VECTORS IN THREE DIMENSIONS

1 CHAPTER 2. BASIC TRIGONOMETRY 1 INSTITIÚID TEICNEOLAÍOCHTA CHEATHARLACH INSTITUTE OF TECHNOLOGY CARLOW VECTORS IN THREE DIMENSIONS 1 Vectors in Two Dimensions A vector is an object which has magnitude and direction. Man phsical quantities, such as velocit, acceleration, force, electric field and magnetic field are eamples of vector quantities. Displacement between points ma also be represented using vectors. We stud some relationship between algebra and geometr. We shall first stud some algebra which is motivated b geometric considerations. We then use the algebra later to better understand some problems in geometr. This mathematics will form the basis of the stud of computer graphics. Vectors are central to the design of an two-dimensional or three-dimensional computer game. The are used to represent points in space, like corners of a door or window or the location of an object in a scene. The are also used to describe a direction, for eample the orientation of a camera or the direction in which a gun is pointing. A vector in two-dimensions R 2 can be described as an ordered pair u = (u 1, u 2 ), where u 1, u 2 R. Definition Two vectors u = (u 1, u 2 ) and v = (v 1, v 2 ) in R 2 are said to be equal, denoted b u = v, if and onl if u 1 = v 1 and u 2 = v 2. Defining vectors in R 2 as ordered pairs of real numbers enables us to state precisel when two vectors are equal it also provides us with the easiest wa of defining addition and various kinds of multiplication. To describe the position of an point in two-dimensions we ma choose two aes and which are mutuall perpendicular and intersect in a point O called the origin, as shown.

2 CHAPTER 2. BASIC TRIGONOMETRY 2 (u 1, u 2 ) u 2 O u α u 1 An point P in two dimensions corresponds the ordered pair (u 1, u 2 ) of real numbers, where u 1 represents the magnitude of the component vector along -ais and u 2 represents the magnitude of the component vector along -ais. The magnitude of the vector u, from Pthagoras theorem, is given as u = u 2 1 + u2 2 The direction of the vector u is defined b α ( ) α = tan 1 u2 u 1 1.1 Addition of Vectors Definition For an two vectors u = (u 1, u 2 ) and v = (v 1, v 2 ) in R 2, we define their sum to be u + v = (u 1, u 2 ) + (v 1, v 2 ) = (u 1 + v 1, u 2 + v 2 ) Similarl, we define their difference to be u v = (u 1, u 2 ) (v 1, v 2 ) = (u 1 v 1, u 2 v 2 ) Addition of vectors ma be pictured using the parallelogram law.

3 CHAPTER 2. BASIC TRIGONOMETRY 3 O P Q P + Q Eample For the following pair of vectors u = (1, 6) and v = ( 5, 2) in R 2, we can calculate u + v = (1, 6) + ( 5, 2) = ( 4, 8) u v = (1, 6) ( 5, 2) = (6, 4) Theorem 1 (VECTOR ADDITION) i For ever u, v R 2, we have u + v R 2. ii For ever u, v, w R 2, we have u + ( v + w) = ( u + v) + w. iii For ever u R 2, we have u + 0 = u where 0 = (0, 0) R 2. iv For ever u R 2, there eists v R 2 such that u + v = 0. v For ever u, v R 2, we have u + v = v + u. 1.2 Scalar Multiplication of Vectors Definition For an vector u = (u 1, u 2 ) in R 2 and an scalar c R, we define the scalar multiple to be c u = c(u 1, u 2 ) = (cu 1, cu 2 )

4 CHAPTER 2. BASIC TRIGONOMETRY 4 Scalar multiplication ma be pictured as follows 2P P Q O Q Eample For the following pair of vectors u = (2, 3) and v = ( 1, 5) in R 2, we can evaluate 2 u + 4 v = 2(2, 3) + 4( 1, 5) = (4, 6) + ( 4, 20) = (0, 26) 2 u v = 2(2, 3) ( 1, 5) = (4, 6) ( 1, 5) = (5, 1) Theorem 2 (SCALAR MULTIPLICATION) i For ever c R and u R 2, we have c u R 2. ii For ever c R and u, v R 2, we have c( u + v) = c u + c v. iii For ever a, b R and u R 2, we have (a + b) u = a u + b u. iv For ever a, b R and u R 2, we have (ab) u = a(b u). v For ever u R 2, we have 1 u = u. Eercise For the following pair of vectors u = (4, 3) and v = (1, 7) in R 2, evaluate i 3 u + 3 v, ii u + 2 v, iii 7 u 3 v.

5 CHAPTER 2. BASIC TRIGONOMETRY 5 Remark There is another wa in which vectors ma be pictured namel as arrows in two dimensions. The vector (u 1, u 2 ) can be pictured b an arrow with initial point O and terminal point (u 1, u 2 ). It is, however, convenient to picture vectors in a more general wa. Consider an arrow with the initial point P = ( 1, 1 ) and terminal point Q = ( 2, 2 ). This arrow is denoted b P Q. We define P Q = ( 2 1, 2 1 ) = Q P We can picture this as follows using the parallelogram law P O P P Q Q P Q Note that this vector is alread represented b the arrow with initial point O and terminal point ( 2 1, 2 1 ). In fact, an one vector ma be represented b infinitel man arrows. We define two arrows to be equivalent whenever their corresponding components are equal it is the components, and not the individual initial and terminal points, which are used to see if two arrows are equivalent. Since components are determined b the length and direction of an arrow we can state that two arrows are equivalent whenever the have the same length and direction. Since one vector is now represented b an one of infinitel man equivalent arrows, we agree to regard these equivalent arrows as equal. The end result is that we ma picture a vector as an arrow which has a given length and lies in a given direction, and ma be positioned between an pair of points provided that the points determine the same length and direction.

6 CHAPTER 2. BASIC TRIGONOMETRY 6 Eample Let P = (2, 5), Q = (3, 2), R = (1, 3) and S = ( 1, 3) be four points in two dimensions, as shown. R O S Q P Then P Q = (1, 3) OR = (1, 3) SO = (1, 3) These arrows represent the same vector, namel, A = (1, 3) and we write P Q = OR = SO = A = (1, 3) Eercise In each case write the vector u in terms of components i u is a vector from the point A(2, 5) to the point B(0, 4), ii u is a vector from the point A( 1, 3) to the point B(5, 2), iii u is a vector from the point A(5, 12) to the point B( 3, 6).

7 CHAPTER 2. BASIC TRIGONOMETRY 7 1.3 Magnitude and Direction Definition For an vector u = (u 1, u 2 ) in R 2, we define the magnitude of u to be the non-negative real number u = u 2 1 + u2 2 The direction of the vector u is defined b α ( ) α = tan 1 u2 u 1 u 2 O (u 1, u 2 ) u α u 1 Remark i Suppose that P = ( 1, 1 ) and Q = ( 2, 2 ) are two points in R 2. To calculate the distance d(p, Q) between the two points, we must first find a vector from P to Q. This is given b ( 2 1, 2 1 ). The distance d(p, Q) is then the magnitude of this vector, so that d(p, Q) = ( 2 1 ) 2 + ( 2 1 ) 2 Hence, the definition of the magnitude (or norm) of a vector u is simpl the distance from O to the point (u 1, u 2 ). ii A vector of magnitude 1 is called a unit vector or normalized vector. determines a unit vector ( ) 1 u u = u1 u, u 2 u An non-zero vector u

8 CHAPTER 2. BASIC TRIGONOMETRY 8 Eample The vector u = (3, 4) has magnitude 5. This vector has direction α = 53 13 o. Eample The vector u = (2, 5) has magnitude 5 385. This vector has direction α = 68 199 o. Eample The vector u = (1, 2) has magnitude 2 236. This vector has direction α = 63 434 o. Eample The vector u = ( 1, 2) has magnitude 2 236. ( 1, 2) 1 φ 2 α From the diagram ( ) 2 φ = tan 1 = 63 43 o 1 Hence α = 180 o 63 43 o = 116 57 o. Note The angle of direction α is the alwas quoted relative to the positive ais. Eample The vector u = ( 2, 3) has magnitude 3 606. This vector has direction α = 123 69 o. Eample The vector u = ( 1, 1) has magnitude 1 414. ( 1, 1) 1 φ α 1

9 CHAPTER 2. BASIC TRIGONOMETRY 9 From the diagram ( ) 1 φ = tan 1 = 45 o 1 Hence α = 180 o + 45 o = 225 o. Eample The vector u = ( 4, 5) has magnitude 6 403. This vector has direction α = 231 34 o. Eample The vector u = ( 9, 12) has magnitude 15. This vector has direction α = 233 13 o. Remark We ma be required to determine the components u 1 and u 2 of the vector u = (u 1, u 2 ) when presented with the magnitude and direction of u onl. If, for eample, the magnitude and direction of two distinct vectors are presented converting each vector to components will allow for simpler addition (subtraction or scalar multiplication) of the vectors. u 2 u α O u 1 (u 1, u 2 ) For u = (u 1, u 2 ) we can write, u 1 = u cos α u 2 = u sin α Eample The vector u with magnitude 5 and direction α = 40 o with positive ais has components u 1 = 5 cos 40 o u 2 = 5 sin 40 o Hence u = (3 83, 3 21).

10 CHAPTER 2. BASIC TRIGONOMETRY 10 Eample The vector u with magnitude 200 and direction α = 210 o with positive ais has components u 1 = 200 cos 210 o u 2 = 200 sin 210 o Hence u = ( 173 21, 100). Eample The vector u has magnitude 10 and direction α = 45 o with positive ais. The vector v has magnitude 15 and direction α = 205 o with positive ais. Determine the magnitude and direction of each of the following vectors: i u + v ii u 2 v iii 2 u 3 v Solution: Let u = (u 1, u 2 ) and v = (v 1, v 2 ). Now u 1 = 10 cos 45 o = 7 07 u 2 = 10 sin 45 o = 7 07 v 1 = 15 cos 205 o = 13 59 v 2 = 15 sin 205 o = 6 34 Hence u = (7 07, 7 07) and v = ( 13 59, 6 34). i u + v u v

11 CHAPTER 2. BASIC TRIGONOMETRY 11 u + v = (7 07, 7 07) + ( 13 59, 6 34) = (7 07 13 59, 7 07 6 34) = ( 6 52, 0 73) The vector u + v has magnitude 6 56. This vector has direction α = 96 39 o. ii u 2 v = (7 07, 7 07) 2( 13 59, 6 34) = (7 07, 7 07) ( 27 18, 12 68) = (7 07 + 27 18, 7 07 + 12 68) = (34 25, 19 75) The vector u 2 v has magnitude 39 54. This vector has direction α = 29 97 o. iii 2 u 3 v = 2(7 07, 7 07) 3( 13 59, 6 34) = (14 14, 14 14) ( 40 77, 19 02) = (14 14 + 40 77, 14 14 + 19 02) = (54 91, 33 16) The vector 2 u 3 v has magnitude 64 15. This vector has direction α = 31 13 o. Remark For computer games and graphics programming the component representation of a vector with round (or square) brackets is used. This notation will be used in all modern computer programming languages. In C#, for eample, the DrawLine() method draws a line from one vector point (1, 1) to a second vector point (2, 2) on the graphics form: g.drawline(p, 1, 1, 2, 2); //Drawline method The following lines of code from C# will draw a line from the point (100, 150) to (300, 400): privatevoidf orm1 P aint(object sender, P ainteventargs e) { Graphics g = e.graphics; //T he graphics class P en p = newp en(color.red); //T he P en class g.drawline(p, 100, 150, 300, 400); //Drawline method }

12 CHAPTER 2. BASIC TRIGONOMETRY 12 2 Vectors in Three Dimensions A vector in three-dimensions R 3 u 1, u 2, u 3 R. can be described as an ordered triple u = (u 1, u 2, u 3 ), where Definition Two vectors u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ) in R 3 are said to be equal, denoted b u = v, if and onl if, u 1 = v 1, u 2 = v 2 and u 3 = v 3. Defining vectors in R 3 as ordered triples of real numbers enables us to state precisel when two vectors are equal it also provides us with the easiest wa of defining addition and various kinds of multiplication, as we will show later. To describe the position of an point in space, we ma choose three aes, and z which are mutuall perpendicular and intersect in a point O called the origin, as shown. u O z (u 1, u 2, u 3 ) An point in space corresponds to the ordered triple (u 1, u 2, u 3 ) of real numbers, where u 1 represents the magnitude of the component vector along -ais, u 2 represents the magnitude of the component vector along -ais and u 3 represents the magnitude of the component vector along the z-ais. The magnitude of the vector u, from Pthagoras theorem, is given as u = u 2 1 + u2 2 + u2 3

13 CHAPTER 2. BASIC TRIGONOMETRY 13 The direction of the vector in three-dimensions R 3 in is defined b three angles θ, θ and θ z the vector makes with the -ais, -ais and z-ais respectivel. For a vector u = (u 1, u 2, u 3 ) in R 3 u 1 = u cos θ u 2 = u cos θ u 3 = u cos θ z 2.1 Addition of Vectors Definition For an two vectors u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ) in R 3, we define their sum to be u + v = (u 1, u 2, u 3 ) + (v 1, v 2, v 3 ) = (u 1 + v 1, u 2 + v 2, u 3 + v 3 ) Similarl, we define their difference to be u v = (u 1, u 2, u 3 ) (v 1, v 2, v 3 ) = (u 1 v 1, u 2 v 2, u 3 v 3 ) Addition of vectors ma be pictured using the parallelogram law. P Q O z P + Q

14 CHAPTER 2. BASIC TRIGONOMETRY 14 Eample For the following pair of vectors u = (8, 4, 3) and v = ( 2, 2, 0) in R 3, we can calculate, for eample u + v = (8, 4, 3) + ( 2, 2, 0) = (6, 6, 3) u v = (8, 4, 3) ( 2, 2, 0) = (10, 2, 3) Theorem 3 (VECTOR ADDITION) i For ever u, v R 3, we have u + v R 3. ii For ever u, v, w R 3, we have u + ( v + w) = ( u + v) + w. iii For ever u R 3, we have u + 0 = u where 0 = (0, 0, 0) R 3. iv For ever u R 3, there eists v R 3 such that u + v = 0. v For ever u, v R 3, we have u + v = v + u. 2.2 Scalar Multiplication of Vectors Definition For an vector u = (u 1, u 2, u 3 ) in R 3 and an scalsr c R, we define the scalar multiple to be c u = c(u 1, u 2, u 3 ) = (cu 1, cu 2, cu 3 ) Scalar multiplication ma be pictured as follows 2P P Q O Q z

15 CHAPTER 2. BASIC TRIGONOMETRY 15 Eample For the following pair of vectors u = (2, 3, 4) and v = ( 1, 3, 8) in R 3, we can evaluate, for eample 2 u + 4 v = 2(2, 3, 4) + 4( 1, 3, 8) = (4, 6, 8) + ( 4, 12, 32) = (0, 18, 24) 2 u v = 2(2, 3, 4) ( 1, 3, 8) = (4, 6, 8) ( 1, 3, 8) = (5, 3, 16) Theorem 4 (SCALAR MULTIPLICATION) i For ever c R and u R 3, we have c u R 3. ii For ever c R and u, v R 3, we have c( u + v) = c u + c v. iii For ever a, b R and u R 3, we have (a + b) u = a u + b u. iv For ever a, b R and u R 3, we have (ab) u = a(b u). v For ever u R 3, we have 1 u = u. Eercise For the following pair of vectors u = (1, 3, 5) and v = (1, 2, 4) in R 3, evaluate i 2 u + 3 v, ii u 5 v, iii 4 u + 3 v. Remark There is another wa in which vectors ma be pictured namel as arrows in three dimensions. The vector (u 1, u 2, u 3 ) can be pictured b an arrow with initial point O and terminal point (u 1, u 2, u 3 ). It is, however, convenient to picture vectors in a more general wa. Consider an arrow with the initial point P = ( 1, 1, z 1 ) and terminal point Q = ( 2, 2, z 2 ). This arrow is denoted b P Q. We define P Q = ( 2 1, 2 1, z 2 z 1 ) = Q P We can picture this as follows using the parallelogram law

16 CHAPTER 2. BASIC TRIGONOMETRY 16 z P P P Q O Q P Q Recall that we ma picture a vector as an arrow which has a given length and lies in a given direction, and ma be positioned between an pair of points provided that the points determine the same length and direction. The following eample illustrates this point. Eample Let P = (2, 5, 4), Q = (3, 2, 6), R = (1, 3, 2) and S = ( 1, 3, 2) be four points in space, as shown. R Then O S Q P z P Q = (1, 3, 2) OR = (1, 3, 2) SO = (1, 3, 2)

17 CHAPTER 2. BASIC TRIGONOMETRY 17 These arrows represent the same vector, namel, A = (1, 3, 2) and we write P Q = OR = SO = A = (1, 3, 2) Eercise In each case write the vector u in terms of components i u is a vector from the point A(1, 5, 4) to the point B(2, 0, 4), ii u is a vector from the point A(1, 2, 3) to the point B(4, 5, 6), iii u is a vector from the point A( 2, 1, 9) to the point B(3, 6, 8). [Solution: u = (1, 5, 0 ), u = (3, 3, 3 ), u = (5, 7, 1 )]. Eercise Let u = (2, 1, 3) v = ( 4, 2 5, 3) w = (1, 1, 2) Write each of the following in terms of components i 2 u ii 3 u 2 v iii v 2 u + 4 w iv 2( u + v) w [Solution: i (4, 2, 6 ), ii (14, 8, 3 ), iii ( 4, 8 5, 11 ), iv ( 5, 2, 14 ) ].

18 CHAPTER 2. BASIC TRIGONOMETRY 18 Eercise Let P = (2, 6, 8), Q = (6, 2, 5), R = (4, 4, 3), S = ( 10, 2, 5) and T = ( 6, 6, 2) be five points in space, as shown. T S O z R Q P Show that P Q = OR = ST. 2.3 Magnitude and Direction Definition For an vector u = (u 1, u 2, u 3 ) in R 3, we define the magnitude of u to be the non-negative real number u = u 2 1 + u2 2 + u2 3 The direction of the vector in three-dimensions R 3 in is defined b three angles θ, θ and θ z the vector makes with the -ais, -ais and z-ais respectivel. For a vector u = (u 1, u 2, u 3 ) in R 3 cos θ = u 1 u cos θ = u 2 u cos θ z = u 3 u

19 CHAPTER 2. BASIC TRIGONOMETRY 19 θ z z (u 1, u 2, u 3 ) θ θ O These angles are difficult to picture, since the are not in the plane onl or the plane onl or the z plane onl. For u = (u 1, u 2, u 3 ) in R 3, we can write, u 1 = u cos θ u 2 = u cos θ u 3 = u cos θ z These equations will determine the components u 1, u 2 and u 3 of the vector u = (u 1, u 2, u 3 ) in R 3 when presented with the magnitude and direction of u onl. Furthermore cos 2 θ + cos 2 θ + cos 2 θ z = 1 Remark A vector of magnitude 1 is called a unit vector or normalised vector. An non-zero vector determines a unit vector 1 u u = ( ) u1 u, u 2 u, u 3 u 1 u u = (cos θ, cos θ, cos θ z ) This normalised form ma be used to convenientl calculate the angles θ, θ, θ z for a given vector u = (u 1, u 2, u 3 ) in R 3. This form is usuall used to describe the direction of an vector in R 3.

20 CHAPTER 2. BASIC TRIGONOMETRY 20 Eample We can write v = (2, 3, 4) as a unit vector (or normalised vector) along u ( ) 1 u u = u1 u, u 2 u, u 3 u Now u = 29. Hence 1 u u = ( 2 29, 3 29, ) 4 29 = (cos θ, cos θ, cos θ ) It follows that cos θ = cos θ = cos θ z = ( ) 2 29 ( ) 3 29 ( ) 4 29 = 0 3714 = 0 5571 = 0 7428 Hence θ = 68 20 θ = 56 15 θ z = 42 03 Eercise For each of the following vectors in R 3, find the angles θ, θ, θ z, i.e., the angle the vector makes with the ais, ais and z ais respectivel. i u = ( 2, 1, 1) ii u = (1, 1, 1) iii u = ( 4, 2, 2) [Solution: i θ = 144 74 o, θ = 65 91 o, θ z = 65 91 o ii θ = 54 74 o, θ = 125 26 o, θ z = 54 74 o iii θ = 144 74 o, θ = 114 09 o, θ z = 65 91 o ]. Remark A ver common problem in games is that an object moves a distance in a particular direction and we have to determine where it ends up, so that we can draw it again in the new position. Suppose that an object is at position A(1, 2, 3) in one frame and it moves 10 units in the direction θ = 75 o, θ = 50 o and θ z = 43 86 o before the net frame. What is the new position in the second frame?

21 CHAPTER 2. BASIC TRIGONOMETRY 21 Let u = (u 1, u 2, u 3 ) with, u 1 = u cos θ = 10 cos 75 o u 2 = u cos θ = 10 cos 50 o u 3 = u cos θ z = 10 cos 43 86 o Hence u = (2 588, 6 43, 7 21). The new position in the second frame will be A + u = (1, 2, 3) + (2 588, 6 43, 7 21) = (3 588, 8 43, 10 21) Eercise For each of the following objects defined in frame one b a vector A, find its new position in frame two: i The object starts from A(2, 3, 1) and moves 7 units in the direction θ = 90 o, θ = 130 o and θ z = 40 o. ii The object starts from A(1, 2, 3) and moves 2 8 units in the direction θ = 120 o, θ = 60 o. iii The object starts from A( 1, 1, 2) and moves 1 5 units in the direction θ = 50 o, θ = 70 o. [Solution: i (2, 1 50, 6 36 ) ii ( 0 4, 3 4, 4 98 ) iii (0 03, 1 96, 2 51 ) ].

22 CHAPTER 2. BASIC TRIGONOMETRY 22 2.4 The Scalar Product Definition Suppose that u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ) are vectors in R 3 and that θ [0, π] represents the angle between them. We define the scalar product u. v of u and v b Alternativel, we can write u. v = u v cos θ u. v = u 1 v 1 + u 2 v 2 + u 3 v 3 Theorem 5 (SCALAR PRODUCT) Suppose that u, v, w R 3 and c R, then i u. v = v. u ii u.( v + w) = ( u. v) + ( u. w) iii c( u. v) = (c u). v = u.(c v) iv u. u 0 v u. u = 0 if and onl if u = 0 Remark i The scalar product is also known as the dot product or the inner product of u and v. ii We sa that two non-zero vectors in R 3 are orthogonal if the angle between them is π 2. It follows immediatel from the definition of scalar product that an two non-zero vectors u, v R 3 are orthogonal if and onl if u. v = 0. iii Using the definition of scalar product we can calculate the angle between u and v since cos θ = u. v u v Eample Suppose u = (2, 4, 6) and v = (1, 2, 3). Then u. v = u 1 v 1 + u 2 v 2 + u 3 v 3 = 2.1 + 4.( 2) + 6.3 = 12

23 CHAPTER 2. BASIC TRIGONOMETRY 23 Eample Suppose u = (2, 0, 0) and v = (1, 1, 2). Then we have u. v = 2. Note now that u = 2 and v = 2. It follows that cos θ = u. v u v = 2 4 = 1 2 Hence ( ) 1 θ = cos 1 = 60 o 2 Eample Suppose u = ( 4, 1, 1) and v = (1, 2, 5). Then we have u. v = 3. Note now that u = 18 and v = 30. It follows that cos θ = u. v u v = 3 = 3 18. 30 23 24 Hence ( ) 3 θ = cos 1 = 82 58 o 23 24 Eample Suppose u = (2, 3, 5) and v = (1, 1, 1). Then we have u. v = 0. It follows that u and v are orthogonal. Eercise Let u = (2, 4, 3) and v = (8, 1, 1). Determine the scalar product u. v and hence determine the angle between u and v. Eercise Let u = (1, 2, 5) and v = (0, 1, 1). determine the angle between u and v. Determine the scalar product u. v and hence Eercise Let u = (5, 6, 9) and v = (1, 1, 1). Determine the scalar product u. v and hence determine the angle between u and v. 2.5 Components and Projections Definition Let u and v be two non-zero vectors and θ the angle between them. The scalar component of u along v is the number u. v v * Since u. v = u v cos θ, the scalar component ma also be written as u cos θ.

24 CHAPTER 2. BASIC TRIGONOMETRY 24 The vector projection of u along v, denoted b proj v ( u), is the vector defined b proj v ( u) = ( ) u. v v v v that is, the scalar multiple of the direction of v b the scalar component of u along v. Remark i In some applications it can be useful to decompose or resolve a vector u into two vectors one parallel to non-zero vector v and the other perpendicular to v. u w v θ proj v ( u) To resolve a given vector u into two vectors one parallel to to a given non-zero vector v and the other perpendicular to the vector v we first calculate proj v ( u), the vector projection of u along v and secondl a perpendicular vector to v, which we will label w and is given as w = u proj v ( u) The resolvent of proj v ( u) and w will ield u. Furthermore, their scalar product is zero. ii As the name suggests, the scalar component is a scalar and the vector projection is a vector. iii The proj v ( u) has the same direction as v if θ is acute, and the opposite direction if θ is obtuse. iv The length or magnitude of proj v ( u) is u. v v that is, the absolute value of the scalar component of u along v.

25 CHAPTER 2. BASIC TRIGONOMETRY 25 Eample Let u = (2, 1, 3) and v = (1, 3, 1). To determine the vector projection of u along v we have proj v ( u) = ( ) u. v v v v Hence proj v ( u) = ( ) 2 + 3 3 (1, 3, 1). = 2 (1, 3, 1) 11 11 11 Note Consider the eample above with the diagram depicting a vector projection. From the vector u we have determined proj v ( u), the vector projection of u along v. u w v θ proj v ( u) We can determine the vector w as follows w = u proj v ( u) = (2, 1, 3) 2 5 (1, 3, 1) = (4, 1, 7) 11 11 The resolvent of proj v ( u) and w should ield u. proj v ( u) + w = 2 ( 5 22 (1, 3, 1) + (4, 1, 7) = 11 11 11, 11 11, 33 ) = (2, 1, 3) 11 Finall, their scalar product is zero. proj v ( u). w = 2 11 (1, 3, 1). 5 40 (4, 1, 7) = 11 11 + 30 11 70 11 = 0 as epected.

26 CHAPTER 2. BASIC TRIGONOMETRY 26 Eercise Let u = (2, 0, 1) and v = (3, 1, 2). Determine the vector projection of u along v Eercise Let u = ( 1, 2, 4) and v = (0, 1, 6). Determine the vector projection of u along v Eercise Let u = (2, 2, 7) and v = (3, 6, 5). Determine the vector projection of u along v Eercise Let u = (2, 1, 2) and v = (6, 1, 0). Resolve the vector u into vectors parallel and perpendicular to the vector v. Eercise Let u = (3, 4, 5) and v = (1, 1, 2). Resolve the vector u into vectors parallel and perpendicular to the vector v. 2.6 The Vector Product We now discuss a product of vectors unique to R 3. The idea of vector products has a wide applications in geometr, phsics and engineering, and is motivated b the wish to find a vector that is perpendicular to two given vectors. Definition Suppose that u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ) are vectors in R 3 and that θ [0, π] represents the angle between them. Let n be a unit vector perpendicular to both u and v. Then the vector product (or cross product) of u and v is the vector denoted b u v and defined b u v = u v sin θ n Alternativel 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 ) Remark The vector product u v ields a vector in R 3. In order to develop this component representation of u v we will switch momentaril from the component representation of a vector u = (u 1, u 2, u 3 ) to its equivalent cartesian form u = u 1 i + u 2 j + u 3 k where the three unit vectors i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1), the unit vectors along the, and z-aes respectivel.

27 CHAPTER 2. BASIC TRIGONOMETRY 27 z O u = u 1 i + u 2 j + u 3 k Suppose u = u 1 i + u 2 j + u 3 k v = v 1 i + v 2 j + v 3 k u v = (u 1 i + u 2 j + u 3 k) (v1 i + v 2 j + v 3 k) = u 1 v 1 i i + u 1 v 2 i j + u 1 v 3 i k +u 2 v 1 j i + u 2 v 2 j j + u 2 v 3 j k +u 3 v 1 k i + u 3 v 2 k j + u 3 v 3 k k Using each of the following facts u v = u v sin θ n u v = ( v u) we can establish each of the following i i = 0 i j = k j i = k j j = 0 j k = i i k = j k k = 0 k i = j k j = i Hence, we have u v = u 1 v 1 (0) + u 1 v 2 k + u1 v 3 ( j) +u 2 v 1 ( k) + u 2 v 2 (0) + u 2 v 3 i +u 3 v 1 j + u 3 v 2 ( i) + u 3 v 3 (0) = (u 2 v 3 u 3 v 2 ) i + (u 3 v 1 u 1 v 3 ) j + (u 1 v 2 u 2 v 1 ) k

28 CHAPTER 2. BASIC TRIGONOMETRY 28 Finall, returning to our component representation we have 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 ) Remark A convenient wa of determining the vector product u v is as follows u v = det Using the cofactor epansion b row 1, we have i j k u 1 u 2 u 3 v 1 v 2 v 3 ( ) ( ) ( ) u2 u 3 u1 u 3 u1 u 2 u v = det i det j + det k v 2 v 3 v 1 v 3 v 1 v 2 = ( ( ) ( ) ( )) u2 u 3 u1 u 3 u1 u 2 det, det, det v 2 v 3 v 1 v 3 v 1 v 2 = (u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ) We will first show that the vector product u v is orthogonal to both u and v. Theorem 6 Suppose that u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ) are vectors in R 3. Then i u.( u v) = 0 ii v.( u v) = 0 Eample Suppose that u = (1, 1, 2) and v = (3, 0, 2). Then u v = det = ( det i j k 1 1 2 3 0 2 ( 1 2 0 2 ), det ( 1 2 3 2 ) ( 1 1, det 3 0 )) = ( 2 + 0, (2 6), 0 + 3) = ( 2, 4, 3) Note that (1, 1, 2).( 2, 4, 3) = 0 and (3, 0, 2).( 2, 4, 3) = 0.

29 CHAPTER 2. BASIC TRIGONOMETRY 29 Eercise For the vectors u = (1, 2, 3) and v = (3, 2, 1) in R 3, evaluate i u v ii v u What comment can ou make about ou answer. Theorem 7 (VECTOR PRODUCT) Suppose that u, v, w R 3 and c R. Then i u v = ( v u); ii u ( v + w) = ( u v) + ( u w); iii ( u + v) w = ( u w) + ( v w); iv c( u v) = (c u) v = u (c v); v u 0 = 0; vi u u = 0. Now to consider an application of vector product to evaluate the area of a parallelogram. To do this we first establish the following result. Theorem 8 Suppose that u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ) are non-zero vectors in R 3, and that θ [0, π] represents the angle between them. Then i u v 2 = u 2 v 2 ( u. v) 2 ii u v = u v sin θ Now consider a parallelogram below. v sin θ O v θ u u

30 CHAPTER 2. BASIC TRIGONOMETRY 30 The base of the parallelogram is given b u, and hence the height of the parallelogram is given as v sin θ. Therefore, from theorem 6 we can sa that the area of the parallelogram is given b u v. Theorem 9 Suppose that u, v R 3. Then the parallelogram with u and v as two of its sides has area u v. Eercise Let u = (1, 1, 4) and v = (4, 1, 7) in R 3. Determine the area of the parallelogram that is defined b u and v. 2.7 Scalar Triple Product Now suppose that u, v, w R 3 that do not lie all on the same plane. What is formed is a parallelepiped, i.e. a solid bod in which each face is a parallelogram, with u, v and w as three of its edges. v w P O u w v The base of the parallelepiped has area v w If the vector OP is perpendicular to the base of the parallelepiped, then OP is in the direction of v w. Now the height of the parallelepiped is equal to the norm of the orthogonal projection of u on v w. In other words, the parallelepiped has height proj v w ( u) = ( ) u.( v w) v w v w v w Hence, we have proj v w ( u) = u.( v w) v w

31 CHAPTER 2. BASIC TRIGONOMETRY 31 Therefore the volume of the parallelepiped is given b V = u.( v w) Theorem 10 Suppose that u, v, w R 3. Then the parallelepiped with u, v and w as three of its edges has volume u.( v w). Definition Suppose that u, v, w R 3. Then u.( v w) is called the scalar triple product of u, v and w. Remark It follows from theorem 9 that three vectors in R 3 are coplanar if and onl if their scalar triple product is zero. Eample Suppose that u = (1, 0, 1), v = (2, 1, 3) and w = (0, 1, 1). Then Hence u, v and w are coplanar. u.( v w) = det 1 0 1 2 1 3 0 1 1 = 0 Eample The volume of the parallelepiped with u = (1, 0, 1), v = (2, 1, 4) and w = (0, 1, 1) as three of its edges are given b We take the absolute value 1. V = u.( v w) = det 1 0 1 2 1 4 0 1 1 = 1 Eercise Let u = (2, 1, 3), v = (4, 1, 0) and w = (2, 0, 1). Show that u, v and w are coplanar. Eercise Let u = (5λ, 2λ, 3), v = (1, 1, 0) and w = (0, 2, 1). For which values of λ are these vectors coplanar?

32 CHAPTER 2. BASIC TRIGONOMETRY 32 2.8 Some Eercises Eercise Consider the following vectors u = (1, 2, 3) and v = (3, 2, 1) in R 3. i Evaluate u 4 v ii Evaluate u. v ii Determine 7 u 2 v iv Determine u v. v Determine the vector projection of u along v, i.e., proj v ( u) Eercise Consider the following vectors u = (1, 0, 1), v = (2, 1, 3) and w = (0, 1, 1) in R 3. i Evaluate u 4 v + 2 w ii Determine 7 u 2 v iii Evaluate u. v iv Determine the angle θ between u and v. v Determine the vector projection of u along v, i.e., proj v ( u) vi Calculate the components a, b and c of some non-zero vector that is orthogonal to u and v. vii Determine the area of the parallelogram that is defined b u and v. viii Determine u.( v w). What comment can ou make about the vectors u, v and w? Eercise Find the interior angles α, β, γ of a triangle ABC whose vertices are the points A( 1, 0, 2), B(2, 1, 1), C(1, 2, 2)

33 CHAPTER 2. BASIC TRIGONOMETRY 33 Contents 1 Vectors in Two Dimensions 1 1.1 Addition of Vectors....................................... 2 1.2 Scalar Multiplication of Vectors................................ 3 1.3 Magnitude and Direction................................... 7 2 Vectors in Three Dimensions 12 2.1 Addition of Vectors....................................... 13 2.2 Scalar Multiplication of Vectors................................ 14 2.3 Magnitude and Direction................................... 18 2.4 The Scalar Product....................................... 22 2.5 Components and Projections................................. 23 2.6 The Vector Product...................................... 26 2.7 Scalar Triple Product..................................... 30 2.8 Some Eercises......................................... 32