Chapter 2: Vector Geometry

Transcription

1 Chapter 2: Vector Geometry Daniel Chan UNSW Semester Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

2 Goals of this chapter In this chapter, we will answer the following geometric Questions 1 2 How do you define and then, compute the angle between 1 1, 1 3? How far is the point 2 from the plane 2x y + z = 3? What is the area of the parallelogram in space with sides 2 and 1? 1 1 We will in fact, generalise many geometric notions such as angle, to R n, by introducing auxiliary gadgets called the dot (or scalar) product and the cross (or vector) product of vectors. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

3 Revise length of vectors Definition The length, modulus or norm of a vector a 1 a 2 a =. Rn a n is defined as a = a a a2 n. In R 2 or R 3, a is the length of the geometric vector with coordinate vector a, the distance from the origin of the point with position vector a. Depending on the context, you should think of a like this in higher dimensions too. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

4 Angles via cosine rule Let a = OA, b = OB R n be non-zero. Let s think about what the angle θ = AOB between a and b is by considering AOB and assuming the cosine rule is valid in R n. Cosine Rule b a 2 = a 2 + b 2 2 a b cos θ. The formula for the length of a vector gives (b i a i ) 2 = ai 2 + i i i b 2 i 2 a b cos θ Cancelling gives 2 a b cos θ = i 2a ib i so i cos θ = a ib i a b We can solve for θ as long as RHS lies in [ 1, 1] (see Cauchy-Scwarz thm later). Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

5 Remarks concerning above thought experiment The distance function d(a, b) = i (b i a i ) 2 gives you information not just about lengths, but angles too. It s actually better not to base our theory on this function, but on the numerator expression for cos θ, i.e. i a ib i. Definition For a, b R n, we define the dot or scalar product of a, b to be a b = i a i b i. We prove later Theorem (Cauchy-Schwarz) a b a b a b. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

6 Angles Cauchy-Schwarz thm = we may now define Definition The angle between non-zero vectors a = OA, b = OB R n is ( ) a b θ = cos 1. a b Of course, this recovers the old definition in R 2, R 3 since the cosine rule is fine there. 1 0 Example. Let a = 1 1 and b = What is the angle between a and b? Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

7 Properties of the dot product Proposition Suppose a, b, c R n. Then 1 a b = b a. 2 a (λb) = (λa) b = λ(a b). 3 a (b + c) = a b + a c. 4 a a = a 2 0. Note the last, means that the dot product gives the length function and thus angles can be written out in terms of dot products alone too. To prove these, just write things out using the definition! Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

8 Proof of the Cauchy-Schwarz thm We prove Theorem (Cauchy-Schwarz) a b a b a b. Proof. The inequality holds when b = 0 so we assume b 0. Consider the real function of (the real variable) λ q(λ) = a λb 2 0. q(λ) = (a λb) (a λb) (1) = a 2 2λa b + λ 2 b 2. (2) The discriminant of this quadratic function of λ must be non-positive, hence 0 ( 2a b) 2 4 a 2 b 2 = 4(a b) 2 4 a 2 b 2 = (a b) 2 a 2 b 2 Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

9 Orthogonality Definition Two vectors a, b R n are said to be orthogonal if a b = 0 i.e. the angle between them is 1 2 Example. a = 1 1 and b = 1 1 are orthogonal. 1 2 Theorem (Pythagoras) If a, b R n are orthogonal then a + b 2 = a 2 + b 2. Proof. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

10 Application: Orthocentre Question Show that the altitudes of ABC are concurrent. A Let P be the intersection of the altitudes through A and B. It suffices to show that PC is an altitude too. We may pick C to be the origin O & let p, a, b, 0 be the position vectors of P, A, B, C. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

11 Orthonormal sets of vectors Definition The vectors v 1,..., v k R n form an orthogonal set if they are mutually orthogonal. If furthermore, they all have length 1, we say they are orthonormal. Equivalently, v 1,..., v k are orthonormal if { 1, if i = j, v i v j = δ ij := 0, if i j. [δ ij is called the Kronecker delta symbol.] The standard basis vector ( e 1, e 2, e 3 obviously ( ) are orthonormal E.g. The vectors v 1 = ), v 1 2 = are orthonormal Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

12 Linear combinations of orthonormal vectors E.g. Express possible). A Possible because ( ) ( ( ) as a linear combination of v 1 1 = ), v 1 2 = 2 1 (if 2 Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

13 Point-normal forms for planes in R 3 We can prescribe a plane P by giving a point a on the plane, and its orientation which is usually done by giving 2 non-parallel vectors giving directions. In R 3 the orientation, can also be given by a normal vector n, i.e. so n is perpendicular to every vector parallel to the plane. The plane P is the set of all point x such that or equivalently n (x a) = 0. (PN) These are called the point-normal form of the plane. We can re-write (3) in Cartesian form where b is the constant n a. n x = n a (3) n 1 x 1 + n 2 x 2 + n 3 x 3 = b Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

14 Example: point-normal form 1 E.g. Find the Cartesian form for the plane in R 3 with normal 1 passing 2 1 through 1. 3 E.g. Find a normal to the plane P 1 : 3x 2y + 5z = 7. Challenge Q What s the angle between P 1 and P 2 : x + y + z = 9? Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

15 Projection: informal heuristics Let b R n and v R n 0. Informally, the projection of b onto v is obtained by dropping a perpendicular from the head of b to the line through O in the direction v. This projection has form c = λv for some λ R. To determine λ, we use trig to see c = b cos θ where θ = angle between b, v. Hence c = b cos θ v v = b v v v v. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

16 Projection: formal treatment The above suggests, Definition For b R n and v R n 0, the projection of b onto v is ( ) b v proj v b = v 2 v. 1 3 E.g. Find proj v b when v = 2, b = Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

17 Properties of the projection Our definition agrees with the procedure of dropping a perpendicular by Proposition proj v b is the unique vector of the form λv such that b λv is orthogonal to v. Proof. 0 = (b λv) v = b v λ v 2 has unique soln λ = b v v 2. Proposition proj v b is the unique point on the line x = λv, λ R, closest to b. Proof. It s best to see this with a picture and use Pythagoras. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

18 Distance between a point and a line E.g. Find the point on the line x = λ 2 1, λ R, 2 closest to b = Find this distance from b to the line. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

19 Distance from a point to a plane E.g. Find the distance between the plane P : x 1 + x 2 + x 3 = 0 and b = A If c gives the point on P which is closest to b, then our argument using Pythagoras thm says that we should have b c is orthogonal to P i.e. b c is parallel to Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

20 Determinants We will look at determinants more fully in chapter 5. Here s what we need for now. Below a i, b i, e i are real (and later complex) scalars. Definition We define the 2 2 determinant by a 1 a 2 b 1 b 2 = a 1b 2 a 2 b 1. E.g. The 3 3 determinant is defined by e 1 e 2 e 3 a 1 a 2 a 3 b 1 b 2 b 3 = e 1 a 2 a 3 b 2 b 3 e 2 a 1 a 3 b 1 b 3 + e 3 a 1 a 2 b 1 b 2. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

21 Determinants and row swaps Proposition If you swap two rows of a determinant, it changes sign e.g. e 1 e 2 e 3 a 1 a 2 a 3 b 1 b 2 b 3 = e 1 e 2 e 3 b 1 b 2 b 3 a 1 a 2 a 3 In particular, the determinant is 0 if 2 rows are the same (swapping them both negates and keeps them the same). Proof For 2 2-matrices b 1 b 2 a 1 a 2 = b 1a 2 b 2 a 1 = a 1 a 2 b 1 b case is harder Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

22 Cross product For a, b R 3, the cross or vector product of a and b is the vector e 1 e 2 e 3 a b = a 1 a 2 a 3 b 1 b 2 a 3 a = e 2 a 3 1 b 2 b 3 e 2 a 1 a 3 b 1 b 3 + e 3 a 1 a 2 b 1 b 2 N.B. The second term is not really well-defined, but is a useful mnemonic. 1 4 E.g. Find N.B. There are higher dimensional versions of the cross product, but they are much more complicated and not as useful. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

23 Arithmetic properties of the cross product Proposition For a, b, c R 3 and λ R we have 1 a a = 0 (by row swapping & determinants). 2 is distributive: a (b + c) = (a b) + (a c) (a + b) c = (a c) + (b c). 3 (λa) b = λ(a b) = a (λb) Proof. Just expand both sides with the defn! Alternately, wait until we ve looked at more properties of determinants in chapter 5. Warning a b = b a (by row swapping of determinants) so is not commutative! The cross product is NOT associative! Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

24 The magnitude of a b Theorem Suppose that a, b R 3 are nonzero, with angle θ between them. Then a b = a b sin θ. Proof. Remember that θ [0, π] in R 3 so sin θ 0. Thus, it is enough to show that a b 2 = a 2 b 2 sin 2 θ. Remembering the definition of θ: ( ( )) ( ) 2 a b a b sin 2 θ = 1 cos 2 cos 1 = 1 a b a b and so Now get MAPLE to expand out a 2 b 2 sin 2 θ = a 2 b 2 (a b) 2. a b 2 ( a 2 b 2 (a b) 2 ) in terms of the coordinates of a, b & check it is zero! Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

25 Areas of parallelograms via cross product Proposition The area of the parallelogram with sides a, b is a b. Proof The area of the parallelogram is Thm previous slide gives the result. A = base x perp height = a b sin θ. E.g. Find the area of the parallelogram with vertices at (1, 1), (4, 2), (2, 3) and (5, 4). Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

26 Scalar triple product Let a, b, e R 3. Proposition-Definition The scalar e 1 e 2 e 3 e (a b) = a 1 a 2 a 3 b 1 b 2 b 3 It is called the scalar triple product of e, a, b. Proof If c = a b then e 1 e 2 e 3 e (a b) = e 1 c 1 + e 2 c 2 + e 3 c 3 = a 1 a 2 a 3 b 1 b 2 b 3. E.g. Find e 1 (e 2 e 3 ). Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

27 Direction of a b Row swapping of determinants give the following Proposition 1 a (b c) = a (c b) = c (a b). 2 a (a b) = 0 = a (b b). Part 2) gives Proposition The vector a b is orthogonal to both a and b. N.B This proposition and our formula for a b determines a b up to a choice of two vectors. The choice of which one is given by the right hand rule. Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

28 Application: Parametric to cartesian form via point-normal There are many geometric problems in R 3 where one needs to find a vector which is orthogonal to two given vectors. Example. Find a point-normal, and hence a Cartesian form for the plane x = 1 + λ λ 2 0, λ 1, λ 2 R A A point on the plane is a = A vector normal to the plane is n = Thus a point-normal form is Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

29 Application: distance between lines Problem. What is the shortest distance between the two lines L 1, L 2 R 3? Our argument via Pythagoras thm shows that the shortest line segment joining the two lines needs to be orthogonal to both the lines, that is orthogonal to the two direction vectors. Proposition Let n be orthogonal to both lines. Then the shortest distance between the lines equals the length of the vector proj n (a 1 a 2 ), where a j is any point on L j. Why? Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

30 Example: distance between lines Problem. What is the shortest distance between the two lines L 1 : x = 0 + λ 2, λ R L 2 : x = 1 + µ 0, µ R? A Here we can take n = 2 0 = 2, and a 1 = 0, a 2 = 1, so a 1 a 2 = 1. 1 The shortest distance between the lines is proj n (a 1 a 2 ) = n (a 1 a 2 ) n 2 n = n (a 1 a 2 ) n = Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

31 Volumes of parallelepipeds A parallelepiped is a 3-dim version of a parallelogram. Consider a parallelepiped P with edges a = OA, b = OB, c = OC. Let n = b c. If the base of P is the parallelogram sides b, c, then the perpendicular height P is the length of the projection of a onto n. Hence Volume of P = area base perp. height = b c proj n a = b c a (b c) b c = a (b c). Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32

32 Example: volume of parallelepiped Example. Find the volume of a parallelepiped with vertices at 0 0, 2 4 2, adjacent to the vertex A Daniel Chan (UNSW) Chapter 2: Vector Geometry Semester / 32