MATH 19520/51 Class 2

Transcription

1 MATH 19520/51 Class 2 Minh-Tam Trinh University of Chicago

2 1 Review dot product. 2 Angles between vectors and orthogonality. 3 Projection of one vector onto another. 4 Cross product and its properties. 5 Computing cross products from determinants. 6 Computing volume from cross products.

3 Review of Dot Product Example In R 3, find a vector of length 650 in the direction of v = (3, 4, 12).

4 Review of Dot Product Example In R 3, find a vector of length 650 in the direction of v = (3, 4, 12). First, find a vector of length 1, or unit vector, in that direction: (1) v v = ( 3 13, 4 13, ). Next, rescale by a factor of 650: The answer is (2) 650 v v = (150, 200, 600).

5 What are the magnitudes of the following vectors? (3) (4, 3), (3, 4), (0, 5), ( 1.4, 4.8), ( 3, 4), ( 4, 3)

6 What are the magnitudes of the following vectors? (3) (4, 3), (3, 4), (0, 5), ( 1.4, 4.8), ( 3, 4), ( 4, 3) Draw them! What are the following dot products? (4) (4, 3) (4, 3) =? (4, 3) (3, 4) =? (4, 3) (0, 5) =? (4, 3) ( 1.4, 4.8) =? (4, 3) ( 3, 4) =? (4, 3) ( 4, 3) =?

7 What are the magnitudes of the following vectors? (3) (4, 3), (3, 4), (0, 5), ( 1.4, 4.8), ( 3, 4), ( 4, 3) Draw them! What are the following dot products? (4) (4, 3) (4, 3) = 25 (4, 3) (3, 4) = 24 (4, 3) (0, 5) = 15 (4, 3) ( 1.4, 4.8) = 8.8 (4, 3) ( 3, 4) = 0 (4, 3) ( 4, 3) = 7

8 Angles If u, v R n are nonzero vectors, then there is a well-defined angle θ u, v [0, π] between the direction of u and the direction of v. It turns out that (5) u v = u v cos θ u, v.

9 Angles If u, v R n are nonzero vectors, then there is a well-defined angle θ u, v [0, π] between the direction of u and the direction of v. It turns out that (5) u v = u v cos θ u, v. Suppose the lengths of u and v are held fixed, but we vary their directions. 1 When is u v maximized? 2 When is u v minimized? 3 When is u v zero?

10 Angles If u, v R n are nonzero vectors, then there is a well-defined angle θ u, v [0, π] between the direction of u and the direction of v. It turns out that (5) u v = u v cos θ u, v. Suppose the lengths of u and v are held fixed, but we vary their directions. 1 When is u v maximized? u, v in the same direction 2 When is u v minimized? u, v in opposite directions 3 When is u v zero?

11 We say that u and v are orthogonal iff u v = 0, i.e., either they re perpendicular or one of them is the zero vector.

12 We say that u and v are orthogonal iff u v = 0, i.e., either they re perpendicular or one of them is the zero vector. Example In R 3, what are the vectors (x, y, z) orthogonal to (1, 0, 0)?

13 We say that u and v are orthogonal iff u v = 0, i.e., either they re perpendicular or one of them is the zero vector. Example In R 3, what are the vectors (x, y, z) orthogonal to (1, 0, 0)? Since (6) (x, y, z) (1, 0, 0) = x, the vectors orthogonal to (1, 0, 0) must take the form (0, y, z). The set of all such vectors is the (y, z)-plane.

14 Projections Given nonzero u, v R n, we can always write u as the sum of: 1 A vector orthogonal to v. 2 A vector in the direction of + v or v, depending on θ u, v. It is called the vector projection of u onto v and denoted proj v u. u u (7) proj v u v proj v u v Since the orthogonal part doesn t contribute to u v, we have u v = (proj v u) v.

15 The scalar projection of u onto v is the scalar, comp v u, such that: (8) proj v u = (comp v u) v v.

16 The scalar projection of u onto v is the scalar, comp v u, such that: (8) proj v u = (comp v u) v v. Taking the dot product with v on both sides and using u v = (proj v u) v, we obtain (9) comp v u = u v. v

17 The scalar projection of u onto v is the scalar, comp v u, such that: (8) proj v u = (comp v u) v v. Taking the dot product with v on both sides and using u v = (proj v u) v, we obtain (9) comp v u = u v. v Therefore, (10) proj v u = ( ) u v v 2 v.

18 Example What are the scalar and vector projections of u = (0, 0, 4) onto v = (1, 1, 1)?

19 Example What are the scalar and vector projections of u = (0, 0, 4) onto v = (1, 1, 1)? We compute u v = 4 and v = 3, so (11) comp v u = u v v = 4 3 = and (12) proj v u = ( ) ( u v 4 v 2 v = 3, 4 3, 4 ). 3

20 Example What is the distance between (0, 0, 4) and the line x = y = z?

21 Example What is the distance between (0, 0, 4) and the line x = y = z? The shortest segment between a point and a line is always perpendicular to the line: (13) (0, 0, 4) u (0, 0, 0) ( 4 3, 4 3, 4 3 ) x = y = z In the notation of the previous example, the answer is (14) u proj v u = (0, 0, 4) ( 4 3, 4 3, 4 3 ) = ( 4 3, 4 3, 8 3 ) =

22 Cross Product The cross product is a special operation that only works in R 3. Given u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ), we define (15) u v = (u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ).

23 Cross Product The cross product is a special operation that only works in R 3. Given u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ), we define (15) u v = (u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ). This operation is useful because it produces a vector orthogonal to the original vectors: (16) (17) u ( u v) = u 1 (u 2 v 3 u 3 v 2 ) + u 2 (u 3 v 1 u 1 v 3 ) + u 3 (u 1 v 2 u 2 v 1 ) = 0. v ( u v) = v 1 (u 2 v 3 u 3 v 2 ) + v 2 (u 3 v 1 u 1 v 3 ) + v 3 (u 1 v 2 u 2 v 1 ) = 0.

24 Example (1, 0, 0) (0, 1, 0) = (0, 0, 1). More generally, the RHR gives the direction of u v: (18) u v u v (In the figure, u and v are meant to point out of the page. The angle between them need not be π/2.)

25 The cross product is anticommutative: (19) u v = v u. It distributes over vector addition: (20) u ( v + w) = u v + u w.

26 The cross product is anticommutative: (19) u v = v u. It distributes over vector addition: (20) u ( v + w) = u v + u w. Warning! Repeated cross products make sense, but is not associative: (21) ( u v) w u ( v w) in general.

27 While the dot product depends on cos θ, the magnitude of the cross product depends on sin θ: (22) u v = u v sin θ u, v.

28 While the dot product depends on cos θ, the magnitude of the cross product depends on sin θ: (22) u v = u v sin θ u, v. Example Since θ u, u = 0, we always have u u = u 2 sin 0 = 0. That is, (23) u u = 0 for all u.

29 Cross Products via Determinants How to remember the formula for? First, define the determinant of a 2 2 matrix by a b (24) c d = ad bc. Then (25) ( ) u u v = 2 u 3 v 2 v 3, u 3 u 1 v 3 v 1, u 1 u 2 v 1 v 2. Notice how the subscripts get permuted cyclically.

30 Cross Products via Determinants How to remember the formula for? First, define the determinant of a 2 2 matrix by a b (24) c d = ad bc. Then (25) ( ) u u v = 2 u 3 v 2 v 3, u 3 u 1 v 3 v 1, u 1 u 2 v 1 v 2. Notice how the subscripts get permuted cyclically. Warning! Order matters: a b c d = b a d c.

31 Area and Volume via Cross Products The angle formula is also useful because, by trigonometry, u v = u v sin θ is the area of the parallelogram (26) v θ u

32 Example Find the area of a parallelogram in R 3 with vertices at (1, 1, 1), (1, 3, 4), (5, 6, 0).

33 Example Find the area of a parallelogram in R 3 with vertices at (1, 1, 1), (1, 3, 4), (5, 6, 0). We need two sides starting at the same vertex. Let (27) u = (1, 3, 4) (1, 1, 1) = (0, 2, 3), v = (5, 6, 0) (1, 1, 1) = (4, 5, 1). We compute ( ) 2 3 u v = 5 1, , 0 2 (28) 4 5 = ( 17, 12, 8) and u v = ( 17) ( 8)

34 By combining the angle formula for dot product with the area formula, one can show that w ( u v) is the volume of the parallelpiped (29) w v u

35 Example If u = (1, 0, 0) and v = (0, 2, 3) and w = (1, 4, 6), then what is the value of w ( u v)?

36 Example If u = (1, 0, 0) and v = (0, 2, 3) and w = (1, 4, 6), then what is the value of w ( u v)? In general, w ( u v) is the volume of (30) (w 1, w 2, w 3 ) (v 1, v 2, v 3 ) (0, 0, 0) (u 1, u 2, u 3 ) But in this case, w = u + 2 v, so the parallelpiped is flat: (31) (It lies within a plane.) The volume is 0.