1 Vectors. c Kun Wang. Math 151, Fall Vector Supplement

Transcription

1 Vector Supplement 1 Vectors A vector is a quantity that has both magnitude and direction. Vectors are drawn as directed line segments and are denoted by boldface letters a or by a. The magnitude of a vector a is its length, denoted by a or a. A two-dimensional vector is an ordered pair a =< a 1, a 2 > of real numbers. The numbers a 1 and a 2 are called the components of a. Given the points A(x 1, y 1 ) and B(x 2, y 2 ), the vector a with initial point A and terminal point B (also written AB) is a =< x2 x 1, y 2 y 1 >. The length of the vector AB from A(x1, y 1 ) to B(x 2, y 2 ) is AB = (x 2 x 1 ) 2 + (y 2 y 1 ) 2. Example: Given the points A(2, 5) and B(4, 1), find the vector AB and its magnitude. Remarks: (1) A vector with initial point at the origin is said to be in standard position. (2) The only vector with length 0 is the zero vector 0 =< 0, 0 >. This vector is also the only vector with no specific direction. (3) Two vectors are equal if they have the same magnitude and direction. So, it doesnt matter where the vector is as long as it has the same magnitude and direction (the same displacement). (4) A vector < a 1, a 2 > with initial point at the origin is called the position vector of the point (a 1, a 2 ). 1

2 Vector Addition: If a =< a 1, a 2 > and b =< b 1, b 2 >, then the vector a + b is defined by a + b =< a1 + b 1, a 2 + b 2 >. Scalar Multiplication: If c is a scalar and a =< a 1, a 2 >, then the vector c a is defined by c a =< ca 1, ca 2 >. Two vectors a and b are parallel if b = c a for some scalar c. Vector Difference: If a =< a 1, a 2 > and b =< b 1, b 2 >, then the vector a b is defined by a b =< a1 b 1, a 2 b 2 >. Example: If a =< 5, 4 > and b =< 1, 2 >, find 2 a b. 2

3 Unit Vector: A vector with length (magnitude) 1 is called a Unit Vector. A unit vector in the direction of a is: Example: If a =< 2, 3 >, find a unit vector in the direction of a. Example: If a =< 2, 3 >, find a vector of length 5 in the same direction as a. Standard Basis Vectors: There are two special unit vectors that we use all the time: i =< 1, 0 >, j =< 0, 1 >. Then i and j are unit vectors pointing in the directions of the positive x- and y-axes. The vectors i and j are called basis vectors because EVERY vector a =< a 1, a 2 > can be written in terms of these unit vectors by a = a 1 i + a 2 j. 3

4 Example: Given the vectors a =< 2, 1 >, b = 3i + 5j, and c = 2i + 6j, find scalars s and t so that sa + tb = c. Direction: The direction of a vector is the positive angle θ formed by the positive x-axis and the vector. Suppose you are given a vector a with magnitude a and direction θ. Then a =< a cos θ, a sin θ >. Example: A vector a has a = 4 and θ = 60. Write a in component form. Example: Find the direction of the vector a = 4i 5j from the positive x-axis. 4

5 Applications The velocity of an object can be modeled by a vector, where the direction of the vector is the direction of motion, and the magnitude of the vector is the speed. Often when dealing with boats or planes, directions are expressed as bearings: Example: A person is swimming due north at 5 mi/h relative to the water. The water is flowing due west at 2 mi/h. Find the true course and speed of the person. Example: Suppose that a wind is blowing in the direction S45E at a speed of 60 km/h. A pilot is steering a plane in the direction N60E at an airspeed (speed in still air) of 150 km/h. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane. (Round your answers to one decimal place.) 5

6 Another application of vectors involves forces. A force has magnitude (measured in pounds or Newtons) and a direction. If several forces are acting on an object, the resultant force is the vector sum of all these forces. Example: Two forces F 1 and F 2 are acting on an object at a point P. F 1 has a magnitude of 40 lbs in a direction of 60 from the positive x-axis. F 2 has a magnitude of 20 lbs in a direction of 330 from the positive x-axis. Find the resultant force F along with both its magnitude and direction. 6

7 2 The Dot Product The dot product of two nonzero vectors a and b, denoted a b, is the number a b = a b cos θ, where θ is the angle between a and b, 0 θ π. If the vectors are given in component form where a =< a 1, a 2 > and b =< b 1, b 2 >, then a b = a 1 b 1 + a 2 b 2. Important: The dot product of two vectors is always a SCALAR, not a vector. For this reason, the dot product is sometimes called the scalar product. Example: Calculate a b in the following scenarios. (1) a =< 4, 7 >, b =< 3, 3 > (2) a = 7, b = 2, and the angle between the two vectors is π/6. Two vectors are orthogonal or perpendicular if if the angle between them is π/2 or 90. Thus, two vectors are orthogonal if and only if a b = 0, since cos π/2 = 0. Example: Find the value(s) of x such that the vectors a =< 4x, 1 > and b = xi - 6xj are orthogonal. Orthogonal Complement: Sometimes it is useful to find a vector that is orthogonal to a given vector with the same length. Given the nonzero vector a =< a 1, a 2 >, the orthogonal complement of a is the vector a =< a 2, a 1 > Example: Find a vector of length 2 that is orthogonal to < 2, 8 >. 7

8 The dot product can also be used to find the angle between two vectors when they are given in component form. Since a b = a b cos θ, then cos θ = a b a b θ = arccos a b a b Example: Find the angle between the vectors a =< 1, 4 > and b =< 2, 5 >. Example: Given a triangle with vertices A(1, 5), B(2, 1), and C(4, 6), find BCA. 8

9 Work: Work is an application of the dot product to physics. The work done in moving an object through a distance d by a force F is W = F d. However, this formula only works if the force is applied in the same direction as the motion. We now generalize to the case where the force may not be applied in the same direction by using vectors. Suppose a force F moves an object from a point P to a point Q. The displacement vector is D = P Q. How much work is done? W = F D = F D cos θ. If the force is measured in Newtons and the displacement is measured in meters, then the unit of work is Nm, which is also call a joule (J). If the force is measured in pounds and the displacement is measured in feet, then the unit of work is the ft lb. Example: A force F = 4i+2j moves an object from the point P (3, 6) to the point Q(2, 9). How much work is done if force is measured in lbs and distance is measured in ft? Example: A horizontal force of 30 N is used to push a box up a 20 m ramp which is inclined at an angle of 15. How much work is done? 9

10 Projections Scalar Projection of b onto a (also called the Component of b along a): Vector Projection of b onto a: Comp a b = a b a Proj a b = ( a b a ) a a = a b a 2 a. Example: Find the scalar and vector projections of the vector < 3, 1 > onto the vector < 2, 5 >. Example: Find the distance from the point (3, 6) to the line y = 1 2 x

11 3 Parametric Equations and Vector Functions Parametric Curves: Sometimes, instead of representing a curve using justx and y, it is more convenient to use parametric equations using a parameter such ast. This means that the values of x and y are defined as functions of this parameter, x(t) and y(t). Example: Sketch the curve represented by the parametric equations x = 3t + 1, y = 2t 1. If given a set of parametric equations, it may be useful to convert back into a Cartesian equation (using x and y only). In order to do this, you must eliminate the parameter. How? 1. If possible, solve one of the parametric equations for t and use substitution. 2. If the parametric equations involve trig functions, use a trig identity, often sin 2 θ + cos 2 θ = 1. Example: Eliminate the parameter from the previous example and write a Cartesian equation for the curve Sometimes, there may be a restriction on the values of t or the values of x and y may have bounds you need to watch out for. Example: Eliminate the parameter to find a Cartesian equation for the following curves, sketch a graph, and denote the direction of motion as t increases. 1. x = t + 1, y = t 3, 2 t 4 11

12 2. x = t, y = 2 t 3. x = 3 + sin t, y = 2 + cos t 4. x = sin(t), y = csc(t), 0 < t < π/2 12

13 Vector Functions: We can define vector functions using these parametric equations by r(t) =< x(t), y(t) >. It is called a vector function because it takes values of t and produces vectors. These vectors are tracing out the curve. Example: Sketch the curve represented by the vector function r(t) = (4 cos t)i + (sin t)j, 0 t π. Example: The position of an object after t seconds, t 0, is given by the vector function r(t) =< t 2, t 2 >. 1. What is the position of the object at time t = 6? 2. At what time, if ever, is the object at position (9, 1)? 3. Does the object pass through the point (49, 7)? 4. Find an equation in x and y whose graph is the path of the object and sketch its path. 13

14 Vector Equation of a Line: If P 0 (x 0, y 0 ) is a point on the line with position vector r 0 and v is a vector parallel to the line, then the vector equation of the line is r(t) = r 0 + tv. Parametric equations of the line that passes through the point P (x 0, y 0 ) and is parallel to the vector < a, b > are given by x = x 0 + at y = y 0 + bt Example: Find a vector equation and parametric equations for the line that passes through the points (2, 5) and (1, 7). Example: Find parametric equations for the line that passes through the point (3, 2) and is parallel to the line y = 2 3 x

15 Example: Find a vector equation of the line that passes through the point (1, 4) and is perpendicular to the line with vector equation r(t) =< 3 5t, 2 + 7t >. Determine whether the lines r 1 (t) = ( 1 2t)i + (2 + t)j and r 2 (s) = (5 + 3s)i + (3 + 6s)j are parallel, perpendicular, or neither. If they are not parallel, find the point of intersection 15