1. Vectors in the Plane

Transcription

1 CHAPTER 10 Vectors and the Geometry of Space 1. Vectors in the Plane We denote the directed line segment from the point P (initial point) to the!! point Q (terminal point) as P Q. The length of P Q is its magnitude, denoted k P! Qk. Definition. A vector is the equivalence class of all directed line segments (or arrows) having the same magnitude and the same direction (an arrow and its parallel translates). Any arrow in the class may be used to name the vector. Example. a = b = c since a, b, c name the same vector. 1

2 2 10. VECTORS AND THE GEOMETRY OF SPACE Definition. A position vector is the arrow in each class with tail (initial point) at the origin. If the tip (terminal point) is at the point A(a 1, a 2 ), we write a = OA! = ha 1, a 2 i. In ha 1, a 2 i, a 1 and a 2 are called the components of a a 1 is the first component, a 2 the second component. From the Pythagorean Theorem, kak = kha 1, a 2 ik = q a a2 2. Note. The magnitude of a is often called the norm or 2-norm of a, and is sometimes denoted kak 2. For a =! OA = ha 1, a 2 i and b =! OB = hb 1, b 2 i, a = b () A = B () a 1 = b 1 and a 2 = b 2. () means if and only if (=) is only if and (= is if ).

3 Basic Operations Addition: 1. VECTORS IN THE PLANE 3 ha 1, a 2 i + hb 1, b 2 i = ha 1 + b 1, a 2 + b 2 i. This is the tip-to-tail or parallelogram law. The sum a+b is called the resultant. Example. h4, 3i + h 1, 4i = h3, 7i

4 4 10. VECTORS AND THE GEOMETRY OF SPACE Subtraction: ha 1, a 2 i hb 1, b 2 i = ha 1 b 1, a 2 b 2 i. For direction of di erence vector, make sure b + (a b) = a. Zero vector: 0 = h0, 0i k0k = 0 and the zero vector has all directions. Scalar multiplication: Thus ca is a vector and ca = cha 1, a 2 i = hca 1, ca 2 i kcak = khca 1, ca 2 ik = p q (ca 1 ) 2 + (ca 2 ) 2 = c 2 a c2 a 2 2 qc = 2 (a 21 + a22 ) = p q c 2 a a2 2 = c kak.

5 1. VECTORS IN THE PLANE 5 Vectors are parallel if one is a scalar multiple of the other. ca k a, in the same direction for c 0, the opposite direction for c apple 0. Additive inverse: hx, yi x, y 2 R, the set of all two-dimensional posi- Definition. V 2 = tion vectors. a = ha 1, a 2 i = ( 1)ha 1, a 2 i = h a 1, a 2 i k ak = k( 1)ha 1, a 2 ik = 1 kha 1, a 2 ik = kak Definition. A vector space is a set of vectors with the same dimension and a set of scalars (real or complex numbers) with addition and scalar multiplication such that for any vectors u, v, and w, and any scalars and, (1) v + w = w + v (2) (u + v) + w = u + (v + w) (3) ( v) = ( )v (4) ( + )v = v + v (5) (v + w) = v + w (6) 1v = v (7) 0v = 0 (8) v + 0 = v (9) For each v there is v such that v + ( v) = 0

6 6 10. VECTORS AND THE GEOMETRY OF SPACE Theorem (1.1). V 2 is a vector space. Proof. (of (2)) (u + v) + w = (hu 1, u 2 i + hv 1, v 2 i) + hw 1, w 2 i = hu 1 + v 1, u 2 + v 2 i + hw 1, w 2 i = h(u 1 + v 1 ) + w 1, (u 2 + v 2 ) + w 2 i = hu 1 + (v 1 + w 1 ), u 2 + (v 2 + w 2 )i = hu 1, u 2 i + hv 1 + w 1, v 2 + w 2 i = hu 1, u 2 i + (hv 1, v 2 i + hw 1, w 2 i) = u + (v + w) Note. For any two distinct points A(x 1, y 1 ) and B(x 2, y 2 ), to the position vector hx 2 x 1, y 2 y 1 i. Any vector can be written in terms of the standard basis vectors For any a 2 V 2, i = h1, 0i and j = h0, 1i.! AB corresponds a = ha 1, a 2 i = ha 1, 0i + h0, a 2 i = a 1 h1, 0i + a 2 h0, 1i = a 1 i + a 2 j. We call a 1 and a 2 the horizontal and vertical components of a Note. kik = kjk = 1. A unit vector is a vector a where kak = 1. Theorem (1.2). For any nonzero vector a, a unit vector having the same direction as a is u = 1 kak a. Proof. a 6= 0 =) kak > 0 =) u is a positive scalar multiple of a =) u has the same direction as a. kuk = 1 kak a = 1 kak kak = 1 kak kak = 1.

7 1. VECTORS IN THE PLANE 7 Example. Find a unit vector in the same directioin as a = h6, 8i. Therefore kak = kh6, 8ik = p = p 100 = 10. u = 1 10 h6, 8i = h3 5, 4 5 i. Resolving a vector a into horizontal and vertical components: a = kak cos i + kak sin j = kak(cos i + sin j) = kakhcos, sin i Note. khcos, sin ik = 1.

8 8 10. VECTORS AND THE GEOMETRY OF SPACE Example. A plane is flying at 500 km/hr. The wind is blowing at 60 km/hr toward the SE. In what direction should the plane head to go due east? What is the plane s speed relative to the ground? Solution Let p = the motion vector for the plane and w = the wind vector. Find and kp + wk. w = kwk cos( 45 )i + kwk sin( 45 )j p p 2 2 = 60 2 i 60 2 j = 30p 2 i 30 p 2 j = 42.43i 42.43j. Since the j-component of p + w is 0, p = xi + 30 p 2j where x is the i-component of p. Since kpk = 500, q 500 = x 2 + (30 p 2) 2 =) = x =) Thus x 2 = 250, = 248, 200 =) x p = i j =) tan = =.085 =) = arctan.085 = 4.86, and since p + w = i + 0j, kp + wk = km/hr.

9 2. VECTORS IN SPACE 9 Maple. See vectorsinplane(10.1).mw or vectorsinplane(10.1).pdf 2. Vectors in Space We use a right-handed coordinate system. The coordinate system has 8 octants. The first octant has x > 0, y > 0, z > 0. Distance Formula d (x 1, y 1, z 1 ), (x 2, y 2, z 2 )} = p (x 2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2. Vectors in R 3 These are similar to vectors in R 2, with one more dimension. V 3 = hx, y, zi x, y, z 2 R is a vector space with a + b = ha 1, a 2, a 3 i + hb 1, b 2, b 3 i = ha 1 + b 1, a 2 + b 2, a 3 + b 3 i, ca = cha 1, a 2, a 3 i = hca 1, ca 2, ca 3 i, and a b = ha 1, a 2, a 3 i hb 1, b 2, b 3 i = ha 1 b 1, a 2 b 2, a 3 b 3 i,

10 VECTORS AND THE GEOMETRY OF SPACE Also, so kak = kha 1, a 2, a 3 ik = q a a2 2 + a2 3, kcak = c kak, 0 = h0, 0, 0i, and a = h a 1, a 2, a 3 i Note. The vector with initial point P (a 1, a 2, a 3 ) and terminal point Q(b 1, b 2, b 3 ) corresponds to the position vector! P Q = hb 1 a 1, b 2 a 2, b 3 a 3 i. The standard basis consists of i = h1, 0, 0, i, j = h0, 1, 0i, k = h0, 0, 1i with kik = kjk = kkk = 1. For any vector a = ha 1, a 2, a 3 i, a = a 1 i + a 2 j + a 3 k. Also, for any vector a = ha 1, a 2, a 3 i 6= 0, a unit vector u in the same direction as a is u = 1 kak a. The sphere of radius r centered at (a, b, c) consists of all points (x, y, z) such that d (x, y, z), (a, b, c)} = p (x a) 2 + (y b) 2 + (z c) 2 or (x a) 2 + (y b) 2 + (z c) 2 = r 2, the standard form of the equation of a sphere.

11 2. VECTORS IN SPACE 11 Problem (Page 711 #28). Identify the geometric shape given by Solution Rewrite as x 2 + x + y 2 y + z 2 = 7 2. (x 2 + x ) + (y 2 y ) + z 2 = 7 2. Complete squares: x 2 + x y 2 y z 2 = =) x y + z 2 = Thus we have the sphere of radius 2 centered at 2, 1 2., 0 Problem (Page 711 #20). Find a vector of magnitude 3 in the same direction as v = 3i + 3j k. Solution Thus kvk = p ( 1) 2 = p 19 u = 1 p 19 v is a unit vector in the same direction of v, so 3u = p 3 v = p 9 i + p 9 j p 19 k has magnitude 3 in the same direction as v. Problem (Page 711 #40). An equation for the x-axis is y 2 + z 2 = 0 or y = z = 0.

12 VECTORS AND THE GEOMETRY OF SPACE Problem (Page 712 #62). Three ropes are attached to a 300-pound crate. Rope A exerts a force of h10, 130, 200i pounds on the crate, and rope B exerts a force of h 20, 180, 160i pounds on the crate. What force must rope C exert on the crate to move it up and to the right with a constant force of h0, 30, 20i pounds. Solution Force of rope A is a = h10, 130, 200i, rope B is b = h 20, 180, 160i, rope C is c = hc 1, c 2, c 3 i, and the weight is w = h0, 0, 300i. We want the net force a + b + c + w = h0, 30, 20i =) h 10 + c 1, 50 + c 2, 60 + c 3 i = h0, 30, 20i =) c 1 = 10, c 2 = 20, c 3 = 40 =) c = h10, 20, 40i =) kck Thus rope C exerts a force of about 45.8 pounds in the direction h1, 2 4i. Maple. See vectorsinspace(10.2).mw or vectorsinspace(10.2).pdf 3. The Dot Product This is also often referred to as the scalar product. In V 2 : In V 3 : a b = ha 1, a 2 i hb 1, b 2 i = a 1 b 1 + a 2 b 2 a b = ha 1, a 2, a 3 i hb 1, b 2, b 3 i = a 1 b 1 + a 2 b 2 + a 3 b 3 Example. If a = h 3, 2, 5i and b = h3, 5, 6i, a b = h 3, 2, 5i h3, 5, 6i = ( 3)(3) + 2(5) + 5( 6) = = 29

13 Note. (1) The dot product of two vectors is a number. 3. THE DOT PRODUCT 13 (2) A vector ha, bi in V 2 can be thought of a vector ha, b, 0i in V 3, so results we prove for V 3 also hold for V 2. Theorem (3.1). For vectors a, b, c and any scalar d, (1) a b = b a (2) a (b + c) = a b + a c (3) (da) b = d(a b) = a (db) (4) 0 a = 0 (5) a a = kak 2 Proof. (2) a (b + c) = ha 1, a 2, a 3 i hb 1, b 2, b 3 i + hc 1, c 2, c 3 i = ha 1, a 2, a 3 i hb 1 + c 1, b 2 + c 2, b 3 + c 3 i = a 1 (b 1 + c 1 ) + a 2 (b 2 + c 2 ) + a 3 (b 3 + c 3 ) = a 1 b 1 + a 1 c 1 + a 2 b 2 + a 2 c 2 + a 3 b 3 + a 3 c 3 = a 1 b 1 + a 2 b 2 + a 3 b 3 + a 1 c 1 + a 2 c 2 + a 3 c 3 = ha 1, a 2, a 3 i hb 1, b 2, b 3 i + ha 1, a 2, a 3 i hc 1, c 2, c 3 i = a b + a c. (5) a a = ha 1, a 2, a 3 i ha 1, a 2, a 3 i = a a a 2 3 = kha 1, a 2, a 3 ik 2 = kak 2. Note. a b = 0 does not imply a = 0 or b = 0, so this di ers from our usual multiplication.

14 VECTORS AND THE GEOMETRY OF SPACE For a 6= 0 and b 6= 0 in V 3, the angle (0 apple apple ) between a and b is the smaller of the two angles in the plane formed by giving both vectors the same initial point. If a and b have the same direction, = 0. If a and b have the opposite direction, =. a and b are orthogonal (or perpendicular) if = 2. We say 0 is orthogonal to every vector. Law of Cosines ka bk 2 = kak 2 + kbk 2 2kakkbk cos Theorem (3.2). Let be the angle between nonzero vectors a and b. Then a b = kakkbk cos. Corollary. For nonzero vectors a and b, cos = a b kakkbk. Example. Find the angle between a = h2, 3, 6i and b = h 2, 1, 4i. 2( 2) + 3(1) + 6(4) cos = p p = p 21 =) 23 = arccos 7 p.7713 R

15 or 3. THE DOT PRODUCT 15 Corollary. a and b are orthogonal (a? b) () a b = 0. Theorem (3.3 Cauchy-Schwarz Inequality). For any vectors a and b, Proof. a 1 b 1 + a 2 b 2 + a 3 b 3 apple a b apple kakkbk q q a a2 2 + a2 3 b b2 2 + b2 3. a b = kakkbk cos = kakkbk cos apple kakkbk. Theorem (3.4 Triangle Inequality). For any vectors a and b, ka + bk apple kak + kbk. Proof. ka + bk 2 = (a + b) (a + b) = a a + a b + b a + b b = kak 2 + 2a b + kbk 2 apple kak 2 + 2kakkbk + kbk 2 = (kak + kbk) 2 Taking square roots then gives the result.

16 VECTORS AND THE GEOMETRY OF SPACE Component of a along b comp b a 0 < < 2 -comp b a 2 < < Note. cos( ) = cos comp b a = kak cos = kakkbk kbk cos = 1 1 kakkbk cos = kbk kbk a b =) comp b a = a b kbk is a number, a directed length, that depends on the length of a since unit vector. b kbk is a

17 Projection of a onto b 3. THE DOT PRODUCT 17 This is a force vector parallel to b having the same component along b as a. proj b a is a vector. 0 < < 2 proj b a = comp b a b kbk = proj b a = a b kbk proj b a! a b kbk b kbk! 2 < < b kbk =)

18 VECTORS AND THE GEOMETRY OF SPACE Problem (Page 721 #63b). Find the component of the weight vector along the road vector for a 2500-pound car on a 15 bank. Let b be the direction of the banked road with kbk = 1. b = hcos 15, sin 15 i. The weight w = h0, 2500i. The component of w in the direction of b is comp b w = w b = h0, 2500i hcos 15, sin 15 i = 2500 sin 15 = 647. kbk Thus the force is to the inside of the track, allowing the car to turn. Problem (Page 720 #56). The orthogonal projection of a vector a along a vector b is defined as orth b a = a proj b a. Sketch a picture showing a, b, proj b a, and orth b a, and explain what is orthogonal about orth b a. Maple. See dotprod(10.3).mw or dotprod(10.3).pdf

19 4. THE CROSS PRODUCT The Cross Product This is a product that exists only in 3 dimensions. We fist need an introduction to determinants for computational purposes. apple a1 a Definition. For a 2 2 ( 2 by 2 ) matrix M = 2, the determinant b 1 b 2 of M is M = a 1 a 2 = a b 1 b 1 b 2 a 2 b 1. 2 Definition. The determinant of a 3 3 matrix M is M = Example. 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 = a 1 b 2 c 3 a 1 b 3 c 2 + a 2 b 3 c 1 a 2 b 1 c 3 + a 3 b 1 c 2 a 3 b 2 c = = 1 Area of the Parallelogram formed by a and b Area = (base) (height) = kakkbk sin

20 VECTORS AND THE GEOMETRY OF SPACE Definition. The cross product a b is the vector orthogonal to a and b by the right-hand rule, whose magnitude is the area of the parallelogram determined by the vectors a and b, i.e., ka bk = kakkbk sin. Theorem (4.4). For nonzero vectors a and b in V 3, if is the angle between a and b (0 < < ), then ka bk = kakkbk sin. Theorem (4.2). For any vectors a and b in V 3, a b? a and a b? b. Definition (Algebraic Definition of Cross Product). a b = i j k a 1 a 2 a 3 b 1 b 2 b 3 = a 2 a 3 b 2 b 3 i a 1 a 3 b 1 b 3 j + a 1 a 2 b 1 b 2 k.

21 Example. h5, 1, 2i h 1, 3, 4i = 4. THE CROSS PRODUCT 21 i j k = (4 + 6)i + (2 20)j + (15 + 1)k = h10, 18, 16i kh10, 18, 16ik = p = p 680 = 2 p Theorem (4.1). For any vector a in V 3, a a = 0 and a 0 = 0. Example. i j k i j = but i j k j i = so the cross product is not commutative. = k, = k, i j = k j i = k j k = i k j = i k i = j i k = j

22 VECTORS AND THE GEOMETRY OF SPACE Theorem (4.3). For any a, b, c 2 V 3, and any scalar d, (1) a b = (b a) (anti-commutativity) (2) (da) b = d(a b) = a (db) (3) a (b + c) = a b + a c (4) (a + b) c = a c + b c (5) a (b c) = (a b) c (scalar triple product) (6) a (b c) = (a c)b (a b)c (vector triple product) Corollary (to Theorem 4.4). Two nonzero vectors a and b in V 3 are parallel if and only if a b = 0. Proof. a k b () (\ between a and b) = 0 or = () sin = 0 () ka bk = kakkbk sin = 0 () a b = 0. Volume of a Parallelepiped having noncoplanar vectors a, b, c as adjacent edges. Volume = (area of base) altitude = ka bk comp a b c = ka bk the absolute value of a scalar triple product. c (a b) ka bk = c (a b)

23 4. THE CROSS PRODUCT 23 Corollary (to Theorem 4.3). The scalar triple product of a, b, c is Note. a (b c) = a a (b c) = b (c a) = c (a b). i j k b 1 b 2 b 3 c 1 c 2 c 3 = ha 1, a 2, a 3 i b 2 b 3 c 2 c 3 i = 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 = a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 b 1 b 3 c 1 c 3 j + b 1 b 2 c 1 c 2 k Example. Find the volume of the parallelepiped with 3 adjacent edges formed by h 2, 1, 2i, h1, 1, 2i, and h2, 1, 3i. V = = = 15!

24 VECTORS AND THE GEOMETRY OF SPACE Example. Find the distance from a point Q to the line through P and R. k! P Q! P Rk = k! P Qkk! P Rk sin = d = k P! Q P! Rk k P! Rk {z} d=k! P Qk sin k! P Rk d =) Problem (Page 732 #20). Find the distance from the point Q = (1, 3, 1) to the line through P = (1, 3, 2) and R = (1, 0, 2).!! P Q = h0, 0, 3i and P R = h0, 3, 0i =) d = kh0, 0, 3i h0, 3, 0ik kh0, 3, 0ik = 1 3 k i j k k = 1 3 k9ik = 1 (9) = 3. 3

25 4. THE CROSS PRODUCT 25 Example (Torque applied by a wrench). F = force applied at end of handle. = torque applied to the bolt. r = position vector for handle. Then = r F =) k k = kr Fk = krkkfk sin Problem (Page 732 #28). If you apply a force of magnitude 40 pounds at the end of an eighteen-inch-long wrench at an angle of to the wrench, find 3 the magnitude of the torqe applied to the wrench. kfk = 40, krk = 18, = 3 =) k k = sin 3 = 720 p 3 2 = 360p Maple. See crossprod(10.4).mw or crossprod(10.4).pdf.

26 VECTORS AND THE GEOMETRY OF SPACE 5. Lines and Planes in Space Lines in 3-space: (1) Vector equation specified by two points or a point and a direction. P = OP! 1 + ta = hx 1, y 1, z 1 i + tha 1, a 2, a 3 i = hx 1 + ta 1, y 1 + ta 2, z 1 + ta 3 i or! P 1 P k a =) P! 1 P = ta =) hx x 1, y y 1, z z 1 i = tha 1, a 2, a 3 i Note. If given two points, use one point with the direction vector determined by the two points. (2) Parametric equations x = x 1 + ta 1, y = y 1 + ta 2, z = z 1 + ta 3 or x x 1 = ta 1, y y 1 = ta 2, z z 1 = ta 3 (3) Symmetric equations (solve for t in the three equations above, and equate the results). x x 1 = y y 1 = z z 1 a 1 a 2 a 3

27 5. LINES AND PLANES IN SPACE 27 Problem (Page 740 #8). Find the vector, parametric and symmetric equations of the line through ( 3, 1, 0) and perpendicular to both h0, 3, 1i and h4, 2, 1i. h0, 3, 1i h4, 2, 1i = i j k = h1, 4, 12i is perpendicular to bothh0, 3, 1i and h4, 2, 1i. Then the vector equation is the parametric equations are hx + 3, y 1, zi = th1, 4, 12i, x + 3 = t, y 1 = 4t, z = 12t, and the symmetric equations are x + 3 = y 1 = z For t = 3, x = 0, y = 13, and z = 36, so (0, 13, 36) is on this line and each quotient of the symmetric equations is 3. Definition. Let l 1 and l 2 be two lines in R 3, with parallel (or direction) vectors a and b, respectively, and let be the angle between a and b. (1) l 1 and l 2 are parallel whenever a and b are parallel. (2) If l 1 and l 2 intersect, then (a) the angle between l 1 and l 2 is. (b) l 1 and l 2 are orthogonal whenever a and b are orthogonal. (3) Nonparallel, nonintersecting lines are called skew lines.

28 VECTORS AND THE GEOMETRY OF SPACE Problem (Page 740 #14). Determine whether the following lines are parallel, skew, or intersect: 8 8 >< x = 1 2t >< x = 3 + 2s y = 2t y = 2 >: z = 5 t >: z = 3 + 2s We have hx 1, y, z 5i = th 2, 2, 1i and hx 3, y + 2, z 3i = sh2, 0, 2i. Since h 2, 2, 1i 6= ch2, 0, 2i, lines are not k. Do lines intersect? Equate x s: Equate y s and substitute: 1 2t = 3 + 2s =) 2t = 2s 2. 2t = 2 =) 2 = 2s 2 =) s = 0 and t = 1. Substitute into z s: = 3 + 0, so the lines do not intersect, and thus we have skew lines.

29 Planes in R 3 5. LINES AND PLANES IN SPACE 29 A plane in space is determined by a vector a = ha 1, a 2, a 3 i that is normal to the plane and a point P 1 (x 1, y 1, z 1 ) lying in the plane. A vector is normal to a plane if it is orthogonal to every vector in the plane. (1) Linear equation: a! P 1 P = ha 1, a 2, a 3 i hx x 1, y y 1, z z 1 i = Simplifying, every equation of the form a 1 (x x 1 ) + a 2 (y y 1 ) + a 3 (z z 1 ) = 0 {z } linear equation for the plane ax + by + cz + d = 0 is the equation of a plane with normal vector ha, b, ci 6= 0. (2) Vector equation: a! P 1 P = 0 Definition. Two planes are parallel when their normal vectors are parallel, and orthogonal when their normal vectors are orthogonal.

30 VECTORS AND THE GEOMETRY OF SPACE Example. Three noncollinear points determine a plane. Find the plane containing P ( 1, 3, 0), Q(3, 2, 4), and R(1, 1, 5). We first find two vectors in the plane by using the points given.!! P Q = h4, 1, 4i and QR = h 2, 3, 1i. We then find a normal vector n by taking their cross product.! P Q QR! = Therefore, the equation is or i j k = h11, 12, 14i. 11(x + 1) 12(y 3) 14(z 0) = 0 11x 12y 14z + 47 = 0. Intercepts are x = 47 11, y = 47 12, z =

31 5. LINES AND PLANES IN SPACE 31 Problem (Page 740 #24). Find the plane containing the point (3, 0, 1) and? to the planes x + 2y z = 2 and 2x z = 1. Normal vectors to the two given planes are n 1 = h1, 2, 1i and n 2 = h2, 0, 1i. Then a normal vector to the sought plane is Then the equation is or n = n 1 n 2 = i j k = h 2, 1, 4i. 2(x 3) y 4(z + 1) = 0 2x + y + 4z = 2.

32 VECTORS AND THE GEOMETRY OF SPACE Distance from a point P 0 to a plane ax + by + cz + d = 0 We measure the distance orthogonal to the plane. (1) Choose any point P 1 in the plane. (2) a = ha, b, ci is a normal vector to the plane. (3) The distance d is then 2. d = comp a! P1 P 0. Problem (Page 741 #40). Find the distance from (1, 3, 0) to 3x+y 5z = P 0 = (1, 3, 0), a normal vector is h3, 1, 5i, and we choose P 1 = (0, 2, 0) since! it satisfies the equation. Then P 1 P 0 = h1, 1, 0i, and d = comp h3,1, 5i h1, 1, 0i = h1, 1, 0i h3, 1, 5i kh3, 1, 5ik = 4 p 35 = 4 p 35 Note. You can see the derivation of the distance formula on page 739 of the text.

33 5. LINES AND PLANES IN SPACE 33 Problem (Page 740 #36 Finding the intersection of 2 planes). Find the intersection of 3x + y z = 2 and 2x 3y + z = 1. (1) Possibilities are that the intersection is a line, a plane, or does not exist (the planes are parallel. Since normal vectors to the planes, h3, 1, 1i and h2, 3, 1i, are not parallel, the intersection here is a line. (2) Solve each equation for the same variable z seems easiest here. (3) Set these equal to each other. z = 3x + y 2 and z = 2x + 3y 1 3x + y 2 = 2x + 3y 1 (4) Solve for a second variable we choose y. 2y = 5x 1 =) y = 5 2 x 1 2 (5) Substitute this into either first equation. 5 z = 3x + 2 x 1 2 =) z = x 5 2 or 5 z = 2x x 1 1 =) z = x 5 2 (6) Set the third variable equal to a parameter t. x = t (7) Then parametric equations for the line are 8 >< x = t y = 5 2 >: t 1 2 z = 11 2 t 5 2 Note. Compare this problem to Example 5.8 on page 738.

34 VECTORS AND THE GEOMETRY OF SPACE Maple. See Lines and Planes(10.5).mw or Lines and Planes(10.5).pdf Also see Matching Lines. 6. Sufaces in Space Maple. See surfaces(10.6).mw or surfaces(10.6).pdf Also see Drawing Surfaces, Drawing 3D Surfaces, and Matching Quadric Surfaces.