Vectors. of real numbers. The numbers are called the components of a. A representation of the vector a a1,

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1 Vectors Vector is used by scientists to indicate a quantity (such as velocity or force) that has both magnitude and direction. A vector is often represented by an arrow or a directed line segment. The length of the arrow represents the magnitude of the vector and the arrow points in the direction of the vector. Figure 1 shows a particle moving along a path in the plane and its velocity vector v at a specific location of the particle. Here the length of the arrow represents the speed of the particle and it points in the direction that the particle is moving. Vector has two characteristics: magnitude (length) and direction. As long as the vectors have same direction and same length, the vectors are the same. In Figure notice that all of the arrows are equivalent in the sense that they have the same length and point in the same direction even though they are in different locations. All vectors in Figure are the same as vector OP which starts at origin (, ) and ends at P (3,. We use <3, > to represents OP. Definition: A two-dimensional vector is an ordered pair a a1, of real numbers. The numbers a1, a are called the components of a A representation of the vector a a1, is a directed line segment AB from any point A( x, y ) to the point B( x a, y a ). A particular representation of vector a is the 1 directed line segment OP from the origin to the point P( a1, a, and a1, is called the position vector of the point P( a1, a (see Figure 3). Theorem 1: Given the points A( x1, y 1) and B( x, y, the vector a with representation AB is a AB x x, y y 1 1

2 Example 1: Find the vector represented by the directed line segment with initial point A (,4) and terminal point B (3,7). Solution: AB x x1, y y1 3,7 4 1,3 The magnitude or length of the vector a a1, is denoted by a or a, by using the distance formula, we have a a a 1 The length of the vector AB from A( x1, y 1) to B( x, y is AB ( x x ) ( y y ) 1 1 The only vector with length is the zero vector,. This vector is also the only vector with no specific direction. A unit vector is a vector whose length is 1. In general, if a, then the unit vector that has the same direction as a is 1 a u a a a Example : Find the unit vector in the direction of the vector <-4, 3>. Solution: the length of vector <-4, 3>, 4,3 ( 4) 3 5 The unit vector in the direction of <-4, 3> is 1 4,3 4, Vector Addition (Triangle Law and Parallelogram Law)

3 By the subtraction a b of two vectors, we mean a b a ( b). Vector Addition (subtraction). If a a1,, b b1, b, then a b a b, a b 1 1 Multiplication of a Vector by a Scalar. If c is a scalar (a real number), and a a1,, then ca ca, ca 1 Properties of Vectors. If a, b and c are vectors, c and d are scalars, then a b b a a ( b c) ( a b) c a a a ( a) c( a b) ca cb ( c d) a c a d a (cd) a c( d a) Example 3: If a 4,, b 3,5. Find vectors a 3 b and 5 b a. Solution: a 3b 4,- 3 3,5 8,-4 9,15 17,11 5 b a 5 3,5-4,- 15, 5-4,- 11, 7 There are two special unit vectors: i 1,, j,1 (See Figure 9). Any vector a a1, can write as a a1, a1 i j Example 4. If a i 3 j, b 3i j, find a b Solution: a b (i 3 j) (3 i j) 4i 6 j 3i j i 7 j t

4 We conclude this section by considering one of the many applications of vectors in physice and engineering. A force is represented by a vector because it has both a magnitude (measured in pounds or newtons, N) and a direction. If several forces are acting on an object, the resultant force experienced by the object is the vector sum of thes forces. Example 5. A 1-lb weight hangs from two wires as shown in Figure 11. Find the tensions (forces) T 1 and T in both wires and their magnitudes. Solution: The resultant force T1 T counterbalances the weight w and so must have (1) T1 T w 1 j We can express T 1 and T in terms of their horizontal and vertical components. From Figure 1, and the Triangle Law we have ( T T cos5 i T sin5 j T T cos3 i T sin 3 j Plugging them into equation (1), we have T cos5 i T sin 5 j+ T cos3 i T sin 3 j 1j 1 1 ( T cos5 T cos3 ) i + ( T sin 5 T sin 3 ) j 1 j 1 1 Equating the components, we have T cos5 T cos3 T sin5 T sin Solving the first of these equations for T and substituting into the second, we get

5 T 1 cos cos3 cos5 sin5 sin3 T sin5 sin3 1 T lb cos3 and T1 cos5 T lb cos3 Thus, substituting these values in (, we have the forces T T cos5 i T sin5 j 85.64cos5 i 85.64sin5 j 55.5 i 65.6 j T T cos3 i T sin 3 j 64.91cos3 i 64.91sin 3 j 55.5 i 34.4 j