Section 9.1 Vectors in Two Dimensions

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1 Section 9.1 Vectors in Two Dimensions Geometric Description of Vectors A vector in the plane is a line segment with an assigned direction. We sketch a vector as shown in the first Figure below with an arrow to specify the direction. We denote this vector by AB. Point A is the initial point, and B is the terminal point of the vector AB. The length of the line segment AB is called the magnitude or length of the vector and is denoted by AB. We use boldface letters to denote vectors. Thus, we write u = AB. Two vectors are considered equal if they have equal magnitude and the same direction. The Figures below illustrate the way we (a) add (b) subtract (c) multiply a vector by a scalar 1

2 Vectors in the Coordinate Plane By placing a vector in a coordinate plane, we can describe it analytically (that is, by using components). We represent a vector v as an ordered pair of real numbers. v = a,b where a is the horizontal component of v and b is the vertical component of v. The vector a, b has many different representations, depending on its initial point. (a) Find the component form of the vector u with initial point ( 2,5) and terminal point (,7). (b) If the vector v =,7 is sketched with initial point (2,4), what is its terminal point? (c) Sketch representations of the vector w = 2, with initial points at (0,0), (2,2), ( 2, 1), and (1,4). (a) The desired vector is u = ( 2),7 5 = 5,2 (b) Let the terminal point of v be (x,y). Then x 2,y 4 =,7 So x 2 = and y 4 = 7, or x = 5 and y = 11. The terminal point is (5,11). (c) Representations of the vector w are sketched in the Figure above. (a) Find the vector u with initial point ( 5,4) and terminal point (4,8). (b) If the vector v = 4,7 is sketched with initial point (4,5), what is its terminal point? 2

3 (a) Find the vector u with initial point ( 5,4) and terminal point (4,8). (b) If the vector v = 4,7 is sketched with initial point (4,5), what is its terminal point? (a) The desired vector is u = 4 ( 5),8 4 = 9,4 (b) Let the terminal point of v be (x,y). Then x 4,y 5 = 4,7 So x 4 = 4 and y 5 = 7, or x = 8 and y = 12. The terminal point is (8,12). Find the magnitude of each vector. (a) u = 2, (b) v = 5,0 (c) w = 5, 4 5 (a) u = 2 2 +( ) 2 = 1 (b) v = = 25 = 5 ( ) 2 ( ) (c) w = + = = 25 = 1 If u = 2, and v = 1,2, find u+v, u v, 2u, v, and 2u+v.

4 If u = 2, and v = 1,2, find u+v, u v, 2u, v, and 2u+v. We have u+v = 2, + 1,2 = 2+( 1), +2 = 1, 1 u v = 2, 1,2 = 2 ( 1), 2 =, 5 2u = 2 2, = 2(2),2( ) = 4, 6 v = 1,2 = ( )( 1),( )2 =, 6 2u+v = 2 2, + 1,2 = 4, 6 +,6 = 4+( ), 6+6 = 1,0 The zero vector is the vector 0 = 0,0. A vector of length 1 is called a unit vector. For instance, in the Example above, the vector w = 5, 4 is a unit vector. Two useful unit 5 vectors are i and j, defined by i = 1,0 j = 0,1 These vectors are special because any vector can be expressed in terms of them. (a) Write the vector u = 5, 8 in terms of i and j. (b) If u = i+2j and v = i+6j, write 2u+5v in terms of i and j. 4

5 (a) Write the vector u = 5, 8 in terms of i and j. (b) If u = i+2j and v = i+6j, write 2u+5v in terms of i and j. (a) u = 5i+( 8)j = 5i 8j (b) The properties of addition and scalar multiplication of vectors show that we can manipulate vectors in the same way as algebraic expressions. Thus 2u+5v = 2(i+2j)+5( i+6j) = (6i+4j)+( 5i+0j) = (6 5)i+(4+0)j = i+4j (a) A vector v has length 8 and direction π/. Find the horizontal and vertical components, and write v in terms of i and j. (b) Find the direction of the vector u = i+j. (a) We have v = a,b, where the components are given by a = 8cos π = = 4 and b = 8sin π = 8 Thus, v = 4,4 = 4i+4 j. 2 = 4 (b) From the Figure on the right we see that the direction θ has the property that tanθ = 1 = Thus, the reference angle for θ is π/6. Since the terminal point of the vector u is in Quadrant II, it follows that θ = 5π/6. A woman launches a boat from one shore of a straight river and wants to land at the point directly on the opposite shore. If the speed of the boat (relative to the water) is 10 mi/h and the river is flowing east at the rate of 5 mi/h, in what direction should she head the boat in order to arrive at the desired landing point? 5

6 A woman launches a boat from one shore of a straight river and wants to land at the point directly on the opposite shore. If the speed of the boat (relative to the water) is 10 mi/h and the river is flowing east at the rate of 5 mi/h, in what direction should she head the boat in order to arrive at the desired landing point? We choose a coordinate system with the origin at the initial position of the boat as shown in the Figure above. Let u and v represent the velocities of the river and the boat, respectively. Clearly, u = 5i, and since the speed of the boat is 10 mi/h, we have v = 10, so v = (10cosθ)i+(10sinθ)j where the angle θ is as shown in the Figure above. The true course of the boat is given by the vector w = u+v. We have w = u+v = 5i+(10cosθ)i+(10sinθ)j = (5+10cosθ)i+(10sinθ)j Since the woman wants to land at a point directly across the river, her direction should have horizontal component 0. In other words, she should choose θ in such a way that 5+10cosθ = 0 cosθ = 1 2 θ = 120 Thus she should head the boat in the direction θ = 120 (or N 0 W). 6