CHAPTER ONE VECTOR GEOMETRY

CHAPTER ONE VECTOR GEOMETRY. INTRODUCTION In this chapter ectors are first introdced as geometric objects, namely as directed line segments, or arrows. The operations of addition, sbtraction, and mltiplication by a scalar (real nmber) are defined for these directed line segments. Two and three dimensional Rectanglar Cartesian coordinate systems are then introdced and sed to gie an algebraic representation for the directed line segments (or ectors). Two new operations on ectors called the dot prodct and the cross prodct are introdced. Some familiar theorems from Eclidean geometry are proed sing ector methods.. SCALARS AND VECTORS Some physical qantities sch as length, area, olme and mass can be completely described by a single real nmber. Becase these qantities are describable by giing only a magnitde, they are called scalars. [The word scalar means representable by position on a line; haing only magnitde.] On the other hand physical qantities sch as displacement, elocity, force and acceleration reqire both a magnitde and a direction to completely describe them. Sch qantities are called ectors. If yo say that a car is traeling at 90 km/hr, yo are sing a scalar qantity, namely the nmber 90 with no direction attached, to describe the speed of the car. On the other hand, if yo say that the car is traeling de north at 90 km/hr, yor description of the car's elocity is a ector qantity since it incldes both magnitde and direction. To distingish between scalars and ectors we will denote scalars by lower case italic type sch as a, b, c etc. and denote ectors by lower case boldface type sch as,, w etc. In handwritten script, this way of distingishing between ectors and scalars mst be modified. It is cstomary to leae scalars as reglar hand written script and modify the symbols sed to represent ectors by either nderlining, sch as or, or by placing an arrow aboe the symbol, sch as r or r.

. Problems. Determine whether a scalar qantity, a ector qantity or neither wold be appropriate to describe each of the following sitations. a. The otside temperatre is 5º C. b. A trck is traeling at 60 km/hr. c. The water is flowing de north at 5 km/hr. d. The wind is blowing from the soth. e. A ertically pwards force of 0 Newtons is applied to a rock. f. The rock has a mass of 5 kilograms. g. The box has a olme of.5 m 3. h. A car is speeding eastward. i. The rock has a density of 5 gm/cm 3. j. A blldozer moes the rock eastward 5m. k. The wind is blowing at 0 km/hr from the soth. l. A stone dropped into a pond is sinking at the rate of 30 cm/sec..3 GEOMETRICAL REPRESENTATION OF VECTORS Becase ectors are determined by both a magnitde and a direction, they are represented geometrically in or 3 dimensional space as directed P Q line segments or arrows. The length of the arrow corresponds to the magnitde of the ector while the direction of the arrow corresponds to the direction of the

3 ector. The tail of the arrow is called the initial point of the ector while the tip of the arrow is called the terminal point of the ector. If the ector has the point P as its initial point and the point Q as its terminal point we will write = PQ. Eqal ectors Two ectors and, which hae the same length and same direction, are said to be eqal ectors een thogh they hae different initial points and different terminal points. If and are eqal ectors we write =. Sm of two ectors The sm of two ectors and, written + is the ector determined as follows. Place the ector so that its initial point coincides with the terminal point of the ector. The ector + is the ector whose initial point is the initial point of and whose terminal point is the terminal point of. Zero ector + The zero ector, denoted 0, is the ector whose length is 0. Since a ector of length 0 does not hae any direction associated with it we shall agree that its direction is arbitrary; that is to say it can be assigned any direction we choose. The zero ector satisfies the property: + 0 = 0 + = for eery ector. Negatie of a ector If is a nonzero ector, we define the negatie of, denoted, to be the ector whose magnitde (or length) is the same as the magnitde (or length) of the ector, bt whose direction is opposite to that of. r If AB is sed to denote the ector from point A to point B, then the ector from point B to point A is denoted by BA r, and BA r r = AB. Difference of two ectors

4 If and are any two ectors, we define the difference of and, denoted, to be the ector + ( ). To constrct the ector we can either (i) constrct the sm of the ector and the ector ; or (ii) position and so that their initial points coincide; then the ector from the terminal point of to the terminal point of is the ector. (i) (ii) Mltiplying a ector by a scalar If is a nonzero ector and c is a nonzero scalar, we define the prodct of c and, denoted c, to be the ector whose length is c times the length of and whose direction is the same as that of if c > 0 and opposite to that of of c < 0. We define c = 0 if c = 0 or if = 0. Parallel ectors The ½ ectors ( ) and c are parallel to each other. Their directions coincide if c > 0 and the directions are opposite to each other if c < 0. If and are parallel ectors, then there exists a scalar c sch that = c. Conersely, if = c and 0, c then and are parallel ectors. Let O, A and B be 3 points in the plane. Let OA = a and let OB = b. Find an expression for the ector BA in terms of the ectors a and b. b B O a A

5 Soltion BA = BO + OA = OB + OA = OA OB = a b. Proe that the line joining the mid points of two sides of a triangle is parallel to and one half the length of the third side of the triangle. Soltion Let ABC be gien. Let M be the mid point of side AC and let N be the mid point of side BC. Then MN = MC + CN = AC + CB = (AC + CB) = AB. This shows that MN is one half the length of AB and also r that MN is parallel to AB [since the two ectors MN and AB r are eqal, they hae the same direction and hence are r r parallel, so MN and AB will also be parallel]. A M C N B Let M be the mid point of the line segment PQ. Let O be a point not on the line PQ. Proe that OM = OP + OQ. Soltion OM = OP + PM = OP + PQ = OP + (PO + OQ) = OP + PO + OQ = OP OP + OQ = OP + OQ P M Q O.3 Problems. For each of the following diagrams, find an expression for the ector c in terms of the ectors a and b.

6 b. b c a. c. b c c b a a a B C. Let OACB be the parallelogram shown. Let a = OA and let b = OB. Find expressions for the diagonals OC and AB in terms of the ectors a and b. O b a A 3. Let ABC be a triangle. Let M be a point on AC sch that the length of AM = ½ length of MC. Let N be a point on BC sch that the length of BN = ½ length of NC. Show that MN is parallel to AB and that the length of MN is 3 the length of AB. 4. Let the point M diide the line segment AB in the ratio t:s with t + s =. Let O be a point not on the line AB. Proe OM = s OA + t OB. 5. Proe that the diagonals of a parallelogram bisect each other. 6. Proe that the medians of a triangle are concrrent..4 COORDINATE SYSTEMS In order to frther or stdy of ectors it will be necessary to consider ectors as algebraic entities by introdcing a coordinate system for the ectors. A coordinate system is a frame of reference that is sed as a standard for measring distance and direction. If we are working with ectors in two dimensional space we will se a two

7 dimensional rectanglar Cartesian coordinate system. If we are working with ectors in three dimensional space, the coordinate system that we se is a three dimensional rectanglar Cartesian coordinate system. To nderstand these two and three dimensional rectanglar coordinate systems we first introdce a one dimensional coordinate system also known as a real nmber line. Let R denote the set of all real nmbers. Let l be a gien line. We can set p a one toone relationship between the real nmbers R and the points on l as follows. Select a point O, which will be called the origin, on the line l. To this point we associate the nmber 0. Select a nit of length and se it to mark off eqidistantly placed points on either side of O. The points on one side of O, called the positie side, are assigned the nmbers,, 3 etc. while the points on the other side of O, called the negatie side are assigned the nmbers,, 3 etc. A one to one correspondence now exists between all the real nmbers R and the points on l. The reslting line is called a real nmber line or more simply a nmber line and the nmber associated with any gien point on the line is called its coordinate. We hae jst constrcted a one dimensional coordinate system. l 3 0 3 O Two dimensional rectanglar Cartesian coordinate system The two dimensional Cartesian coordinate system has as its frame of reference two nmber lines that intersect at right angles. The y horizontal nmber line is called the x axis and the y axis ertical nmber line is the y axis. The point of intersection of the two axes is called the origin and is denoted by O. To each point P in twodimensional space we associate an ordered pair of real nmbers (x, y) called the coordinates of the point. The nmber x is called the x coordinate of P( x, y) the point and the nmber y is the y coordinate of the point. The x coordinate x is the horizontal distance origin x axis of the point P from the y axis while the y coordinate y is the ertical distance of the point P from the x axis. The set of all ordered pairs of real nmbers is denoted R. O x y x

8 Three dimensional rectanglar Cartesian coordinate system The three dimensional Cartesian coordinate system has as its frame of reference three nmber lines that intersect at right angles at a point O called the origin. The nmber lines are called the x axis, the y axis and the z axis. To each point P in three dimensional space we associate an ordered triple of real nmbers (x, y, z) called the coordinates of the point. The nmber x is the distance of the point P from the yz coordinate plane. The nmber y is the distance of the point P from the xz coordinate plane. The nmber z is the distance of the point P from the xy coordinate plane. The set of all ordered triples of real nmbers is denoted by R 3. When the coordinate axes are labeled as shown in the xz coord plane z O y yz coord plane y z O P x y x xy coord plane following diagrams, the coordinate system is said to be a right handed Cartesian coordinate system. x point P(x,y,z) z Right handed Cartesian coordinate system A right handed Cartesian coordinate system is one in which the coordinate axes are so labeled that if we crl the fingers on or right hand so as to point from the positie x axis towards the positie y axis, the thmb will point in the direction of the positie z axis. [If the thmb is pointing in the direction opposite to the direction of the positie z axis, the coordinate system is a left handed coordinate system.] y.4 Problems. Draw a right handed three dimensional Cartesian coordinate system, and plot the following points with the gien coordinates. a. P (,, 3) b. Q (3, 4, 5) c. R (,, ) d. S (0,, )

9. A cbe has one ertex at the origin, and the diagonally opposite ertex is the point with coordinates (,, ). Find the coordinates of the other ertices of the cbe. 3. A rectanglar parallelepiped (box) has one ertex at the origin and the diagonally opposite ertex at the point (, 3, ). Find the coordinates of the other ertices. 4. A pyramid has a sqare base located on the xy coordinate plane. Diagonally opposite ertices of the sqare base are located at the points with coordinates (0, 0, 0) and (,, 0). The height of the pyramid is nits. Find the coordinates of the other ertices of the pyramid. [Assme that the top of the pyramid lies directly aboe the centre of the sqare base.] 5. A reglar tetrahedron is a solid figre with 4 faces, each of which is an eqilateral triangle. If a reglar tetrahedron has one face lying on the xy coordinate plane with ertices at (0, 0, 0) and (0,, 0), find the coordinates of the other two ertices if all coordinates are nonnegatie tetetrahedron.5 DEFINING VECTORS ALGEBRAICALLY Since a ector is determined solely by its magnitde and direction, any gien ector may be relocated with respect to a gien coordinate system so that its initial point is at the origin O. Sch a ector is said to be in standard position. When a gien ector is in standard position there exists a niqe terminal point P sch that = OP. This one to one relationship between the ector and the terminal point P enables s to gie an algebraic definition for the ector. If is a ector in two dimensional space and P(a, b) is the niqe point P sch that = OP, then we will identify the ector with the ordered pair of real nmbers ( a, b ) and write = (a, b). Similarly if is a ector in three dimensional space and P(a, b, c) is the niqe point P sch that = OP, then we will identify with the ordered triple of real nmbers ( a, b, c) and write = (a, b, c). The two dimensional ector = (a, b) is said to hae components a and b and the three dimensional ector = ( a, b, c) is said to hae components a, b and c. O y P(a,b) x

0 To aoid confsion, when dealing with the components of seeral ectors at the same time it is cstomary to denote the components of a gien ector by sbscripted letters that agree with the letter sed to designate the ector. Ths we will write = (, ) if is a ector in R and = (,, 3 ) if is a ector in R 3. Eqal ectors If eqal ectors and are located so that their initial points are at the origin, then their terminal points will coincide, and hence the corresponding components of and mst be eqal to each other. Ths = in R if and only if = and = while for ectors in R 3, = if and only if =, = and 3 = 3. Sm of two ectors Let = (, ) and = (, ) be two ectors in R. If the ectors are located so that their initial points are at the origin, then their terminal points are the points with coordinates (, ) and (, ). If is now placed so that its initial point is at (, ), which is the terminal point of, then the terminal point of is the point with coordinates ( +, + ). Hence + = ( +, + ). A similar argment for the ectors = (,, 3 ) and = (,, 3 ) in R 3 gies + = ( +, +, 3 + 3 ). y + x Let = (,, 3) and = (4,, 5). Then + = ( + 4, +, 3 + 5) = (5, 3, 8). Mltiplying a ector by a scalar If = (, ) is a ector in R that has its initial point at the origin, then the terminal point of is the point with coordinates (, ). If c>0, then the ector c has the same direction as and is c times as long as so its terminal point is the point with coordinates (c, c ). A similar argment applies if c<0, except in this case the direction is reersed. In either case we hae c = (c, c ). c c c If instead is a ector in R 3, then a similar argment will show that c = (c, c, c 3 ).

If = (3,, ), then 5 = (5 3, 5, 5 ) (5, 5,0) =. Difference of two ectors The ector is defined to be eqal to the ector sm + ( ). If = (, ) and = (, ) are two ectors in R, then = (, ) + ( ) (, ) = (, ) + (, ) = (, ). Similarly, in R 3 we hae = (,, 3 3 ). If = (4,5, ) and = (,,3) then = 4, 5 ( ), 3 = (, 6, ). ( ) Vector representation of a directed line segment Let = AB where A is the point with coordinates (a, a ) and B is the point with coordinates (b, b ). Then = AB = AO + OB r r r r = OA + OB = OB OA = (b, b ) (a, a ) = (b a, b a ). A B In R 3, if A = (a, a, a 3 ) and B = (b, b, b 3 ) then AB = (b a, b a, b 3 a 3 ). O If A = (,, 3) and B = (4, 6, 9), then AB = (4, 6, 9 3) = (3, 4, 6) Length of a ector If = (, ) then the length of is eqal to the length of the directed line segment from the origin (0, 0) to the point (, ). We will se the symbol (, )

to represent the length of the ector. Using Pythagoras theorem for right triangles we can calclate that length to be + and so we hae the formla = +. A similar argment for a ector = (,, 3 ) in R 3, sing Pythagoras theorem twice, gies = 3 + +. Theorem If c is a scalar and is a ector in R or R 3, then c = c. Proof The following proof is for in R. The proof for in R 3 is similar. c = ( c, c ) = ( c ) + ( c ) = c ( + ) = c + = c. Unit ector If = we say is a nit ector. Becase the length of a ector is a positie qantity, the length of the ector c is ector, mltiply the ector by the scalar c. To find a nit ector in the direction of a gien. The reslting ector ector in the direction of. A nit ector in the direction opposite to is. is a nit If = (,, ), then the length of is = + + = 4 + 4 + = 9 = 3 and a nit ector in the direction of is = (,,) =,,. A nit ector in the 3 3 3 3 direction opposite to that of is,, 3 3 3..5 Problems

3 Let = (,, 3), = (3,, ) and w = (4,, ).. Find the following ectors. a. + b. c. w d. 3 e. + 3w f. + 3 w. Find the following lengths. a. b. c. w d. + e. f. w 3. Find components of the ector eqal to the directed line segment PQ. a. P = (,, 3) Q = (, 4, 7) b. P = (3,, 4) Q = (5, 7, ) c. P = (, 5, ) Q = (4, 3, ) d. P = (0, 3, ) Q = (, 0, 5) 4. Let = AB. If and A are as gien below, find the coordinates of B. a. = (3, 5, 4) A = (, 3, ) b. = (, 5, 4) A = (,, ) 5. Let = AB. If and B are as gien below, find the coordinates of A. a. = (3, 5, 4) B = (, 5, 6) b. = (, 5, 4) B = (4,, 7). 6. Let be the gien ector. Find a nit ector in the direction of and find a nit ector in the direction opposite to that of. a. = (,, )b. = (3, 0, 4) c. = (,, 3) d. = (, 3, 4). 7. If = (3a, 4a, 5a) and = 0, find the ale of a..6 THE DOT PRODUCT (SCALAR PRODUCT) The dot prodct is a method for mltiplying two ectors. Becase the prodct of the mltiplication is a scalar, the dot prodct is sometimes referred to as the scalar prodct. The dot prodct will be sed to find an angle between two ectors and will hae applications in finding distances between points and lines, points and planes, etc. If = (, ) and = (, ) are two ectors in R, we define their dot prodct, denoted, as follows: = +.

4 If = (,, 3 ) and = (,, 3 ) are two ectors in R 3, we define their dot prodct to be = + + 3 3. Let = (,, 3) and = (4, 5, 6). Then = ()(4) + ()(5) + (3)(6) = 4 + 0 + 8 = 3. The following theorem relates the length of a ector to the dot prodct of the ector with itself. Theorem For any ector in R or in R 3, =. Proof The following proof is for R. The proof for R 3 is similar. Let = (, ). Then = (, ) (, ) = + = Taking sqare roots gies =.. The next theorem lists some algebraic properties of the dot prodct. Theorem Let, and w be ectors in R or R 3, and let c be a scalar. Then (a) = (b) c ( ) = (c) = (c) (c) ( + w) = + w (d) 0 = 0. Proof (a) Let = (, ) and = (, ) be any two ectors in R. Then = + = + =. The proof for R 3 is similar The proofs for parts (b), (c) and (d) are similar straightforward comptations. The following theorem shows how the dot prodct of two ectors and is related to the angle between the ectors.

5 Theorem Let and be two ectors in R or R 3. Let θ be the angle between and. Then = cos θ. Proof Let and to be a pair of adjacent sides of a triangle whose third side is. Using the cosine law for triangles we get = + cos θ ( ) ( ) = + + = + cos θ cos θ θ = = cos θ cos θ Angle between two ectors The preceding theorem proides a method for finding the cosine of the angle between two ectors and hence finding the angle between the two ectors. Soling = cos θ for cos θ gies the formla cos θ =. Find the cosine of the angle between the ectors = (3,, ) and = (, 4, 3). Soltion (3,, ) (, 4, 3) 3+ 4 + 6 3 3 3 cosθ = = = = = (3,, ) (, 4, 3) 9 + + 4 + 6 + 9 4 6 7 3 7 3 Haing fond the cosine of the angle θ, we can find the angle θ= cos 47 7 =. Orthogonal ectors Vectors and are said to be orthogonal or perpendiclar to each other if they meet at right angles. If and are orthogonal, then = cos( π ) = 0. [Since cos (π ) = 0.] Conersely, if = 0 we mst hae either = 0 or = 0 or. Since the zero ector 0 can hae any direction, we will agree that 0 is orthogonal to any ector. Hence we say that and are orthogonal if and only if = 0.

6 Show that the ectors = (,, ) and = (,, ) are orthogonal ectors. Soltion = (,, ) (,, ) = + 4 = 0. Hence. Normal ector If l is a line in R or in R 3 and n is a ector that is orthogonal to the line l, we call n a normal ector to the line l. Theorem Let ax + by = c be the eqation of a line l in R. Then the ector n = (a, b) is a normal ector to the line l. Proof First select two points P and Q on l. Select P = (c/a, 0) and Q = (0, c/b), then the ector PQ lies on l. Bt PQ = (0, c/b) (c/a, 0) = ( c/a, c/b). To show that n PQ we take the dot prodct. n PQ = (a, b) ( c/a, c/b) = c + c = 0. This proes that the ector n is a normal ector to the line l. Find a ector that is normal to the line x + 3y =5. n=(a,b) P l: ax+by=c Q Soltion From the preios theorem the ector n = (, 3) is normal to the gien line x + 3y =5 since the coefficients of x and y are and 3. Projections Let and be two gien ectors with 0. The projection of along, denoted proj is the ector p fond as follows. Drop a perpendiclar from the terminal point of that intersects the line throgh at the point P. Then proj = p = OP. O p P p We find p as follows. Since p lies along, there is a scalar k sch that p = k. Now p is orthogonal to so ( p) = 0. Bt

7 ( p) = 0 p = 0 k = 0 k = =. Hence proj = p = k = =. Let = (8,, 4) and let = (,, ). Find proj. Soltion proj = (8,, 4) (,, ) 8 + + 8 = (,, ) = (,, ) = (,, ) = (, 4, 4) (,, ) (,, ) + 4 + 4 Distance between a point and a line in R To find the distance D between a point P and a line l in R, we select a point Q on the line l, then the distance D is the length of the projection of QP on n, a normal ector to the line l. r r QP n D = projnqp = n n n r r r QP n QP n PQ n = n = = n n n n r r Note that QP n = PQ n and so either of the last two forms for the distance D can be sed interchangeably. Find the distance between the point P = (9, ) and the line 3x + 4y = 6. Soltion The point Q = (, 0) lies on the line 3x + 4y = 6 so QP = (9, ) (, 0) = (7, ). QP n Since n = (3, 4), the distance is D = = n n ( 7, ) (3, 4) = (3, 4) l D + 4 = 9 + 6 Q P 5 = 5 5.6 Problems

8 In problems to 3 below, let = (,, ), = (3,, 4) and w = (,, 3).. Calclate the following dot prodcts. a. b. w c. w d. ( + w) e. ( + 3w). Find the length of each of each of the following ectors. a. b. c. w d. + e. 3 3. Find the cosine of the angle between the following pairs of ectors. a. and b. and w c. and w d. + and 4. Show that the following pairs of ectors are orthogonal. a. (,, 3) and (,, ) b. (, 3, 5) and (,, ) c. (4, 5, ) and (,, 3) d. (, 0, ) and (0,, 0) 5. Find a ector n which is normal to the gien line in R. a. x + 3y = 5 b. x y = 3 c. 3x + y = 4 d. x + 3y = 6. Find proj for each of the following pairs of ectors and. a. = (,, ) and = (3,, 0) b. = (3,, 4) and = (,, ) c. = (5, 4, 3) and = (3,, ) d. = (,, ) and = (3, 4, ) 7. Find the distance between the point P and the line l in R. a. P = (, 3) l: 3x + 4y = b. P = (5, ) l: 3x 4y = c. P = (5, 3) l: 5x + y = d. P = (3, 4) l: x + y = 3 8. Proe Pythagoras theorem: The sqare on the hypotense of a right triangle eqals the sm of the sqares on the other two sides. 9. Proe that the angle inscribed in a semi circle is a right angle. 0. Proe that the sm of the sqares of the diagonals of a parallelogram eqals the sm of the sqares of its sides.. Proe that the diagonals of a rhombs (parallelogram with eqal sides) are perpendiclar.

9. Proe that the mid point of the hypotense of a right triangle is eqidistant from the three ertices of the triangle 3. Proe that the altitdes of a triangle are concrrent. 4. Let a and b be nit ectors in the xy plane making angles α y and β respectiely with the x axis. Let i and j be the ectors (, 0) and (0, ) respectiely. a. Show that i i =, i j = 0 and j j =. b. Show that a = cosα i + sinα j and b = cos β i + sin β j c. Proe that cos( α β) = cosαcosβ + sin αsinβ. j α β i a b x.7 THE CROSS PRODUCT (VECTOR PRODUCT) In the preios section we were introdced to the dot prodct of two ectors. The reslt of taking the dot prodct of two ectors is a scalar qantity. We now introdce a second method of mltiplying two ectors from R 3 that reslts in a ector qantity. The symbol sed to denote this prodct is a cross, hence the name "cross prodct". Becase the reslt is a ector, the term "ector prodct" is sometimes sed for this prodct. The cross prodct has a nmber of applications. We will se the cross prodct to find the areas of triangles and parallelograms. It will also be sed to calclate the olme of a parallelepiped and later to find the distance between a point and a line in R 3. Cross prodct (ector prodct) If = (,, 3 ) and = (,, 3 ) are two ectors in R 3, the cross prodct is the ector in R 3 defined as follows. = ( 3 3, 3 3, ). Let = (3,, ) and let = (4, 6. 5). Then = ( 5 6, 4 3 5, 3 6 4) = ( 7, 7, 4).

0 Althogh the definition of the cross prodct as gien aboe may be difficlt to remember, the concept of a determinant can be sed to simplify the process. a b a b Consider the array of nmbers. The determinant of c d, written, c d a b a b det or, is defined to be the nmber ad bc. Then the cross prodct of c d c d = (,, 3 ) and = (,, 3 ), sing determinants, can be written as the ector = 3 3, 3 3, We remember the components of as follows. 3 ) Form the 3 rectanglar array where the first row consists of the 3 components of the ector and the second row consists of the components of ector.. ) To find the first component of, delete the first colmn and take the determinant of the remaining array; to find the second component of, delete the second colmn and take the negatie of the determinant of the remaining array; to find the third component of, delete the third colmn and take the determinant of the remaining array. Find if = (, 3, 4) and = (5, 6, 7). Soltion Constrct the rectanglar array 5 3 6 4 7. Then 3 4 =, 6 7 5 = (3 7 4 6, = ( 3, 6, 3) 4 3, 7 5 6 ( 7 4 5), = ( 4, (4 0), 5) 6 3 5)

Theorem = Proof = ( 3 3, 3 3, ) = ( 3 3, 3 3, ) = ( 3 3, 3 3, ) = Theorem is orthogonal to both and. Proof We show that is orthogonal to by showing that the dot prodct of and is eqal to zero. The proof that is orthogonal to is similar. ( ) = ( 3 3, 3 3, ) (,, 3 ) = ( 3 3 ) + ( 3 3 ) + ( ) 3 = 3 3 + 3 3 + 3 3 = 0 Since ( ) = 0, and are orthogonal. arrows indicate canceling pairs Find a ector orthogonal to both = (, 3, ) and = (4, 0, ). Soltion The ector is orthogonal to both and, so we calclate. 3 = 0, 4, 4 3 = (3 0, 0 ( 8), 0 ) = (3, 7, )

The next theorem is a sefl reslt that can be applied to calclate the area of a triangle and the area of a parallelogram. It is also sed to calclate the olme of a parallelepiped in R 3 and to find the distance between a point and a line in R 3. Theorem = sin θ where θ is the angle between and. Proof The proof consists of steps. () We first show = ( ) sides separately and showing that they are eqal to each other. by compting the left and right hand = ( 3 3 ) + ( 3 3 ) + ( )....(i) 3 ( ) = ( + + )( + + ) ( + + 3 3 3 )...(ii) A lengthy comptation shows right hand sides of (i) and (ii) are eqal and so we conclde () Starting with = ( ). = ( ) = = = ( ( cos sin Taking sqare roots gies the reqired reslt: θ cos θ) θ) we expand the dot prodct on the right = sin θ. The next theorem lists seeral properties of the cross prodct. The properties are established by straightforward comptations and so the proofs are omitted. Theorem properties. Let, and w be ectors in R 3. Then, and w satisfy the following (a) ( + w) = + w (b) ( + ) w = w + w (c) 0 = 0 = 0 (d) = 0

3 The area of a parallelogram Let and be the adjacent sides of a parallelogram. The area of a parallelogram is length of base height. From the adjoining diagram we hae that the length of the base is and the height is h. From trigonometry we get h = sin θ. Therefore the area A is gien by A = base height = sinθ = h = sinθ so θ h The area of a triangle Let and be the adjacent sides of a triangle. Since the area of the triangle is one half the area of the parallelogram with and as its adjacent sides, the area of the triangle is A =. θ h Find the area of the parallelogram haing adjacent sides = (, 3, ) and = (4, 0 ). Soltion 3 3 =,, = (6, 0, ) 0 4 4 0 Area = = (6, 0, ) = 36 + 0 + 44 = 80 = 36 5 = 36 5 = 6 5 Find the area of the triangle whose ertices are A = (,, ), B = (3, 4, 5) and C = (5, 6, 4) Soltion

4 Let = AB r = (3, 4, 5) (,, ) = (,, 3) and r let = AC = (5, 6, 4) (,, ) = (4, 4, ). 3 3 Then =,, = ( 8, 8, 0) 4 4 4 4 Area of triangle ABC = = 8, 8, 0) = 64 + 64 + 0 = 4 ( The olme of a parallelepiped A parallelepiped is a solid (3 dimensional) figre haing six faces with opposite pairs of faces being congrent parallelograms. A parallelepiped can be specified by giing 3 ectors, and w that form the 3 edges emanating from a common ertex. The olme of the parallelepiped is the area of the base height. The area of the base is the area of the parallelogram with and as adjacent sides and is eqal to height is the length of the projection of w onto = Bt proj w. w w ( ) w ( ) w ( ) proj = ( ) = =. ( ) ( ) w ( ) Ths the olme of the parallelepiped is V = = w ( ). θ w. The Find the olme of the parallelepiped haing the following three ectors as edges. = (, 3, ), = (3, 4, 3) and w = (4, 5, 6) Soltion 3 3 =,, = (5, 3, ) 4 3 3 3 3 4 w ( ) = ( 4, 5, 6) (5, 3, ) = 0 5 6 = Volme = w ( ) = =.7 Problems

5 For problems to 5 let = (4, 3, ), = (5,, 3) and w = (,, 4).. Find a. b. w c. w d. ( + w). Find a. ( w) b. ( ) w c. ( ) 3. Find a ector orthogonal to a. and b. and w 4. Find the area of the parallelogram whose adjacent sides are a. and b. and w c. and w 5. Find the area of the triangle whose adjacent sides are a. and b. and w c. and w 6. Find the area of the triangle whose ertices are gien. a. (,, 3), (, 4, 5), (4, 5, 8) b. (,, ), (4, 3, 5), (5, 6, 7) A 7. Proe the law of sines for triangles. sin A sin B sin C = =. a b c B c a b C 3 8. Let and be two nonzero ectors in R. a. Proe that if and are parallel ectors, then = 0. b. Proe that if = 0, then and are parallel ectors..8 STANDARD BASIS VECTORS FOR R 3 The following three nit ectors i = (, 0, 0), j = (0,,0) and k = (0, 0,) play a special role in R 3 They are called the standard basis ectors for R 3. Eery ector in R 3 can be written as a niqe combination of these three ectors as follows. Let = (a, b, c) be an arbitrary ector in R 3. Then we can write = ( a, b, c) = ( a,0,0) + (0, b,0) + (0,0, c) = a(,0,0) + b(0,,0) + c(0,0,) = ai + bj + ck. If = (,3,5), then = i + 3j+ 5k.

6 For the dot prodct of the standard basis ectors with each other, we hae the following reslts, which can be erified by a direct comptation. i i = j j = k k = and i j = j i = j k = k j = k i = i k = 0. For the cross prodct of the standard basis ectors with each other, we hae the following reslts which can also be erified by a direct comptation. i i = j j = k k = 0 and i = j k j = k i k= i j j= i k k = j i i = k j. The reslts for the cross prodcts of any two of the three standard basis ectors can be remembered by sing the adjoining diagram. The prodct of any two sccessie ectors in the diagram, when moing clockwise, is the third ector in the diagram. The prodct of any two sccessie ectors in the diagram, when moing conterclockwise, is the negatie of the third ector in the diagram. k ii j.8 Problems. Write each of the following ectors as a combination of the three standard basis ectors i, j, and k. a. = (4, 3, 7) b. = (3,, ) c. w = (, 5, 6) d. r = (, 0, ). Verify the following reslts for the standard basis ectors i = (, 0, 0), j = (0,, 0), and k = (0, 0, ). a. i i = b. j j = c. i j = 0 d. j k = 0 e. i i = 0 f. i j = k g. i k = j h. j k = i 3. Compte the following dot prodcts. a. (i + 3j + k) (3i j + 5k) b. (3i + j + 4k) (i 5j + 6k) c. (i 5j + 3k) (4i + j 3k) d. (i 3j + k) (6i + j 3k) 4. Compte the following cross prodcts. a. (i + 3j + k) (3i j + 5k) b. (3i + j + 4k) (i 5j + 6k) c. (i 5j + 3k) (4i + j 3k) d. (i 3j + k) (6i + j 3k) 5. Find a if the following pairs of ectors are orthogonal.

7 a. = ai + aj+ k, = i 3j 5k b. = 3ai + j 3 k, = i 6j+ ak.9 VECTORS IN R m We hae already seen that the set of all real nmbers R can be identified with a one dimensional nmber line; the set of all ordered pairs of real nmbers R can be identified 3 with a two dimensional plane and that the set of all ordered triples of real nmbers R can be identified with three dimensional space. Contining in this manner wold sggest that the set of all ordered for tples cold be identified with a for dimensional space and more generally the set of all ordered m tples cold be identified with an m dimensional space. We se the symbol R m to denote the set of all ordered m tples =,,,, ). ( 3 m We will refer to the m tples as ectors in the space R m and the entries,, etc. as the components of the ector. Two ectors from R m are said to be eqal if their corresponding components are eqal to each other. That is = if and only if =, =, etc. We define the sm of and by + = +, +,, m + ). ( m We define mltiplication of a ector by a scalar c as c = c, c, c ). ( m The length of the ector is denoted and is defined by + + m = +. The dot prodct is defined to be = + + + m. m If = 0 we say that the ectors and are orthogonal to each other. Note that there is no cross prodct defined for R m when m 3.

8 Let = (, 3,, 4) and = (,, 4, 3) be two ectors in R 4. Then + = (, 3,, 4) + (,, 4, 3) = (+, 3, +4, 4+3) = (3,, 6, 7) = (, 3,, 4) (,, 4, 3) = (, 3+, 4, 4 3) = (, 4,, ) = (,3,,4) (,,4,3) = ()() + (3)( ) + ()(4) + (4)(3) = 3+ 8 + = 9 3 = 3(, 3,, 4) = ( 3, 3 3, 3, 3 4) = (3, 9, 6, ) = + 3 + + 4 = + 9 + 4 + 6 = 30.9 Problems For qestions to 6, let = (, 3,, 4), = (5, 3,0, ), and w = ( 3,,, 4).. Find a. + b. 3 c. + w. Find a. b. w c. ( + w) 3. Find a. b. + w c. 4. Find a nit ector in the direction of a. b. c. w 5. Show that the following pairs of ectors are orthogonal by showing that their dot prodct is 0. a. (,, 3, ) (3,,, 8) b. (, 0, 3, ) (5, 6,, 4) c. (,, 3, 4, 5) (4, 4, 3,, ) d. (, 3, 5,, 4) (3, 4,, 3, ) 6. Show that the following sets of ectors are mtally orthogonal by showing that each ector in the set is orthogonal to all the other ectors in the set. a. (,, 0, 0) (,,, 3) (,,, ) b. (,,, 4) (3,, 0, ) (,,, ) c. (,,, ) (,, 3, 3) (3, 3,, ) d. (, 0,, ) (, 3,, 0) (6, 5, 3, 0) 7. Consider the for nit ectors e = (,0,0,0), e = (0,,0,0), e3 = (0,0,,0) and e 4 = (0,0,0,) 4 in R. Write each 4 of the following ectors from R as a combination of the ectors e, e, e3 and e 4. a. (, 3, 5, 4) b. (3,, 0, ) c. (5, 7,, 3)

9 8. Proe the following reslts for e, e, e3 and e 4. a. e = b. e e = 0 c. ( e e ) ( e e ) + = 0