8. at 07º Answer F cos = F F = Fcos = 8. cos

10 Page 0 Math 85 Vectors and, similarl, F = F sin = 8. sin 07º Therefore, F = 5.3 i j. Find a unit vector u in the same direction as the vector A = i + 5j. Answer: Let s first sketch vector A: A =5 A A = The length of the hpotenuse can be found b the Pthagorean theorem: c = a + b A = A + A ( ) = + 5 = 69 A= 3 The unit vector u will be in the same direction as A, but will be onl one unit long. So u will have a triangle similar to A s triangle but with a hpotenuse of. Therefore, if we scale A s triangle down b dividing each side b 3, the length of A s hpotenuse, then we ll get the similar triangle: u = 5/3 u = u = /3 So u = /3 i + 5/3 j.

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