Chapter 2: The Dot Product

Transcription

1 Haberman MTH 11 Section IV: Vectors Chapter : The Dot Product In the previous chapter we studied how to add and subtract vectors and how to multiply vectors by scalars. In this chapter we will study how to multiply one vector by another using an operation called the dot product. Since we are focusing on two-dimensional vectors in this course, we will define the dot product in terms of two-dimensional vectors: DEFINITION: If u u1, u denoted uv, and v v1, v, then the dot product of u and v, is defined as follows: uv u v u v. 1 1 Thus, to compute the dot product of two vectors, we simply multiply the horizontal components of the two vectors and the vertical components of the two vectors and then add the results. It is important to note that the dot product produces a scalar. EXAMPLE 1: If a 3, 9 and b 6, 1, find a b. ab 3, 9 6, 1 36 ( 9) ( 1) Notice that the result is a scalar, not a vector. Now let s state some properties of the dot product. Afterwards, we will justify each of the properties.

2 Properties of the Dot Product If u, v, and w are vectors then the following properties hold true: 1. uv vu (commutative property). u v w u v u w (distributive property) 3. vv v 4. uv u v cos( ) where is the angle between u and v. The fourth property in our list is probably the most interesting since it suggests that the dot product can be used to measure the alignment of two vectors. Before discussing this property, let s justify the other properties in our list. 1. uv vu (commutative property) To justify this statement, let s compute uv for some generic vectors u and that the result is equal to v u. Let u u1, u and v v1, v. Then uv u, u v, v uv 1 1 uv 1 1 vu vu vu 1 1 (since scalar multiplication is commutative) v and show

3 3. u v w u v u w (distributive property) To justify this statement, let s compute uv w for some generic vectors u, v, and w and show that the result is equal to u v u w. Let u u1, u v v, v, 1, and w w1, w. Then u v w u u v v w w 1, 1, 1, u, u v w, v w u ( v w ) u ( v w ) uv uw uv u w uv uv uw u w (since scalar multiplication is distributive) (since scalar addition is commutative) u1, u v1, v uv uw u, u w, w vv v To justify this statement, let s compute vv for some generic vector v and show that the result is equal to v. Let v v1, v. Then vv v, v v, v vv 1 1 v v 1 1 v1 v v (since v v v ) 1

4 4. uv u v cos( ) where is the angle between u and v. 4 To justify this statement, let s first draw two vectors u and v so that is the angle between the vectors; see Figure 1. v u Figure 1 Now let s construct the vector w v u. Recall from the previous chapter that we can obtain the vector v u by drawing u an arrow that starts at the tip of u and ends at the tip of v. We ve drawn w v in Figure. v w u v u Figure If we think of the arrows as being line segments instead of arrows, we have a triangle with side lengths u, v, and w and angle opposite the side w ; see Figure 3. v w u Figure 3

5 5 Now we can use the Law of Cosines to obtain an equation that relates the magnitudes of the vectors and angle. w u v u v cos( ) 1 We can use this equation to obtain the statement given in fourth property in the table above. First, let s find w. Recall that w v u. So w ww (using the property 3 from the table above) v uv u v v u v v u u u (using the property from the table above) v v u v u v u u (using the property 1 from the table above) v u v u (using the property 3 from the table above) We can now substitute v u v u for w in the equation labeled 1, above, and simplify to obtain equation given in the forth property in the table: u v u v cos( ) v uv u uv u v cos( ) uv u v cos( ) We can use the fact that uv u v cos( ) related to the alignment of the vectors u and v. to see that the dot product is intimately

6 6 EXAMPLE : If the angle between u and v is 90 (i.e., if u and v are perpendicular), find u v. So if u and uv u v cos u v cos90 u v 0 0 v are perpendicular, uv 0. EXAMPLE 3: If u and v are non-zero vectors and uv 0, what can you say about the angle between vectors u and v. What if uv 0? Recall that uv u v cos( ) depends on the sign of cos( ).. Since u 0 and v 0, we see that the sign of uv If cos( ) 0 then uv 0 ; since cos( ) 0 when 0 90, we can conclude that uv 0 when the angle between the vectors is acute (i.e., less than 90 ). If cos( ) 0 then uv 0 ; since cos( ) 0 when , we can conclude that uv 0 when the angle between the vectors is obtuse (i.e., greater than 90 ). In the two examples above we see that the dot product can be used to learn about the alignment of two vectors. In fact, the dot product can be used to find the angle between two vectors; see Example 4, below.

7 7 EXAMPLE 4: Find the angle between the vectors a 3, 9 and b 6, 1 from Ex. 1. To find the angle between the two vectors we will use the fact that ab a b cos( ). Recall from Example 1 that ab 7. Let s find a and b : a (3) ( 9) and b (6) ( 1) Thus, ab a b cos( ) cos( ) cos( ) cos 6.1 So the angle between a and b is about 6.1 ; see Figure b a Figure 4