Math 425 Lecture 1: Vectors in R 3, R n

Math 425 Lecture 1: Vectors in R 3, R n Motiating Questions, Problems 1. Find the coordinates of a regular tetrahedron with center at the origin and sides of length 1. 2. What is the angle between the faces of the tetrahedron? 3. What is the area of the faces of the tetrahedron? 4. What is the olume of the tetrahedron? Vectors in R 2, R 3 Let A, B be points in R 3. The directed line segment with tail at A and head at B is the ector denoted AB. Let C, D be two other points in R 3. The ectors AB and CD are the same proided the directed line segments point in the same direction and hae the same length. Let be a ector in R 3. We can find B so that AB = and A is the origin. If B has coordinates (b 1, b 2, b 3 ), we write = (b 1, b 2, b 3 ). We can add two ectors, u R 3. There are two ways of doing this: We can add geometrically. Place the tail of at the head of u. Then the directed line segment from the tail of u to the head of is u +. Note that u + is the same as + u. We can add using coordinates. If u = (u 1, u 2, u 3 ), = ( 1, 2, 3 ), then u + = (u 1 + 1, u 2 + 2, u 3 + 3 ). We can multiply ectors by scalars. There are two ways of iewing this: We can iew this geometrically. When we multiply a ector by a positie scalar λ, we stretch it by the factor λ. If we multiply it by 1 we completely turn the ector around. We can proceed algebraically. If = ( 1, 2, 3 ), then λ = (λ 1, λ 2, λ 3 ). Applications: We can use this to parametrize a line in R 3. If the line L passes through the point P = (p 1, p 2, p 3 ) and has direction = ( 1, 2, 3 ), then x p 1 1 y = p 2 + t 2 z p 3 3 is a parametrization of L. 1

Using the same idea we can parametrize planes in R 3. Assume that a plane passes through a point P = (p 1, p 2, p 3 ) and contains ectors u = (u 1, u 2, u 3 ) and = ( 1, 2, 3 ), u 0, 0 Assume that u i not a scalar multiple of. Then we can parametrize this plane by x = p + t u + s, s, t R.. When we say that a ector is in a plane we mean that if we put the tail on the plane the head is also in the plane. Dot Product, Distance, Angles Definition 1. Let u = (u 1, u 2,, u n ), = ( 1, 2,, n ) R n. We define the dot product ( or scalar product or inner product ) of u with to be u =< u, >= n u i i. 1 It has the following properties: 1. It is symmetric, so < u, >=<, u >, u, R n. 2. It is bilinear, so < au + b, w >= a < u, w > +b <, w > and < u, a + bw >= a < u, > +b < u, w >, a, b R, u, R n. The magnitude or length of a ector u = (u 1, u 2, u n ) R n is u 2 = n u 2 i. 1 Thus we can express length in terms of the dot product: < u, u >= u 2 is the square of the length. We can also express angles using the dot product. Lemma 1. Let u, R n. Then < u, >= 0 u. Proof. We proe =. If u, then α = β ( See figure 1 for notation).thus the two triangles are congruent. Hence u has the same length as u +. We compute: while u 2 =< u, u >= 2 + u 2 2 <, u >, u + 2 =< u, u >= 2 + u 2 + 2 <, u >. We see that if u 2 = u + 2, then <, u >= 0. 2

Figure 1: For Lemma 1 To proe =. We use the aboe calculation in the reerse order. We see that < u, >= 0 implies that u 2 = u + 2. From this we can show that the two triangles are congruent which implies that α = β which, in turn, implies that the ectors meet at right angles. Algorithm Let u, R n. We show how to write = au + w, a R, w u. We say that au is the component of in the direction of u, or au is the orthogonal projection of onto u. We say that w is the component of orthogonal to u. To find these projections dot the aboe equation with u. Since < u, w >= 0 we obtain We conclude that and hence Theorem 2. Let u, R 2. Then where θ is the angle between u and. <, u >=< au, u >. a =<, u > / < u, u > w = au. < u, >= u cos(θ) Proof. Assume that u = = 1, so there are angles α, β so that u = (cos(α), sin(α)), = (cos(β, sin(β)). The dot product is cos(α) cos(β) + sin(α) sin(β) = cos(β α) = u cos(β α). This says that the theorem is true in the case where the length of both ectors is 1. We do the general case. Let u, be arbitrary. Let θ be the angle between these ectors. We can write u = au 1, = b 1 so that a, b R, a, b > 0 and the length of u 1, 1 R 2 is 1. We hae < u, >=< au 1, b 1 >= ab < u 1, 1 >= ab cos(θ) = u cos(θ). 3

Area, Volume, Determinants, Fluid Flow, Cross Products Gien an n n matrix M we can associate to it a number its determinant. We can also think of the determinant of n rows of ectors from R n. Or as n columns of ectors from R n. We outline how to compute the determinant. Here are the first two cases. If A = (a) is a 1 1 matrix with entry a R, then det(a) = a. If then det(a) = ad = bc. ( ) a b A =, c d Method One: Writing A = (a ij ) means that we hae a matrix A and the i th row, j th column entry of A is a ij. Notation: The ij-minor of a square matrix A is denoted A ij and is the matrix gotten from A by remoing the i th row and the j th column of A. We choose a row of A, say the i th row. We can choose any row; the resulting alue of the determinant will be the same no matter which row we choose. We could also use columns. The determinant of the n n matrix A is j=n det(a) = ( 1) i+j a ij det(a ij ). j=1 This method is of geometric and theoretical use. For large matrices the second method we gie is far more efficient. Method Two: We gie this method using row operations. Exactly the same method works using column operations. It uses the following properties of the determinant: I. If you interchange the position of two rows, then the determinant changes sign. II. If you add a row to a different row, then the determinant does not change. III. If you multiply a row by a constant, the resulting determinant is multiplied by that same constant. We also need the basic Fact: The determinant of an upper or lower triangular matrix is the product of its diagonal entries. 4

We do an example using the second method. Let 0 2 4 A = 1/2 1 1. 1 1 1 1/2 1 1 A 1 = 0 2 4, det(a 1 ) = det(a) Property I 1 1 1 1 2 2 A 2 = 0 2 4, det(a 2 ) = 2det(A 1 ) = 2det(A) Property III 1 1 1 1 2 2 A 3 = 0 2 4, det(a 3 ) = det(a 2 ) = 2det(A) Property II 0 1 3 1 2 2 A 4 = 0 1 3, det(a 4 ) = det(a 3 ) = 2det(A) Property I 0 2 4 1 2 2 A 5 = 0 1 3, det(a 5 ) = det(a 4 ) = 2det(A) Property I 0 0 10 By our basic fact we hae det(a 5 ) = 10. Hence det(a) = 5. Determinants, Area, Volume ( ) ( ) a1 a Let a = (a 1, a 2 ), b = (b 1, b 2 ). We write area 2 a = area for the area of the b 1, b 2 b parallelogram ( spanned ) by( the) ectros a, b. a a Claim: Area = det. b b For notation please refer to the diagram with caption Area and Determinants. We write A for the area of region A in the diagram. We hae A = b 1 a 2 B =(a 1 a 2 )/2 C = (b 1 b 2 )/2 D = b 1 a 2 F =(a 1 a 2 )/2 G = (b 1 b 2 )/2 Area of E = Whole Area (A + B + C + D + F + G) Area of E =(a + c)(b + d) (bc + ab/2 + cd/2 + bc + ab/2 + cd/2) =(ad bc). 5

Figure 2: Area and determinants Claim: Let a = (a 2, a 2, a 3 ), b = (b 1, b 2, b 3 ), c = (c 1, c 2, c 3 ) R 3. Then a (Volume of the parallelepiped spanned by(a, b, c) ) = det b. c Proof. Write V (a, b, c) for the olume of the parallelepiped spanned by the ectors a, b, c. Write D(a, b, c) for the absolute alue of the determinant of the matrix with rows a, b, c of the matrix. Both of these functions are unchanged by the operations: 1. Interchanging two rows. 2. Adding a multiple of one row to another different row. The functions are equal in the special case where the matrix proof now follows from the diagram: a b is triangular. The c V (a, b, c) Row Ops V (T riangular Matrix)?= D(a, b, c) Row Ops = D(T riangular M atrix) 6

0.0.1 Cross Product Notation: Occasionally we denote the ectors (1, 0, 0), (0, 1, 0), (0, 0, 1) by i, j, k. Definition 2. Let u = (u 1, u 2, u 3 ), = ( 1, 2, 3 ) R 3. We define the cross product i j k u = det u 1 u 2 u 3. 1 2 3 Obsere that the cross product is an element of R 3. Obsere that u = u since interchanging rows changes the sign of a determinant. Obsere that the cross product is bilinear since the determinant has this property. Obsere that we define the cross product only in R 3. Indeed we can not define the cross product in R n for n 3. In the following we indulge in some abuse of notation. It turns out to be useful. We take the dot product of two ectors in R 3 written in two different notations. ( 1 i + 2 j + 3 k) (u 1, u 2, u 3 ) = 1 u 1 + 2 u 2 + 2 u 3. Remark: In this notation the dot product is obtained by replacing i, j, k with u 1, u 2, u 3. We apply this remark. We hae i j k w (u ) (w 1, w 2, w 3 ) = det u (w 1, w 2, w 3 ) = det u. This is called Lagrange s identity. Corollary 1. The cross product u is orthogonal to u and. Proof. From Lagrange s identity we hae u (u ) u = det u = 0. We now gie a geometrical interpretation of u. We first interpret the magnitude of u up to ±1. Then we deal with the sign. Obsere that if u, hae the same direction and the magnitude of u is 1, then < u, >=. Let w be a ector that is 7

orthogonal to both u and and is of unit length. Then up to ±1 the ector w has the same direction as u. Hence < w, u >= ± u. We compute < w, u > in a second way. It is w det u which is the the olume of the parallelepiped spanned by w, u,. View the base of this figure to be the parallelogram spanned by u and. From the figure we see that the height of the parallelepiped is 1. Hence the olume of the parallelepiped is olume of the parallelepiped = area of the base height = area of the base 1. Combining these we hae w olume of the parallelepiped = area of the base = ±det u = ± u. We conclude that the magnitude of u is the area of the parallelogram spanned by u,. We know the magnitude of u b and qwe know that u is orthogonal to both u and. There re two ectors that satisfy these conditions. Which is u. Answer: We use the right hand rule. Here is an example: i j = k and i j k. 8