Three-Dimensional Space; Vectors

Transcription

1 Chapter 3 Three-Dimensional Space; Vectors 3.1 Rectangular Coordinates in 3-Space; Spheres Rectangular Coordinate Sstems To begin, consider three mutuall perpendicular coordinate lines, called the -ais, the -ais, and the -ais, positioned so that their origin coincide. O The three coordinate aes form a three-dimensional rectangular coordinate sstem (or Cartesian coordinate sstem). The point of intersection of the coordinate aes is called the origin of the coordinate sstem. The coordinate aes, taken in pairs, determine three coordinate planes: the plane, the -plane, and the -plane, which divide space into eight octants. plane plane O plane 44

2 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 45 To each point P in 3-space corresponds to ordered triple of real numbers (a,b,c) which measure its directed distances from the three planes. We call a, b, and c the -coordinate, -coordinate, and -coordinate of P, respectivel, and we denote the point P b (a,b,c) or b P(a,b,c). Distance in 3-Space An important concept in 3-space is that of distance between two points. If two points, P 1 and P 2 have the coordinates of ( 1, 1, 1 ) and ( 2, 2, 2 ), then the distance is given b the following formula. d = ( 2 1 ) 2 +( 2 1 ) 2 +( 2 1 ) 2 Eample 3.1 Find the distance d between the points (2, 3,4) and ( 3,2, 5). Spheres Recall that thestandard equation of a circle in2-space that has center ( 0, 0 ) and radius r is ( 0 ) 2 +( 0 ) 2 = r 2. Analogousl, standard equation of the sphere in 3-space that has center ( 0, 0, 0 ) and radius r is ( 0 ) 2 +( 0 ) 2 +( 0 ) 2 = r 2 (3.1) If the terms in (3.1) are epanded and like terms are collected, then the resulting equation has the form Eample G+H +I +J = 0. (3.2) Find the center and radius of the sphere = 0.

3 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 46 In general, completing the squares in (3.2) produces an equation of the form ( 0 ) 2 +( 0 ) 2 +( 0 ) 2 = k. If k > 0, then the graph of this equation is a sphere with center ( 0, 0, 0 ) and radius k. If k = 0, then thesphere hasradius ero, so thegraphis thesingle point ( 0, 0, 0 ). If k < 0, the equation is not satisfied b an values of,, and, so it has no graph. Theorem 3.1 An equation of the form G+H +I +J = 0. represents a sphere, a point, or has no graph. Eercise A cube of side 4 has its geometric center at the origin and its faces parallel to the coordinate planes. Sketch the cube and give the coordinates of the corners. 2. Suppose that a bo has its faces parallel to the coordinate planes and the points (4,2, 2) and ( 6,1,1) are endpoints of a diagonal. Sketch the bo and give the coordinates of the remaining si corners. 3. Interpret the graph of = 1 in the contets of (a) a number line (b) 2-space (c) 3-space 4. Find the center and radius of the sphere that has (1, 2,4) and (3,4, 12) as endpoints of a diameter. 5. Show that (4,5,2), (1,7,3), and (2,4,5) are vertices of an equilateral triangle. 6. (a) Show that (2,1,6), (4,7,9), and (8,5, 6) are the vertices of a right triangle. (b) Which verte is at the 90 angle? (c) Find the area of the triangle. 7. Find equations of two spheres that are centered at the origin and are tangent to the sphere of radius 1 centered at (3, 2,4) Describe the surface whose equation is given = = 0

4 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana = = = = If a bug walks on the sphere = 0 how close and how far can it get from the origin? 15. The distance between a point P(,, ) and the point A(1, 2, 0) is twice the distance between P and the point B(0,1,1). Show that the set of all such points is a sphere, and find the center and radius of the sphere. Answers to Eercise (4,2, 2),(4,2,1),(4,1,1),(4,1, 2),( 6,1,1),( 6,2,1),( 6,2, 2),( 6,1, 2) 3. (a) point (b) line parallel to the -ais (c) plane parallel to the -plane 4. radius 74, center (2,1, 4) 6. (b) (2,1,6) (c) area 49 ( 1 8. sphere, center ( 5, 2, 1), radius sphere, center 2, 3 ) 4, 5, radius no graph 14. largest distance, 3+ 6; smallest distance, Vectors Some phsical quantities such as length, area, volume, and mass can be completel described b a single real number. Because these quantities are describable b giving onl a magnitude, the are called scalars. On the other hand phsical quantities such as displacement, velocit, force, and acceleration require both a magnitude and a direction to completel describe them. Such quantities are called vectors. To distinguish between scalars and vectors we will denote scalars b lower case italic tpe such as a, b, c, etc. and denote vectors b lower case boldface tpe such as u, v, w, etc. Because vectors are determined b both a magnitude and a direction, the are represented geometricall in 2- or 3-space as directed line segments or arrows. The length of the arrow corresponds to the magnitude of the vector while the direction of the arrow corresponds to the direction of the vector. The tail of the arrow is called the initial point of the vector while the tip of the arrow is called the terminal point of the vector. If the vector v has the point P as its initial point and the point Q as its terminal point we will write v = PQ.

5 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 48 P Two vectors u and v, which have the same length and same direction, are said to be equal vectors (or equivalent) even though the have different initial points and different terminal points. If u and v are equal vectors we write u = v. The three vectors in the following figure are equal, even though the are in different positions. v Q If the initial and terminal points of a vector coincide, then the vector has length ero; we call this the ero vector and denote it b 0. The ero vector does not have a specific direction, so we will agree that it can be assigned an convenient direction in a specific problem. Definition 3.1 If u and v are vectors, then the sum u+v is the vector from the initial point of u to the terminal point of v when the vectors are positioned so the initial point of v is at the terminal point of u. v u u+v Definition 3.2 For a scalar k and a nonero vector v, the scalar multiple of v b k, denoted b kv, is the vector whose magnitude is k times the length of v, points in the same direction as v if k > 0, points in the opposite direction as v if k < 0. We define kv = 0 if k = 0 or v = 0. v 2v 1 v 2 ( 1)v ( 3 2 )v

6 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 49 Observe that if k and v are nonero, then the vectors v and kv lie on the same line if their initial points coincide and lies on parallel or coincident lines if the do not. Thus, we sa that v and kv are parallel vectors. Observealsothatthevector( 1)vhasthesamelengthasvbutisoppositeldirected. We call ( 1)v the negative of v and denote it b v. In particular, 0 = ( 1)0 = 0. Vector subtraction is defined in terms of addition and scalar multiplication b u v = u+( v) In the special case where u = v, their difference is 0; that is, Vectors in Coordinate Sstems u+( u) = u u = 0 If a vector v is positioned with its initial point at the origin of the rectangular coordinate sstem, then the terminal point will have coordinates of the form (v 1,v 2 ) or (v 1,v 2,v 3 ), depending on whether the vector is in 2-space or 3-space. We call these coordinates the components of v, and we write v in component form using the bracket notation v = v 1,v 2 or v = v 1,v 2,v 3 (v 1,v 2 ) (v 1,v 2,v 3 ) v v In particular, the ero vectors in 2-space and 3-space are respectivel. Theorem 3.2 v = 0,0 and v = 0,0,0 Two vectors are equivalent if and onl if their corresponding components are equal. For eample, if and onl if a = 2, b = 3, and c = 5. a,b,c = 2,3,5

7 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 50 Arithmetic Operations on Vectors Theorem 3.3 If u = u 1,u 2 and v = v 1,v 2 are vectors in 2-space and k is an scalar, then u+v = u 1 +v 1,u 2 +v 2 u v = u 1 v 1,u 2 v 2 ku = ku 1,ku 2 Similarl, if u = u 1,u 2,u 3 and v = v 1,v 2,v 3 are vectors in 3-space and k is an scalar, then u+v = u 1 +v 1,u 2 +v 2,u 3 +v 3 u v = u 1 v 1,u 2 v 2,u 3 v 3 ku = ku 1,ku 2,ku 3 Eample 3.3 If u = 1,2, 3 and v = 2,0, 4, then u+v = 3u = 2v = v 3u = Vectors With Initial Point Not At The Origin Suppose that P( 1, 1 ) and Q( 2, 2 ) are points in 2-space and we interested in finding the components of the vector PQ. As illustrated in the following figure, we can write this vector as PQ = OQ OP = 2, 2 1, 1 = 2 1, 2 1 P( 1, 1 ) OP PQ OQ Q( 2, 2 ) O

8 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 51 Theorem 3.4 If PQ is a vector in 2-space with initial point P( 1, 1 ) and terminal point Q( 2, 2 ), then PQ = 2 1, 2 1 Similarl, if PQisavectorin3-spacewithinitialpointP( 1, 1, 1 )andterminal point Q( 2, 2, 2 ), then PQ = 2 1, 2 1, 2 1 Eample 3.4 In 2-space the vector from P(3,2) to Q( 1,4) is PQ = 1 3,4 2 = 4,2 and in 3-space the vector from A(1, 2,0) to B( 3,1,2) is AB = 3 1,1 ( 2),2 0 = 4,3,2 Rules of Vector Arithmetic Theorem 3.5 For an vectors u, v, and w and an scalars k and l, the following relationships hold: (a) u+v = v+u (b) (u+v)+w = u+(v+w) (c) u+0 = 0+u = u (d) u+( u) = 0 (e) k(lu) = (kl)u (f) k(u+v) = ku+kv (g) (k +l)u = ku+lu (h) 1u = u Norm of a Vector The distance between the initial and terminal points of a vector v is called the length, the norm, or the magnitude of v and is denoted b v. This distance does not

9 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 52 change if the vector is translated. The norm of a vector v = v 1,v 2 in 2-space is given b v = v 2 1 +v2 2 and the norm of a vector v = v 1,v 2,v 3 in 3-space is given b v = v 2 1 +v2 2 +v2 3 Eample 3.5 Find the norm of v = 4, 2 and w = 1,3,5. For an vector v and scalar k, the length of kv is For eample, Unit Vectors kv = k v 5v = 5 v = 5 v 3v = 3 v = 3 v v = 1 v = v A vector of length 1 is called a unit vector. v Notice that for an nonero vector v, the vector is a unit vector which points in the v same direction as v, since 1 > 0 and v v v = v = 1. v Dividing a nonero vector v b v is often called normaliing v. Eample 3.6 Find the unit vector that has the same direction as v = 2i j 2k.

10 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 53 There are specific unit vectors which we will often use, called the basis vectors: i = 1,0 and j = 0,1 in R 2 ; i = 1,0,0, j = 0,1,0, and k = 0,0,1 in R 3. These are useful for several reasons: the are mutuall perpendicular, since the lie on distinct coordinate aes; the are all unit vectors: i = j = k = 1; 1 1 j i 1 i k j 1 1 ever vector can be written as a unique scalar combination of the basis vectors: v = a,b = ai+bj in R 2, v = a,b,c = ai+bj+ck in R 3. Eample space 3-space 3, 4 = 3i 4j 2,3, 5 = 2i+3j 5k 5,0 = 5i+0j = 5i 0,0,3 = 3k 0,0 = 0i+0j = 0 0,0,0 = 0i+0j+0k = 0 (3i 2j)+(i+4j) = 2i+2j (2i j+3k)+(3i+2j k) = i+j+2k 3(2i 4j) = 6i 12j 2(3i+4j k) = 6i+8j 2k 3i+4j = = 5 2i j+3k = 2 2 +( 1) = 14 v 1 i+v 2 j = v1 2 +v2 2 v 1,v 2,v 3 = v1 2 +v2 2 +v2 3

11 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 54 Vectors Determined b Length and Angle If v is a nonero vector with its initial point at the origin of an -coordinate sstem, and if θ is the angle from the positive -ais to the radial line through v, then the -component of v can be written as v cosθ and the -component as v sinθ; v v sinθ θ v cosθ and hence v can be epressed in trigonometric form as v = v cosθ,sinθ or v = v cosθi+ v sinθj (3.3) In the special case of a unit vector u this simplifies to u = cosθ,sinθ or u = cosθi+sinθj (3.4) Eample 3.8 (a) Find the vector of length 3 that makes an angle of π/3 with the positive -ais. (b) Find the angle that the vector v = i 3j makes with the positive -ais.

12 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 55 Vectors determined b length and a vector in the same direction In man applications, a direction in 2-space or 3-space is determined b some known unit vector u, and it is interest to find the components of a vector v that has the same direction as u and some specified length v. This can be done b epressing v as v = v u and then reading off the components of v u. Eample 3.9 Find the vector v of length 5 and has the same direction as the vector from the point A(0,0,4) to the point B(2,5,0). Resultant of Two Concurrent Forces The effect that a force has on an object depends on the magnitude and direction of the forces and the point at which it is applied. If two forces F 1 and F 2 are applied at the same point on an object, then the two forces have the same effect on the object as the single force F 1 +F 2 applied at the point. F 2 F 1 +F 2 F 1 Phsicists and engineers call F 1 +F 2 the resultant of F 1 and F 2, and the sa that the forces F 1 and F 2 are concurrent to indicate that the are applied at the same point.

13 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 56 Eample 3.10 Suppose that two forces are applied to an ee bracket, as show in Figure below. Find the magnitude of the resultant and the angle θ that it makes with the positive -ais. F 2 = 300N 40 F 1 = 200N 30 Eercise Sketch the vectors with their initial points at the origin. 1. (a) 2,5 (b) 5, 4 (c) 2,0 (d) 5i+3j (e) 3i 2j (f) 6j 2. (a) 3,7 (b) 6, 2 (c) 0, 8 (d) 4i+2j (e) 2i j (f) 4i

14 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana (a) 1, 2,2 (b) 2,2, 1 (c) i+2j+3k (d) 2i+3j k 4. (a) 1,3,2 (b) 3,4,2 (c) 2j k (d) i j+2k 5 6 Find the components of the vector P 1 P (a) P 1 (3,5), P 2 (2,8) (b) P 1 (7, 2), P 2 (0,0) (c) P 1 (5, 2,1), P 2 (2,4,2) 6. (a) P 1 ( 6, 2), P 2 ( 4, 1) (b) P 1 (0,0,0), P 2 ( 1,6,1) (c) P 1 (4,1, 3), P 2 (9,1, 3) 7. (a) Find the terminal point of v = 3i 2j if the initial point is (1, 2). (b) Find the terminal point of v = 3,1,2 if the initial point is (5,0, 1). 8. (a) Find the terminal point of v = 7,6 if the initial point is (2, 1). (b) Find the terminal point of v = i+2j 3k if the initial point is ( 2,1,4) Perform the stated operations on the vectors u, v, and w. 9. u = 3i k, v = i j+2k, w = 3j (a) w v (b) 6u+4w (c) v 2w (d) 4(3u+v) (e) 8(v+w)+2u (f) 3w (v w) 10. u = 2, 1,3, v = 4,0, 2, w = 1,1,3 (a) u w (b) 7v+3w (c) w+v (d) 3(u 7v) (e) 3v 8w (f) 2v (u+w) Find the norm of v. 11. (a) v = 1, 1 (b) v = i+7j (c) v = 1,2,4 (d) v = 3i+2j+k 12. (a) v = 3,4 (b) v = 2i 7j (c) v = 0, 3,0 (d) v = i+j+k 13. Let u = i 3j+2k, v = i+j, and w = 2i+2j 4k. Find (a) u+v (b) u + v (c) 2u +2 v 1 (d) 3u 5v+w (e) w w (f) 1 w w

15 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Find the unit vectors that satisf the stated conditions. 14. (a) Same direction as i+4j. (b) Oppositel directed to 6i 4j+2k. (c) Same direction as the vector from the point A( 1,0,2) to the point B(3,1,1). 15. (a) Oppositel directed to 3i 4j. (b) Same direction as 2i j 2k. (c) Same direction as the vector from the point A( 3,2) to the point B(1, 1) Find the vectors that satisf the stated conditions. 16. (a) Oppositel directed to v = 3, 4 and half the length of v. (b) Length 17 and same direction as v = 7,0, (a) Same direction as v = 2i+3j and three times the length of v. (b) Length 2 and oppositel directed to v = 3i+4j+k. 18. Ineach part, findthe component formofthe vector v in2-space that hasthe stated length and makes the stated angle θ with the positive -ais. (a) v = 3; θ = π/4 (b) v = 2; θ = 90 (c) v = 5; θ = 120 (d) v = 1; θ = π 19. Find the component form of v + w and v w in 2-space, given that v = 1, w = 1, v makes an angle of π/6 with the positive -ais, and w makes an angle of 3π/4 with the positive -ais. 20. Let u = 1,3, v = 2,1, and w = 4, 1. Find the vector that satisfies 2u v+ = 7+w. 21. Let u = 1,1, v = 0,1, and w = 3,4. Find the vector that satisfies u 2 = w+3v. 22. Find u and v if u+2v = 3i k and 3u v = i+j+k. 23. Find u and v if u+v = 2, 3 and 3u+2v = 1, In each part, find two unit vectors in 2-space that satisf the stated condition. (a) Parallel to the line = 3+2 (b) Parallel to the line + = 4 (c) Perpendicular to the line = 5+1

16 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 59 Answers to Eercise (a) 1,3 (b) 7,2 (c) 3,6,1 7. (a) (4, 4) (b) (8, 1, 3) 9. (a) i+4j 2k (b) 18i+12j 6k (c) i 5j 2k (d) 40i 4j 4k (e) 2i 16j 18k (f) i+13j 2k 11. (a) 2 (b) 5 2 (c) 21 (d) (a)2 3 (b) (c) (d)2 37 (e)(1/ 6)i+(1/ 6)j (2 6)k (f) (a) ( 1/ 17)i+(4/ 17)j (b) ( 3i+2j k)/ 14 (c) (4i+j k)/(3 2) 16. (a) 32,2 1 (b) 7,0, (a) 3 2/2,3 3/2 (b) 0,2 (c) 5/2,5 3/2 (d) 1, ,1 22. u = 5 7 i+ 2 7 j+ 1 7 k, v = 8 7 i 1 7 j 4 7 k 24. (a) 1/ 10,3/ 10, 1/ 10, 3/ 10 (b) 1/ 2, 1/ 2, 1/ 2,1/ 2 (c) ± ,1 3.3 Dot Product; Projection Definition 3.3 Let u = u 1,u 2,u 3 and v = v 1,v 2,v 3 be vectors in 3-space. The dot product or scalar product of u and v, denoted b u v, is given b: u v = u 1 v 1 +u 2 v 2 +u 3 v 3 Similarl, for vectors u = u 1,u 2 and v = v 1,v 2 in 2-space, the dot product is: u v = u 1 v 1 +u 2 v 2 Note that the dot product of two vectors is a scalar. For eample, 4, 3 3,2 = (4)(3)+( 3)(2) = 6 1,2, 3 4, 1,2 = (1)(4)+(2)( 1)+( 3)(2) = 4

17 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 60 Algebraic Properties of the Dot Product Theorem 3.6 If u, v, and w are vectors in 2-space or 3-space and k is a scalar, then: (a) u v = v u (b) u (v+w) = u v+u w (c) k(u v) = (ku) v = u (kv) (d) v v = v 2 (e) 0 v = 0 Definition 3.4 The angle between two nonero vectors with the same initial point is the smallest angle between them. We will alwas choose the smallest nonnegative angle θ between them, so that 0 θ π. u θ v u θ v u θ v u θ v Theorem 3.7 Let u and v be nonero vectors, and let θ be the angle between them. Then cosθ = u v u v (3.5) Eample 3.11 Find the angle between (a) u = 4, 3, 1 and v = 2, 3,5 (b) u = 4,+5,1 and v = 2,3, 7 (c) u = i 2j+2k and v = 3i+6j 6k

18 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 61 Two nonero vectors are perpendicular if the angle between them is 90. Since cos90 = 0, we have the following important corollar to Theorem 3.7: Corollar 3.1 Two nonero vectors u and v are perpendicular if and onl if u v = 0. We will write u v to indicate that u and v are perpendicular. Since cosθ > 0 for 0 θ < 90 and cosθ < 0 for 90 < θ 180, we also have: Corollar 3.2 If θ is the angle between nonero vectors u and v, then > 0 for 0 θ < 90 u v is = 0 for θ = 0 < 0 for 90 < θ 180 u θ v u θ v u θ v u v > 0 u v < 0 u v = 0 Direction Angles In both 2-space and 3-space the angle between a nonero vector v and the vectors i, j, andkarecalled the direction angles of v, andthe cosines ofthese angles arecalled the direction cosines of v. Formulas for the direction cosines of a vector can be obtained form Formula (3.5). For eample, if v = v 1 i+v 2 j+v 3 k, then cosα = cosβ = cosγ = v i v i = v 1 v, v j v j = v 2 v, v k v k = v 3 v.

19 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 62 j v k v β α i γ α β j i Theorem 3.8 The direction cosines of a nonero vector v = v 1 i+v 2 j+v 3 k are cosα = v 1 v, cosβ = v 2 v, cosγ = v 3 v. The direction cosines of a vector v = v 1 i+v 2 j+v 3 k can be computed b normaliing v and reading off the components of v/ v, since v v = v 1 v i+ v 2 v j+ v k v k = (cosα)i+(cosβ)j+(cosγ)k Moreover, we can show that the direction cosines of a vector satisf the equation cos 2 α+cos 2 β +cos 2 γ = 1 Eample 3.12 Find the direction angles of the vector v = 4i 5j+3k.

20 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 63 Decomposing Vectors into Orthogonal Components Our net objective is to develop a computational procedure for decomposing a vector into sum of orthogonal vectors. For this purpose, suppose that e 1 and e 2 are two orthogonal unit vectors in 2-space, and suppose that we want to epress a given vector v as a sum v = w 1 +w 2 so that w 1 is a scalar multiple of e 1 and w 2 is a scalar multiple of e 2. w 2 e 2 v e 1 w 1 That is, we want to find scalars k 1 and k 2 such that v = k 1 e 1 +k 2 e 2 (3.6) We can find k 1 b taking the dot product of v with e 1. This ields Similarl, v e 1 = (k 1 e 1 +k 2 e 2 ) e 1 = k 1 (e 1 e 1 )+k 2 (e 2 e 1 ) = k 1 e = k 1 v e 2 = (k 1 e 1 +k 2 e 2 ) e 2 = k 1 (e 1 e 2 )+k 2 (e 2 e 2 ) = 0+k 2 e 2 2 = k 2 Substituting these epressions for k 1 and k 2 in (3.6) ields v = (v e 1 )e 1 +(v e 2 )e 2 (3.7) In this formula we call (v e 1 )e 1 and (v e 2 )e 2 the vector components of v along e 1 and e 2, respectivel; and we call v e 1 and v e 2 the scalar components of v along e 1 and e 2, respectivel. If θ denote the angle between v and e 1, then v e 1 = v cosθ and v e 2 = v sinθ and the decomposition (3.6) can be epressed as v = ( v cosθ)e 1 +( v sinθ)e 2 (3.8)

21 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 64 Eample 3.13 Let 1 v = 2,3, e 1 = 2, 1 2, e 2 = 1 1, 2 2 Find the scalar components of v along e 1 and e 2 and the vector components of v along e 1 and e 2. Eample 3.14 A rope is attached to a 100-lb block on a ramp that is inclined at an angle of 30 with the ground. 30 How much force does the block eert against the ramp, and how much force must be applied to the rope in a direction parallel to the ramp to prevent the block from sliding down the ramp?

22 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 65 Orthogonal Projections The vector components of v along e 1 and e 2 in (3.7) are also called the orthogonal projections of v on e 1 and e 2 and are denoted b proj e1 v = (v e 1 )e 1 and proj e2 v = (v e 2 )e 2 In general, if e is a unit vector, then we define the orthogonal projection of v on e to be proj e v = (v e)e (3.9) The orthogonal projection of v on an arbitrar nonero vector b can be obtained b normaliing b and then appling Formula (3.9); that is, ( )( ) b b proj b v = v b b which can be rewritten as proj b v = v b b 2 b (3.10) Moreover, if we subtract proj b v from v, then the resulting vector v proj b v will be orthogonal to b; we call this the vector component of v orthogonal to b. v proj b v v v v proj b v b proj b v proj b v b Acute angle between v and b Obtuse angle between v and b Eample 3.15 Find the orthogonal projection of v = i +j+k on b = 2i+2j, and then find the vector component of v orthogonal to b.

23 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 66 Work Recall that we define the work W done on the object b a constant force of magnitude F acting in the direction of motion over the distance d to be W = Fd = force distance (3.11) If we let F denote a force vector of magnitude F = F acting in the direction of motion, then we can write (3.11) as W = F d Moreover, if we assume that the object moves along a line from point P to point Q, then d = PQ, so that the work can be epressed entirel in vector form as W = F PQ The vector PQ is called the displacement vector for the object. P F F Work = F PQ Q In the case where a constant force F is not in the direction of motion, but rather makes an angle θ with the displacement vector, then we define the work W done b F to be W = ( F cosθ) PQ = F PQ (3.12) F F P θ F cosθ Q Work = ( F cosθ) PQ Eample 3.16 A force F = 8i+5j in pound moves an object from P(1,0) to Q(7,1), distance measured in feet. How much work is done?

24 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 67 Eample 3.17 Awagonispulledhoriontallbeerting aconstant forceof10lbonthehandle at an angle of 60 with the horiontal. How much work is done in moving the wagon 50 ft? Eercise Ineachpart, findthedotproduct ofthevectors andthecosineoftheanglebetween them. (a) u = i+2j, v = 6i 8j (b) u = 7, 3, v = 0,1 (c) u = i 3j+7k, v = 8i 2j 2k (d) u = 3,1,2, v = 4,2, 5 2. In each part use the given information to find u v. (a) u = 1, v = 2, the angle between u and v is π/6. (b) u = 2, v = 3, the angle between u and v is In each part, determine whether u and v make an acute angle, an obtuse angle, or are orthogonal. (a) u = 7i+3j+5k, v = 8i+4j+2k (b) u = 6i+j+3k, v = 4i 6k (c) u = 1,1,1, v = 1,0,0 (d) u = 4,1,6, v = 3,0,2 4. Does the triangle in 3-space with vertices ( 1,2,3), (2, 2,0), and (3,1, 4) have an obtuse angle? Justif our answer. 5. The accompaning figure shows eight vectors that are equall spaced around a circle of radius 1. Find the dot product of v 0 with each of the other seven vectors. v 3 v 2 v 1 v 4 v 0 v 5 v 6 v 7

25 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana The accompaning figure shows si vectors that are equall spaced around a circle of radius 5. Find the dot product of v 0 with each of the other five vectors. v 2 v 1 v 3 v 0 v 4 v 5 7. (a) Use vectors to show that A(2, 1,1), B(3,2, 1), and C(7,0, 2) are vertices of the right triangle. At which verte is the right angle? (b) Use vectors to find the interior angles of the triangle with vertices ( 1,0), (2, 1), and (1,4). 8. (a) Show that if v = ai+bj is a vector in 2-space, then the vectors are both orthogonal to v. v 1 = bi+aj and v 2 = bi aj (b) Use the result in part (a) to find two unit vectors that are orthogonal to the vector v = 3i 2j. Sketch the vectors v, v 1, and v Eplain wh each of the following epressions makes no sense. (a) u (v w) (b) (u v)+w (c) u v (d) k (u+v) 10. True or false? If u v = u w and if u 0, then v = w. Justif our conclusion. 11. Verifpart(b)and(c)ofTheorem3.6forthevectorsu = 6i j+2k,v = 2i+7j+4k, w = i+j 3k and k = Let u = 1,2, v = 4, 2, and w = 6,0. Find (a) u (7v+w) (b) (u w)w (c) u (v w) (d) ( u v) w 13. Find r so that the vector from the point A(1, 1,3) to the point B(3,0,5) is orthogonal to the vector from A to the point P(r,r,r). 14. Find two unit vectors in 2-space that make an angle of 45 with 4i+3j Find the direction cosines of v. 15. (a) v = i+j k (b) v = 2i 2j+k 16. (a) v = 3i 2j 6k (b) v = 3i 4k 17. In each part, find the vector component of v along b and the vector component of v orthogonal to b.

26 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 69 (a) v = 2i j, b = 3i+4j (b) v = 4,5, b = 1, 2 (c) v = 3i 2j, v = 2i+j 18. In each part, find the vector component of v along b and the vector component of v orthogonal to b. (a) v = 2i j+3k, b = i+2j+2k (b) v = 4, 1,7, b = 2,3, Epress the vector v as the sum of a vector parallel to b and a vector orthogonal to b. 19. (a) v = 2i 4j, b = i+j (b) v = 3i+j 2k, b = 2i k (c) v = 4i 2j+6k, b = 2i+j 3k 20. (a) v = 3,5, b = 1,1 (b) v = 2,1,6, b = 0, 2,1 (c) v = 1,4,1, b = 3, 2,5 21. Find the work done b a force F = 3j (pounds) applied to a point that moves on the line from (1,3) to (4,7), Assume that distance is measured in feet. 22. A force F = 4i 6j+k newtons is applied to a point that moves a distance of 15 meters in the direction of the vector i+j+k. How much work is done? 23. A boat travels 100 meters due north while the wind that applies a force of 500 newtons toward the northwest. How much work does the wind do? 24. A bo is dragged along the floor b a rope that applies a force of 50 lb at an angle of 60 with the floor. How much work is done moving the bo 15 ft? Answers to Eercise (a) 10;cosθ = 1/ 5 (b) 3;cosθ = 3/ 58 (c) 0;cosθ = 0 (d) 20;cosθ = 20/(3 70) 3. (a) obtuse (b) acute (c) obtuse (d) orthogonal 5. 2/2,0, 2/2, 1, 2/2,0, 2/2 7. (a) verte B (b) 82,60, r = 7/5 15. (a) α = β 55,γ 125 (b) α 48,β 132,γ (a) 3, 4 3, 4 4, 3 3, 7 3, 5 (b) , , , 49 49, 62 49, (a) 1,1 + 4,4 (b) 0, 8 5, 4 + 2, , 26 5 (c) v = 1,4,1 is orthogonal to b J ft lb

27 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Cross Product In Section3.3 we defined the dot product, which gave a wa of multipling two vectors. The resulting product, however, was a scalar, not a vector. In this section we will define a product of two vectors that does result in another vector. This product, called the cross product, is onl defined for vectors in 3-space. Before we define the cross product, we need to define the notion of determinant. Determinants of a Square Matri Definition 3.5 The determinant of a 2 2 matri of real number is defined b a 1 a 2 b 1 b 2 = a 1b 2 a 2 b 1. Definition 3.6 The determinant of a 3 3 matri of real number is defined as a combination of three 2 2 determinants, as follows: a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 b 2 b 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 +a 3 b 1 b 2 c 1 c 2 (3.13) Note that Equation (3.13) is referred to as an epansion of the determinant along the first row. The Cross Product of Two Vectors Definition 3.7 Let u = u 1,u 2,u 3 and v = v 1,v 2,v 3 be vectors in 3-space. The cross product of u and v, denoted b u v, is the vector in 3-space given b u v = u 2 u 3 v 2 v 3 i u 1 u 3 v 1 v 3 j+ u 1 u 2 v 1 v 2 k (3.14) or, equivalentl, u v = (u 2 v 3 u 3 v 2 )i (u 1 v 3 u 3 v 1 )j+(u 1 v 2 u 2 v 1 )k (3.15)

28 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 71 Observe that the right side of Formula (3.14) can be written as i j k u v = u 1 u 2 u 3 v 1 v 2 v 3 (3.16) Eample 3.18 Let u = 1,2,3 and v = 4,5,6. Find u v. Algebraic Properties of the Cross Product Theorem 3.9 If u, v, and w are an vectors in 3-space and k is an scalar, then: (a) u v = (v u) (b) u (v+w) = (u v)+(u w) (c) (u+v) w = (u w)+(v w) (d) k(u v) = (ku) v = u (kv) (e) u 0 = 0 u = 0 (f) u u = 0 The following cross products occur so frequentl that it is helpful to be familiar with them: i j = k j k = i k i = j (3.17) j i = k k j = i i k = j These results are eas to obtain; for eample, i j k i j = = i j k However, rather than computing these cross products each time ou need them, ou can use the diagram in Figure below.

29 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 72 k i j Geometric Properties of the Cross Product The following theorem shows that the cross product of two vectors is orthogonal to both factors. Theorem 3.10 If u and v are vectors in 3-space, then: (a) u (u v) = 0 (u v is orthogonal to u) (b) v (u v) = 0 (u v is orthogonal to v) u v v u v u Eample 3.19 Let u and v be vectors as in Eample Show that u v is orthogonal to both u and v.

30 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 73 Theorem 3.11 Let u and v be nonero vectors in 3-space, and let θ be the angle between these vectors when the are positioned so their initial points coincide. (a) u v = u v sinθ (b) u v is the area of the parallelogram formed b u and v. (c) u v = 0 if and onl if u and v are parallel vectors, that is, if and onl if the are scalar multiples of one another. Eample 3.20 Find the area of the parallelogram with two adjacent sides formed b the vectors u = 1,2, 2 and v = 3,0,1. Eample 3.21 Find the area of the triangle that is determined b the points P 1 (2,2,0), P 2 ( 1,0,2), and P 3 (0,4,3).

31 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 74 Scalar Triple Products Definition 3.8 Let u = u 1,u 2,u 3, v = v 1,v 2,v 3, and w = w 1,w 2,w 3 be vectors in 3-space. The scalar triple product of u, v, and w is the scalar u (v w). This value can be obtained directl from the formula u 1 u 2 u 3 u (v w) = v 1 v 2 v 3 w 1 w 2 w 3 (3.18) Eample 3.22 Calculate the scalar triple product u (v w) of the vectors u = 3i 2j 5k, v = i+4j 4k, w = 3j+2k Theorem 3.12 (Cclic rotation formula for scalar triple product) u (v w) = w (u v) = v (w u) Theorem 3.13 Let u, v and w be nonero vectors in 3-space. (a) u (v w) is the volume of the parallelepiped that has u, v and w as adjacent edges. (b) u (v w) = 0 if and onl if u, v and w lie in the same plane. Eample 3.23 Find the volume of the parallelepiped with three adjacent edges formed b the vectors u = 7,8,0, v = 1,2,3 and w = 4,5,6.

32 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 75 Eercise (a) Use a determinant to find the cross product i (i+j+k) (b) Check our answer in part (a) b rewriting the cross product as i (i+j+k) = (i i)+(i j)+(i k) and evaluate each term. 2. In each part, use the two methods in Eercise1 to find (a) j (i+j+k) (b) k (i+j+k) 3 6 Find u v and check that it is orthogonal to both u and v. 3. u = 1,2, 3, v = 4,1,2 4. u = 3i+2j k, v = i 3j+k 5. u = 0,1, 2, v = 3,0, 4 6. u = 4i+k, v = 2i j 7. Let u = 2, 1,3, v = 0,1,7, and w = 1,4,5. Find (a) u (v w) (b) (u v) w (c) (u v) (v w) (d) (v w) (u v) 8. Find two unit vectors that are orthogonal to both u = 7i+3j+k, v = 2i+4k 9. Find two unit vectors that are normal to the plane determined b the points A(0, 2,1), B(1, 1, 2), and C( 1,1,0). 10. Find two unit vectors that are parallel to the -plane and are orthogonal to the vector 3i j+2k Find the area of the parallelogram that has u and v as adjacent sides. 11. u = i j+2k, v = 3j+k 12. u = 2i+3j, v = i+2j 2k Find the area of the triangle with vertices P, Q, and R. 13. P(1,5, 2), Q(0,0,0), R(3,5,1) 14. P(2,0, 3), Q(1,4,5), R(7,2,9)

33 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Find u (v w). 15. u = 2i 3j+k, v = 4i+j 3k, w = j+5k 16. u = 1, 2,2, v = 0,3,2, w = 4,1, u = 2,1,0, v = 1, 3,1, w = 4,0,1 18. u = i, v = i+j, w = i+j+k Use a scalar triple product to find the volume of the parallelepiped that has u, v, and w as adjacent edges. 19. u = 2, 6,2, v = 0,4, 2, w = 2,2, u = 3i+j+2k, v = 4i+5j+k, w = i+2j+4k 21. In each part, use a scalar triple product to determine whether the vectors lie in the same plane. (a) u = 1, 2,1, v = 3,0, 2, w = 5, 4,0 (b) u = 5i 2j+k, v = 4i j+k, w = i j (c) u = 4, 8,1, v = 2,1, 2, w = 3, 4, Suppose that u (v w). Find (a) u (w v) (c) w (u v) (e) (u w) v (b) (v w) u (d) v (u w) (f) v (w w) Answers to Eercise (a) j+k 3. 7,10,9 5. 4, 6, 3 7. (a) 20, 67, 9 (b) 78,52, 26 (c) 0, 56, 392 (d) 0,56, ± 1 6 2,1, / (a) es (b) es (c) no

34 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Parametric Equation of Lines Lines Determined b a Point and a Vector A line in 2-space or 3-space can be determined uniquel b specifing a point on the line and a nonero vector parallel to the line. L L v P( 0, 0 ) (a,b) P 0 ( 0, 0, 0 ) v (a,b,c) For eample, consider a line L in 3-space that passes through the point P 0 ( 0, 0, 0 ) and is parallel to the nonero vector v = a,b,c. Then L consists precisel of those point P(,,) for which the vector P 0 P is parallel to v. P 0 ( 0, 0, 0 ) v L P(,,) (a,b,c) In other words, the point P(,,) is on L if and onl if P 0 P is a scalar multiple of v, sa P 0 P = tv This equation can be written as which implies that 0, 0, 0 = ta,tb,tc 0 = ta, 0 = tb, 0 = tc Thus, L can be described b the parametric equations = 0 +at, = 0 +bt, = 0 +ct A similar description applies to lines in 2-space.

35 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 78 Theorem 3.14 (a) The line in 2-space that passes through the point P 0 ( 0, 0 ) and is parallel to the nonero vector v = a,b = ai+bj has parametric equations = 0 +at, = 0 +bt (3.19) (b) The line in 3-space that passes through the point P 0 ( 0, 0, 0 ) and is parallel to the nonero vector v = a,b,c = ai+bj+ck has parametric equations = 0 +at, = 0 +bt, = 0 +ct (3.20) Eample 3.24 Find parametric equations of the line passing through the point (1,5,2) and parallel to the vector v = 4,3,7. Also, determine where the line intersects the -plane. Eample 3.25 Find parametric equations of the line L passing through the point P(1,2, 1) and Q(5, 3,4).

36 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 79 Eample 3.26 Let L 1 and L 2 be the lines L 1 : = 1+4t, = 5 4t, = 1+5t L 2 : = 2+8t, = 4 3t, = 5+t (a) Are the lines parallel? (b) Do the lines intersect? Two lines in 3-space that are not parallel and do not intersect are called skew lines.

37 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 80 Line Segments Sometimes one is not interested in an entire line, but rather some segment of a line. Parametric equations of a line segment can be obtained b finding parametric equation for the entire line, then restricting the parameter appropriatel so that onl the desired segment is generated. Eample 3.27 Find parametric equations describing the line segment joining the points P(1,2, 1) and Q(5, 3,4). Vector Equations of Lines We will now show how vector notation can be used to epress the parametric equations of a line. Because two vectors are equal if and onl if their components are equal, (3.19) and (3.20) can be written in vector form as or, equivalentl, as, = 0 +at, 0 +bt,, = 0 +at, 0 +bt, 0 +ct, = 0, 0 +t a,b (3.21),, = 0, 0, 0 +t a,b,c (3.22) For the equation in 2-space we define the vectors r, r 0 and v as r =,, r 0 = 0, 0, v = a,b (3.23) and for the equation in 3-space we define them as r =,,, r 0 = 0, 0, 0, v = a,b,c (3.24) Substituting (3.23) and (3.24) in (3.21) and (3.22), respectivel, ields the equation r = r 0 +tv (3.25) in both case. We call this the vector equation of a line in 2-space or 3-space. In this equation, v is a nonero vector parallel to the line, and r 0 is a vector whose components are the coordinates of a point on the line.

38 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 81 tv L r 0 P 0 v r tv Eample 3.28 The equation is of form (3.25) with,, = 1,0,2 +t 1,5, 4 r 0 = 1,0,2, v = 1,5, 4 Thus, the equation represents the line in 3-space that passes through the point ( 1,0,2) and is parallel to the vector 1,5, 4. Eample 3.29 Find an equation of the line in 3-space that passes through the points P 1 (2,4, 1) and P 2 (5,0,7).

39 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 82 Eercise Find parametric equations for the line through P 1 and P 2 and also for the line segment joining those points. 1. (a) P 1 (3, 2), P 2 (5,1) (b) P 1 (5, 2,1), P 2 (2,4,2) 2. (a) P 1 (0,1), P 2 ( 3, 4) (b) P 1 ( 1,3,5), P 2 ( 1,3,2) 3 4 Find parametric equations for the line whose vector equation is given. 3. (a), = 2, 3 +t 1, 4 (b) i+j+k = k+t(i j+k) 4. (a) i+j = (3i 4j)+t(2i+j) (b),, = 1,0,2 +t 1,3,0 5 6 Find a point P on the line and a vector v parallel to the line b inspection. 5. (a) i+j = (2i j)+t(4i j) (b),, = 1,2,4 +t 5,7, 8 6. (a), = 1,5 +t 2,3 (b) i+j+k = (i+j 2k)+tj 7 8 Epress the given parametric equations of a line using bracket notation and also using i, j, k notation. 7. (a) = 3+t, = 4+5t (b) = 2 t, = 3+5t, = t 8. (a) = t, = 2+t (b) = 1+t, = 7+3t, = 4 5t 9 16 Find parametric equations of the line that satisfies that stated conditions. 9. The line through ( 5,2) that is parallel to 2i 3j. 10. The line through (0,3) that is parallel to the line = 5+t, = 1 2t. 11. The line that is tangent to the circle = 25 at the point (3, 4). 12. The line that is tangent to the parabola = 2 at the point ( 2,4). 13. The line through ( 1,2,4) that is parallel to 3i 4j+k.

40 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana The line through (2, 1,5) that is parallel to 1,2, The line through ( 2,0,5) that is parallel to the line = 1 + 2t, = 4 t, = 6+2t. 16. The line through the origin that is parallel to the line = t, = 1+t, = Where does the line = 1+3t, = 2 t intersect (a) the -ais (c) the parabola = 2? (b) the -ais 18. Where does the line, = 4t,3t intersect the circle = 25? Find the intersections of the lines with -plane, the -plane, and the -plane. 19. = 2, = 4+2t, = 3+t 20. = 1 2t, = 3+t, = 4 t 21. Where does the line = 1+t, = 3 t, = 2t intersect the clinder = 16? 22. Where does the line = 2 t, = 3t, = 1+2t intersect the plane 2+3 = 6? Show that the line L 1 and L 2 intersect, and find their point of intersection. 23. L 1 : = 2+t, = 2+3t, = 3+t L 2 : = 2+t, = 3+4t, = 4+2t 24. L 1 : +1 = 4t, 3 = t, 1 = 0 L 2 : +13 = 12t, 1 = 6t, 2 = 3t Show that the line L 1 and L 2 are skew. 25. L 1 : = 1+7t, = 3+t, = 5 3t L 2 : = 4 t, = 6, = 7+2t 26. L 1 : = 2+8t, = 6 8t, = 10t L 2 : = 3+8t, = 5 3t, = 6+t Determine whether the line L 1 and L 2 are parallel. 27. L 1 : = 3 2t, = 4+t, = 6 t L 2 : = 5 4t, = 2+2t, = 7 2t 28. L 1 : = 5+3t, = 4 2t, = 2+3t L 2 : = 1+9t, = 5 6t, = 3+8t Determine whether the point P 1, P 2, and P 3 lie on the same line. 29. P 1 (6,9,7), P 2 (9,2,0), P 3 (0, 5, 3) 30. P 1 (1,0,1), P 2 (3, 4, 3), P 3 (4, 6, 5)

41 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 84 Answers to Eercise (a) = 3+2t, = 2+3t; line segment: 0 t 1 (b) = 5 3t, = 2+6t, = 1+t; line segment: 0 t 1 3. (a) = 2+t, = 3 4t (b) = t, = t, = 1+t 5. (a) P(2, 1), v = 4i j (b) P( 1,2,4), v = 5i+7j 8k 7. (a) 3,4 +t 1,5 ; 3i+4j+t(i+5j) (b) 2, 3,0 +t 1,5,1 ; 2i 3j+t( i+5j+k) 9. = 2+2t, = 2 3t 11. = 3+4t, = 4+3t 13. = 1+3t, = 2 4t, = 4+t 15. = 2+2t, = t, = 5+2t 17. (a) = 7 (b) = 7 3 (c) = 1± 85, = ( 2,10,0); ( 2,0, 5); the line does not intersect the -plane. 21. (0,4, 2); (4,0,6) 23. (1, 1,2) 27. The lines are parallel. 29. The points do not lie on the same line. 3.6 Plane in 3-Space Plane Parallel to the Coordinate Planes The graph of the equation = a in an -coordinate sstem consists of all points of the form (a,,), where and are arbitrar. One such point is (a,0,0), and all others are in the plane that passes through this point and is parallel to the -plane. Similarl, the graph of = b is the plane through (0,b,0) that is parallel to the -plane, and the graph of = c is the plane through (0,0,c) that is parallel to the -plane. (a,0,0) = a = b (0,b,0) (0,0,c) = c

42 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 85 Planes Determined b a Point and a Normal Vector A plane in 3-space can be determined uniquel b specifing a point in the plane and a vector perpendicular to the plane. A vector perpendicular to the plane is called a normal to the plane. Suppose that we want to find an equation of the plane passing through P 0 ( 0, 0, 0 ) and perpendicular to the vector n = a,b,c. Define the vectors r 0 and r as r 0 = 0, 0, 0 and r =,, n P 0 ( 0, 0, 0 ) r 0 r r 0 P(,,) r O It should be evident from the above Figure that the plane consists precisel of those points P(,,) for which the vector r r 0 is orthogonal to n; or, epressed as an equation, n (r r 0 ) = 0 (3.26) If preferred, we can epress this vector equation in terms of components as from which we obtain a,b,c 0, 0, 0 = 0 (3.27) a( 0 )+b( 0 )+c( 0 ) = 0 (3.28) This is called the point-normal form of the equation of a plane. Formula (3.26) and (3.27) are vector version of this formula. Eample 3.30 Find an equation of the plane passing through the point (1,2,3) and perpendicular to the vector n = 4,5,6.

43 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 86 Observethatifwemultiploutthetermsin(3.28)andsimplif, weobtainanequation of the form a+b +c +d = 0 For eample, equation in Eample 3.30 can be rewritten as Theorem = 0 If a, b, c, and d are constants, and a, b, and c are not all ero, then the graph of the equation a+b +c +d = 0 (3.29) is a plane that has the vector n = a,b,c as a normal. Equation (3.29) is called the general form of the equation of a plane. Eample 3.31 Determine whether the plane = 0 and = 0 are parallel.

44 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 87 Eample 3.32 Find an equation of the plane through the points P 1 (1,2,2), P 2 (2, 1,4), and P 3 (3,5, 2). Eample 3.33 Determine whether the line = 3+8t, = 4+5t, = 3 t is parallel to the plane 3 +5 = 12.

45 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 88 Eample 3.34 Find the intersection of the line and the plane in Eample Eample 3.35 Find an equation for the plane through the point (1,4, 5) and parallel to the plane defined b = 12.

46 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 89 Intersecting Planes Two distinct intersecting plane determine two positive angles of intersection an (acute) angle θ that satisfies the condition 0 θ π/2 and the supplement of that angle. n 1 θ n 2 θ θ 180 θ If n 1 and n 2 are normals to the planes, then depending on the directions of n 1 and n 2, the angle θ is either the angle between n 1 and n 2 or the angle between n 1 and n 2. In both case, we have the following formula for the acute angle θ between the planes: cosθ = n 1 n 2 n 1 n 2 (3.30) Eample 3.36 Find the acute angle of intersection between the two planes = 6 and = 4 Eample 3.37 Find an equation for the line L of intersection of the planes in Eample 3.36.

47 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 90 Distance Problems Involving Planes Net we will consider three basic distance problems in 3-space: Find the distance between a point and a plane. Find the distance between two parallel planes. Find the distance between two skew lines. The three problems are related. If we can find the distance between a point and a plane, then we can find the distance between parallel planes b computing the distance between one of the planes and an arbitrar point P 0 in the other plane. Moreover, we can find the distance between two skew lines b computing the distance between parallel planes containing them. P 0 Theorem 3.16 The distance D between a point P 0 ( 0, 0, 0 ) and the plane a+b+c+d = 0 is D = a 0 +b 0 +c 0 +d a2 +b 2 +c 2 (3.31) Eample 3.38 Find the distance D between the point (1, 4, 3) and the plane = 1

48 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 91 Eample 3.39 The planes +2 2 = 3 and = 7 are parallel since their normals, 1,2, 2 and 2,4, 4, are parallel vectors. Find the distance between these planes. Eample 3.40 It was shown in Eample 3.26 of Section2.5 that the line L 1 : = 1+4t, = 5 4t, = 1+5t L 2 : = 2+8t, = 4 3t, = 5+t are skew. Find the distance between them.

49 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 92 Eercise Find an equation of the plane that passes through the point P and has the vector n as normal. 1. P(2,6,1); n = 1,4,2 2. P( 1, 1,2); n = 1,7,6 3. P(1,0,0); n = 0,0,1 4. P(0,0,0); n = 2, 3, Find an equation of the plane indicated in the figure Find an equation of the plane that passes through the given point. 9. ( 2,1,1), (0,2,3), and (1,0, 1) 10. (3,2,1), (2,1, 1), and ( 1,3,2) Determine whether the planes are parallel, perpendicular, or neither. 11. (a) = = 0 (c) +3 2 = 0 2+ = (a) = = 7 (b) = = 4 (b) = = (c) = = 0

50 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Determine whether the line and planes are parallel, perpendicular, or neither. 13. (a) = 4+2t, = t, = 1 4t; = 0 (b) = t, = 2t, = 3t; +2 = 5 (c) = 1+2t, = 4+t, = 1 t; = (a) = 3 t, = 2+t, = 1 3t; = 0 (b) = 1 2t, = t, = t; = 1 (c) = t, = 1 t, = 2+t; + + = Determine whether the line and planes intersect; if so, find the coordinates of the intersection. 15. (a) = t, = t, = t; = 0 (b) = 2 t, = 3+t, = t; 2+ + = (a) = 3t, = 5t, = t; = 0 (b) = 1+t, = 1+3t, = 2+4t; +4 = Find the acute angle of intersection of the planes. 17. = 0 and = = 5 and = Find an equation of the plane that satisfies the stated conditions. 19. The plane through the origin that is parallel to the plane = The plane that contains the line = 2 + 3t, = 4 + 2t, = 3 t and is perpendicular to the plane 2 + = 5.

51 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana The plane through the point ( 1,4,2) that contains the line of intersection of the planes = 0 and = Theplane through( 1,4, 3)that isperpendicular totheline 2 = t, +3 = 2t, and = t. 23. The plane through (1,2, 1) that is perpendicular to the line of intersection of the planes 2+ + = 2 and +2 + = The plane through the points P 1 ( 2,1,4), P 2 (1,0,3) that is perpendicular to the planes 4 +3 = The plane through ( 1,2, 5) that is perpendicular to the planes 2 + = 1 and + 2 = The plane that contains the point (2,0,3) and the line = 1 + t, = t, and = 4+2t. 27. The plane whose points are equidistant from (2, 1,1) and (3,1,5). 28. The plane that contains the line = 3t, = 1 + t, = 2t and is parallel to the intersection of the planes + = 1 and 2 + = Find parametric equations of the line through the point (5, 0, 2) that is parallel to the planes 4 +2 = 0 and = Let L be the line = 3t+1, = 5t, = t. (a) Show that L lies in the plane 2+ = 2. (b) Show that L is parallel to the plane ++2 = 0. Is the line above, below, or on this plane? Find the distance between the point and the plane. 31. (1, 2,3); = (0,1,5); = Find the distance between parallel planes. 33. (a) 2+ + = = (b) + + = = Find the distance between the given shew lines. 35. = 1+7t, = 3+t, = 5 3t = 4 t, = 6, = 7+2t 36. = 3 t, = 4+4t, = 1+2t = t, = 3, = 2t

52 MA112: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Find an equation of the sphere with center (2,1, 3) that is tangent to the plane 3 +2 = Locate the point of intersection of the plane 2+ = 0 and the line through (3,1,0) that is perpendicular to the plane. Answers to Eercise = = 0 5. = = = (a) parallel (b) perpendicular (c) neither 13. (a) parallel (b) neither (c) perpendicular 15. (a) point of intersection is ( 5, 5, ) 5 (b) no intersection = = = = = = 5 2t, = 5t, = 2+11t / / ( 2) 2 +( 1) 2 +( +3) 2 = Quadric Surfaces Quadric Surfaces In the discussion of Formula(1.2) in Section 1.2 we noted that a second-degree equation A 2 +B +C 2 +D+E +F = 0 represents a conic section. The analog of this equation in an -coordinate sstem is A 2 +B 2 +C 2 +D +E +F +G+H +I +J = 0 (3.32) which is called a second-degree equation in,, and. The graphs of such equations are called quadric surfaces or sometimes quadrics. Si common tpes of quadric surfaces are ellipsoids, hperboloids of one sheet, hperboloids of two sheet, elliptic cones, elliptic paraboloids, and hperbolic paraboloids. Techniques for Graphing Quadric Surfaces Ellipsoids A sketch of an ellipsoids can be obtained b 2 a b + 2 = 1 (a > 0,b > 0,c > 0) (3.33) 2 c2 first plotting the intersections with the coordinate aes, then