Precalc Chapter 6 Notes Section 6.1: Vectors in the Plane In the real world, measurements are not simply about value or magnitude; direction is often a factor of a measurement. For example, force, velocity, and acceleration have components of magnitude and direction. As a result, these kinds of measurements must be measured in terms of both magnitude and direction. This is where vectors, also referred to as directed line segments, come into play. The direction of the vector shown to the left is indicated from the initial point P (where the vector begins) to the terminal point Q (where the vector terminates). It is denoted by PQ and is referred to as “vector PQ”. To figure out the magnitude of the vector, one must figure out the “distance” of the vector using the distance formula. The magnitude of vector PQ is denoted as PQ . Distance formula:
PQ =
(x 2 - x 1 )2 + (y 2 - y 1 )2 To figure out if two vectors are equivalent, one must first realize that vectors are made up of magnitude and direction, meaning that equivalent vectors must have the same magnitude and direction. Since vectors are directed line segments, one can show that vectors are equivalent if they have the following:
· the same magnitude (calculated by the distance formula)
y - y · the same slope (calculated by m = 2 1 ) x 2 - x 1 Example: p. 488, #1: Show that the directed line segments RS containing endpoints R = (–4, 7) and S = (–1, 5) and OP containing endpoints O = (0, 0) and P = (3, –2) are equivalent.
RS =
=
=
m RS =
(x 2 - x 1 )2 + (y 2 - y 1 )2 (- 1 - (- 4 ))2 + (5 - 7 )2 (3 )2 + (- 2 )2 = 9 + 4 =
5 - 7 - 2 2 =
=- 1 - (- 4 )
3 3 OP =
=
13 =
m OP =
(x 2 - x 1 )2 + (y 2 - y 1 )2 (3 - 0 )2 + (- 2 - 0 )2 (3 )2 + (- 2 )2 = 9 + 4 =
13 - 2 - 0 - 2 2 =
=3 - 0 3 3 Since the magnitudes of the vector are the same and the slopes are the same, the vectors are equivalent. Vectors, as stated before, are a combination of distance and direction. As you saw in the previous problem, two­dimensional vectors are made up of two more basic mathematical components: the change in x values and the change in y values. As a result, vectors are normally expressed in terms of x and y components with the assumption that the initial point of the vector is (0, 0):
Component Form of a Vector: For some vector v with initial point (0, 0) and terminal point (v1, v2), the component form of the vector v is v = v 1 , v 2 . Notice that, for any vector, v1 is the change in x for the vector’s endpoints and v2 is the change in y for the vector’s endpoints, meaning that if for vector PQ with initial point P(x1, y1) and terminal point Q(x2, y2), v = x 2 - x 1 , y 2 - y 1 . Figuring out the magnitude is the same as figuring out the magnitude as a vector from before. Since v1 = x2 – x1 and v2 = y2 – y1, and PQ =
(x 2 - x 1 )2 + (y 2 - y 1 )2 , PQ
2 2 = v 1 + v 2 . Magnitude of a Vector: For some vector v = PQ with initial point P(x1, y1) and terminal point Q(x2, y2), 2 2 | v | = v 1 + v 2 =
(x 2 - x 1 )2 + (y 2 - y 1 )2 Example: Find the component form and magnitude of the vector QR containing endpoints Q = (3, 4) and R = (–2, 5). Component form: Magnitude: QR = x 2 - x 1 , y 2 - y 1 = - 2 - 3 , 5 - 4 QR = v 1 + v 2 =
QR = - 5, 1 QR = 26 2 2 (- 5)2 + (1 )2 = 25 + 1 One nice thing about using vectors, aside from properties of both magnitude and direction, is that they can undergo basic mathematical operations to get resulting vectors. For example, vectors can be added and subtracted, and they can also undergo what is referred to as scalar multiplication, which is essentially the same as using the distributive property. As you can see, vectors can be added since the x components and y components of vectors can be added (as shown above and to the left). In other words, to add or subtract, one simply needs to add or subtract the components of one vector to the components of the other vector. Scalar multiplication, what was previously referred to as the “distributive property” of vectors, can also work since one can multiply x and y components of vectors by the same number to lengthen or shorten it.
To summarize: Vector Operation Vector Addition Vector Subtraction Scalar Multiplication For vectors u = <u1, u2> and v = <v1, v2>, Vector Representation How to do the operation u + v u + v = <u1 + v1, u2 + v2> u – v u – v = <u1 – v1, u2 – v2> ku ku = k<u1, u2> = <ku1, ku2> Example: Let u = <–1, 3>, v = <2, 4>, and w = <2, –5>. Find the component form of the vector. u + v: u + v = <–1, 3> + <2, 4> = <–1 + 2, 3 + 4> = <1, 7> u – w: u – w = <–1, 3> – <2, –5> = <–1 – 2, 3 – (–5)> = <–3, 8> –2u – 3v: –2u – 3v = –2<–1, 3> – 3<2, 4> = <2, –6> – <6, 12> = <2 – 6, –6 – 12> = <–4, –18> There is a special kind of vector known as a unit vector. A unit vector is a vector with length of one. In other words, if there is some vector u, vector u is a unit vector if | u | = 1. If a vector is not originally a unit vector, it can be represented by a unit vector by dividing the vector by its magnitude. In other words, v 1
u =
= v v v Vectors can also be written in other forms. In physics, vectors are often written as a linear combination of standard unit vectors. The two separate vectors are i = <1, 0> and j = <0, 1>, meaning that if one were to combine the vectors, the resulting vector v = i + j would be a vector with an x­component of 1 unit and a y­component of 1 unit. However, any vector v can be written as a linear combination of these standard unit vectors: v = <a, b> v = <a, 0> + <0, b> v = a<1, 0> + b<0, 1> v = ai + bj
Example: Find the unit vector in the direction of the vector u = <–4, –5>. Write your answer in (a) component form and (b) as a linear combination of the standard unit vectors i and j. (a) Component form: | u | = u1 2 + u 2 2 =
Unit vector = (b) (- 4)2 + (- 5 )2 = 16 + 25 = 41 u - 4, - 5 1
- 4 - 5 4 41 5 41 = = - 4 , - 5 = , = , u 41 41 41 41 41 41 Linear combination of the standard unit vectors i and j: u = – 4 41 5 41 i – j 41 41 At the beginning of this section, direction was generalized with the use of slope. A more accurate way of expressing the direction of a vector is by analyzing its vertical and horizontal components. Notice in the diagram that a (from v = ai + bj) corresponds with the x­ component of the vector as b corresponds with the y­component of the vector. Based on knowing either component of the vector as well as the magnitude of the vector, one can figure out q based on one of two trigonometric functions. cos q =
a
v æ a ö
÷
ç v ÷
è ø
q = cos -1 ç
sin q =
b
v æ b ö
÷
ç v ÷
è ø
q = sin -1 ç
If given a magnitude of a vector and the angle of the vector, one can use these to actually write the component form or linear combination of unit vectors of the vector. cos q =
a
v sin q =
a = v cos q
b
v b = v sin q
v = ai + bj v = (|v|cos qi + |v|sin qj) v = |v|(cos qi + sin qj) or <|v|cos q, |v|sin q>
Example: Find the magnitude and direction angle of the vector. <3, 4>: Magnitude: 3 2 + 4 2 = 9 + 16 = 25 = 5 æaö
÷ = cos -1 æç 3 ö÷ » 53 . 1 °
ç v ÷
è 5 ø
è ø
This angle is correct since it is in the first quadrant. q = cos -1 ç
Angle:
3i – 4j: Magnitude:
2 3 2 + (- 4 ) = 9 + 16 = 25 = 5 æaö
÷ = cos -1 æç 3 ö÷ » 53 . 1 °
ç v ÷
è 5 ø
è ø
However, with the x­component being positive and the y­component being negative, the vector is in the fourth quadrant, meaning the angle must be subtracted from 360° to get the right angle.
q = cos -1 ç
Angle:
q = 360° – 53.1° = 306.9°
7(cos 135°i + sin 135°j): Magnitude: The magnitude is 7 since cos 135°i + sin 135°j is a unit vector. The reason for this: 7(cos 135°i + sin 135°j) = 7 cos 135°i + 7 sin 135°j
(7 cos 135 °)2 + (7 sin 135 °)2 =
(
49 cos 2 135 ° + 49 sin 2 135 °
)
= 49 cos 2 135 ° + sin 2 135 ° = 49 (1 ) = 49 = 7 Angle: The angle is already shown: 135°. Example: Find the component form of the vector v with magnitude 47 and direction angle 108°. v = <|v|cos q, |v|sin q> = <47 cos 108°, 47 sin 108°> » <–14.524, 44.700> There are many applications of vectors, including piloting, captaining a ship, etc. Example: A ship is heading due north at 12 mph. The current is flowing southwest at 4 mph. Find the actual bearing and speed of the ship. Ship’s heading: Current’s heading: <12 cos 90°, 12 sin 90°> = <0, 12> <4 cos 225°, 4 sin 225°> » <–2.83, –2.83> Actual velocity vector: <0, 12> + <–2.83, –2.83> = <–2.83, 9.17> Actual speed:
Actual direction angle:
(- 2 . 83 )2 + (9 . 17 )2 » 9 . 60 mph æ a ö
æ - 2 . 83 ö
cos -1 ç ÷ = cos -1 ç
÷ » 107 . 15 °
ç v ÷
9 . 60 è
ø
è ø
Since the actual direction angle is 107.15°, the actual bearing is 342.15°.