Vector Geometry - UMass Amherst
Chapter 5Vector Geometryvector in Rn as a position vector as described in section1.3 of Lay’s textbook. A position vector is just a pointerto a certain location in Rn . When using position vectorsit is not necessary to make a firm distinction between avector and its endpoint. For example, when we say that aline is a set of vectors we mean that the endpoints of thevectors lie on the line. If we want to stress the directionof the vector we will usually represent it as an arrow. Ifwe want to stress the particular location that the vectoris pointing to we will usually represent it by a point.In this chapter we will look more closely at certain geometric aspects of vectors in Rn . We will first develop anintuitive understanding of some basic concepts by lookingat vectors in R2 and R3 where visualization is easy, thenwe will extend these geometric intuitions to Rn for any n.The basic geometric concepts that we will look at involvemeasurable quantities such as length, angle, area and volume. We also take a closer look at the two main types ofequations covered in this course: parametric-vector equations and linear equations.We begin with a reminder. We defined a vector in Rnas an n-tuple, i.e., as an n 1 matrix. This is an algebraicdefinition of a vector where a vector is just a list of numbers. The geometric objects we will look at in this chaptershould be seen as geometric interpretations of this algebraic definition. One difficulty that students encounter atthis stage is that there are many different geometric interpretations that can be given to a vector. For example,a vector in Rn can be interpreted geometrically asEXAMPLE 5.1. If A (x1 , x2 , . . . , xn ) and B (y1 , y2 , . . . , yn ) are two points then the vector from A to B (represented by AB ) is defined as follows23y 1 x16 6 y2 x2 77AB 67.45.y n xn an arrow starting at the origin.You can think of this as letting A be the origin of a new coordinate system and then the entries in AB give the location of B relative to A. Or you can imagine translatingboth A and B by subtracting A from both points so thatA is translated to the origin. Finally, you can think of AB as an arrow from A to B.So, for example,we haveP (1, 5, 2) and Q(7, 7, 0)23 if 237 16 4then P Q 7 55 4 2 5. The entries in this vector0 2 2indicate that when you travel from P to Q you move 6units in the x1 direction, 2 units in the x2 direction and 2units in the negative x3 direction. These entries express the location of Q relative to P . If P Q is drawn with the an arrow with a certain length and direction but nofixed location. a point (or more exactly, the coordinates of a pointrelative to some reference point). a directed line segment between two points. a displacement (i.e., a translation).This multiplicity of interpretations is a strength of thevector concept not a weakness. Vectors have many applications and depending on the application one geometricinterpretation may be more relevant than another but nomatter what geometric interpretation is chosen the underlying vector algebra remains the same. We will interpret a1
2Chapter 5. Vector Geometry1intial point at the origin then the terminal point would be(6, 2, -2).–11234560We will usually represent a vector as an n 1 matrixbut there is another standard way of representing vectorsthat is frequently used. In R2 we define» –» –10i j 01–1–22It then follows that any vector in R can be written as» – » – » –aa0 ai bjb0bR3 we define2 32 301i 4 05j 4 1500Similarly inYou should realize that in R2 the vectors i and j are justthe vectors which we have called e1 and e2 , the standardbasis of R2 . Similarly in R3 the vectors i, j and k are thestandard basis of R3 .Distance and LengthThe first geometric concept we want to look at is thethe length of a vector. We define this to be the usualEuclidean distance from the intial point (the origin) tothe end point of the vector. The length any vector vin Rn will be represented by kvk. This quantity is alsoreferred to as» the– magnitude or norm of v.u1be a vector in R2 . The length of thisLet u u2vector would be the distance from the origin (0, 0) to thepoint (u1 , u2 ) and this is given by the Pythagorean Theorem asqkuk u21 u22»–5. Figure 5.1 shows u 3and by the Pythagorean Theorem we can find the normof u asq kuk 52 ( 3)2 34EXAMPLE 5.2. Let u (5,–3)–42 30k 4 051and then any vector in R3 can be written2 3a4 b 5 ai bj ckc5.1–3Figure 5.1.In R3 a similar argument based on the PythagoreanTheorem givesqkuk u21 u22 u232 3u1for any vector u 4u2 5.u3We can extend the above formulas to Rn by definingqkuk u21 u22 · · · u2n3u16 u2 76 7Notice that if u 6 . 7 is any vector in Rn then4 . 5un2 3u16 u2 7ˆ 76uT u u1 u2 · · · un 6 . 7 u21 u22 · · · u2n4 . 5un2We then have the following concise formula which isvalid for vectors in Rn for all nkuk2 uT uEXAMPLE 5.3. Let u be any vector inscalar then“”kkuk2 kuT (ku) k2 uT u k2 kuk2Rn and k be a
5.1. Distance and Length3Taking square roots then giveskkuk k kukThis shows that multiplying any vector in Rn by a scalark scales the length of the vector by k . We will sometimesmake a distinction between the sense of a vector and thedirection of a vector. When a vector is multiplied by anegative scalar the reversal of the arrow is described bysaying the sense has been reversed but the direction hasstayed the same.Definition 5.1. The distance between two vectors uand v in Rn is defined as ku vk.EXAMPLE 5.4. The distance between u i k andv j k isq ku vk ki j 2kk 12 ( 1)2 22 6Unit VectorsA unit vector is a vector whose length is 1.1u is a unitIf u is any non-zero vector in Rn thenkukvector. This can be seen by applying the formula kvk2 1vT v to the vectoru. This gives:kuk„«T „«111uu uT ukukkukkuk21kuk2 kuk2 1The process of multiplying a vector by the reciprocal ofits length to obtain a unit vector is called normalization.Notice that this procedure doesn’t alter the direction orsense of the vector.326 2 77EXAMPLE 5.5. Normalize the vector v 64 0 5. 1 We have kvk 4 4 0 1 9 3 so2 3 2322/3767166 2 7 6 2/3 734 0 5 4 0 5 1 1/32is a unit vector parallel to v. Note: Just to avoid any possible confusion, when we say that two non-zero vectors,u and v, are parallel we mean that they have the samedirection. Each one is a scalar multiple of the other.
4Chapter 5. Vector GeometryProblems 1. If A (4, 2) and AB »–3what is B? 12 3 6 2. If B (5, 4, 7) and AB 4 2 5 what is A?23. Find the length of the following vectors:»–» –cos θ3d.a.sin θ 22 3321cos(s) sin(t)b. 4 4 5e. 4cos(s) cos(t)5 1sin(s)2 3f. i j k46 377g. 4i j 3kc. 64 25 1 2t2 i tj tkh.12 316 176 74. Let v 6 . 7 be the vector in Rn all of whose4 . 51entries are 1. What is kvk?5. Find the lengths of the sides of triangle ABC wherethe vertices are given bya. A(0, 0), B(3, 3), C(5, 1)b. A( 1, 2), B(1, 5), C(3, 1)d.c. A(1, 0, 0),B(0, 1, 0), a. If AB »d. 3i 5j 2k» –1e.t»–cos t sin tf.cos t sin t 9. If kABk 5 and kBCk 3 what are the possible values for kACk?»–»–cos(s)cos(t)10. Let u and v . These are twosin(s)sin(t)2unit vectorsp in R . Show that the distance from uto v is 2 2 cos(s t)2 3v111. Prove that in R3 the length of 4v2 5 is given byv3qv12 v22 v32 .12. True or False:a. kuk2 uuTb. kuk2 uT uc. k2uk2 4uT ud. ku vk2 uT u vT vC(0, 0, 1)e. If kuk kvk then ku vk kuk kvk.A(3, 1, 2),B(4, 1, 2),C( 2, 0, 1)f. kAuk2 uT AT Au–» – 32, BC , and A is the 15point (3, 7) what is C? Draw a diagram illustating this problem.2 32 332 b. If P Q 415, QR 4 15, and R is the01point ( 3, 5, 2) what is P ?2 3321k7. Let u 425 and v 4k 15. Use calculus to4k 2find the value of k for which the distance from u tov is a minimum.8. Find a unit vector parallel to each of the followingvectors:6.» –342 33b. 44552 316 1 77c. 64 15 1a.13. Under what conditions will ku vk kuk kvk?14. Suppose A is an n n matrix such that AT A I.Let v be any vector in Rn . Show kAvk kvk–»cos θ sin θ15. Let A . Show that if v is anysin θcos θvector in R2 then kAvk kvk.
5.2. The Dot Product5.25The Dot Productand so we have the nice formulau · v uT vThe Dot Product in R2»Suppose we have two vectors in» –v1as illustrated in Figure 5.2v2R2 , u uu1–and v 2(This formula uses the standard convention of interpretingthe 1 1 matrix uT v as a scalar.)EXAMPLE 5.6. If u » –» –13and v then313u · v (1)(3) (3)(1) 6andu-vkuk kvk 2vuand the angle between u and v is arccos(3/5) 51.13 .θ010It then follows that if θ is the angle between u and v wehave63cos θ 510 101–1 1234–1Figure 5.2.These two vectors determine a triangle whose third sidewould be u v translated. If we let θ be the angle between u and v then we can apply the law of cosines tothe triangle. This givesThe dot product is often useful in geometric problemsinvolving angles. If the problem is stated in terms ofpoints then it should be “translated” into vector terminology before using the dot product. Look at the followingexample:EXAMPLE 5.7. Draw the parabola y x2 and let Pbe the point (1, 1) on this parabola. If O is the originfind another point Q on the parabola such that the anglebetween OP and OQ is 30 .kuk2 kvk2 2kuk kvk cos θ ku vk2 (u1 v1 )2 (u2 v2 )215.0Q kuk2 kvk2 2u1 u2 2v1 v212.5Cancelling out common factors and terms we get10.0kuk kvk cos θ u1 v1 u2 v27.5The expression on the right hand side of the last lineis given a special name. It is called the dot product ofu and v and is written u · v. Thus we have the followingtwo formulasu · v u1 v 1 u2 v 2andu · v kuk kvk cos θwhere θ is the angle between u and v. Since θ is one angleof a triangle we have 0 θ 180 . This means that θis the smallest positive angle between u and v.There» –is another»way– of indicating the dot product. Ifu1v1u and v thenu2v2» –ˆ v1uT v u1 u2 u1 v 1 u2 v 2v25.02.5P0.0 4 2024xFigure 5.3.» – 1In terms of vectors we have OP and OQ 1» – x. We also have OP · OQ x x2 , kOP k 2, andx2
6Chapter 5. Vector Geometry kOQk x2 x4 . If the angle between these vectors is30 the dot product formula gives px x2 2 x2 x4 cos 30 Substituting cos 30 32and squaring both sides gives 3x2 2x3 x4 2 x2 x44We omit the algebra but you should be able to solve this equation and find two values of x that work x 2 3and x 2 3.The Dot Product in RnThe dot product can be generalized to vectors inDefinition 5.2. Let u and v be vectors indot product is defined byRn .3x16 x2 76 7this let x 6 . 7. Then x·x x21 x22 · · · x2n and this4 . 5xnvalue can never be negative since it is the sum of squares.Next it says that the inner product of the zero vectorwith itself is 0. This is easy: if x 0 then x · x 0 0 · · · 0 0.Finally part (d) claims that if the dot product of avector with itself is 0 then the vector must be the zerovector. To see this suppose x · x 0. Then x21 x22 · · · x2n 0. Since the left hand side of this equation hasno negative terms the only way the terms can add up to0 is if each term is 0. So x1 0, x2 0, . . . , xn 0 andwe have x 0.This last property says that in Rn the only vector withlength 0 is the zero vector.2Rn then theiru · v uT vOrthogonal Vectors3u16 u2 76 7The above definition implies that if u 6 . 7 and4 . 5un2 3v16 v2 76 7v 6 . 7 then4 . 52Definition 5.4. Two vectors in Rn are said to be orthogonal if their dot product is 0.vnu · v u1 v 1 u2 v 2 · · · un v nThe fundamental properties of the dot product aresummarized by the following theorem.Theorem 5.3. If x, y, and z are vectors inis a scalar thenThe formula u · v kuk kvk cos θ relates the dot productto the angle between vectors 1 . If u and v are non-zerovectors then the right hand side of this expression is positive if 0 θ 90 and negative if 90 θ 180 .More importantly the dot product is 0 if θ 90 . Thismeans that two non-zero vectors are perpendicular if andonly if their dot product is 0. This leads to the followingdefinition.Rn and ca. x · y y · xb. x · (y z) x · y x · zc. (cx) · y c (x · y)d. x · x 0 and x · x 0 if and only if x 0.Proof. The proof of part (a) is left as an exercise.Since u · v uT v parts (b) and (c) follow from thecorresponding properties of matrix multiplication.The proof of part (d) is more complicated because itmakes several claims. First it claims that the inner product of a vector with itself can never be negative. To seeThis definition implies that the zero vector inorthogonal to every vector in Rn .Rn isEXAMPLE 5.8. Show that the triangle with verticesA( 2, 3), B(5, 5), and C(0, 4) is a right triangle.We want to interpret the sides of this triangle as vectors. If we treat A as the originthenvectorfrom– the »–» 5 ( 2)7A to B would be AB . Sim5 32–» –» 0 ( 2)2ilarly AC . Now notice that 4 3 7 AB · AC 7(2) 2( 7) 0. Since AB · AC 0 we have kABk kACk cos θ 0. We can then conclude that1Although this formula was proved only for vectors inR2 it is applicable in all Rn . Justification for this will begiven shortly.
5.2. The Dot Product7cos θ 0 and so θ 90 .Proof. If v is the zero vector then both sides of the inequality would be zero and the theorem would be true sowe will assume that v is not the zero vector.Part d of Theorem 5.3 tells us that“u·v ”u·v ” “u v · u v 0v·vv·vEXAMPLE5.9. »For what» –– value(s) of k are the vectorskk 1u and v orthogonal?2kk 1We just have to determine any values of k which makeu · v 0.u · v k(k 1) 2k(k 1) k2 k 2k2 2k 3k2 k k(3k 1)since the dot product of any vector with itself is alwaysgreater than or equal to 0.Simplifying the left hand side of the above gives“u·v ” “u·v ”u·v(u · v)2u v · u v u·u 2u·v v·vv·vv·vv·v(v · v)2It should then be clear that the vectors are orthogonalfor k 0 and k 1/3.We end this section with three theorems which statesome important properties of vectors.Theorem 5.5. (Pythagorean Theorem) Let u and vbe vectors in Rn then ku vk2 kuk2 kvk2 if and onlyif u and v are orthogonal. u·u (u · v)2v·vReplacing this last expression in the original inequalitygives(u · v)2kuk2 0kvk2Rearranging these terms giveskuk2 kvk2 (u · v)2Taking the square root of both sides givesProof. Note that the statement of this theorem is another“if and only if” statement. This means that the theoremis making two claims. These will be proved separatelybelow.First we must show that if u and v are orthogonal thenku vk2 kuk2 kvk2 . The argument is as followsku vk2 (u v) · (u v)(5.2) u·u 0 v·v(5.3) kuk2 kvk2(5.4)Equation 5.2 is a consequence of the distributive and commutative properties of the dot product. Equation 5.3 is aconsequence of u and v being orthogonal (their dot product is 0).Next we must show that if ku vk2 kuk2 kvk2 thenu and v are orthogonal. The argument goes as followsu · u 2u · v v · v u · u v · v(5.5)(5.6)2u · v 0(5.7)u·v 0(5.8)The last line tells us that u and v are orthogonal.Theorem 5.6. ( Cauchy-Schwarz Theorem) If u andv are vectors in Rn then u · v kukkvkThe Cauchy-Schwarz Theorem guarantees that 1 (5.1) u · u 2u · v v · vku vk2 kuk2 kvk2 u · v kukkvku·v 1kuk kvkfor any non-zero vectors u and v in Rn . This allows usto define the angle θ between any non-zero vectors u andv in Rn by the formulacos θ u·vkuk kvk(where 0 θ 180 ).2316 177EXAMPLE 5.10. Find the angle between u 64 1 5 12 316 277and v 64 35 .4 We have u · v 1 2 3 4 2, kuk 1 1 1 1 4 2, and kvk 1 4 9 16 30. Thuscos θ u·v 2 .1826kuk kvk2 30
8Chapter 5. Vector Geometrywhich gives θ 100.52 That is, the entries in any unit vector are the directioncosines of that vector.Theorem 5.7. ( Triangle Inequality) If u and v arevectors in Rn then2(5.12)306 1 77EXAMPLE 5.11. Let u 64 15. If we normalize u 23206 1/2 7 1 17we get 64 1/25. Now cos (0) π/2, cos (1/2) π/3, 2/2 cos 1 ( 1/2) 2π/3, and cos 1 ( 2/2) π/4. So u lies at an angle of 90 from e1 , at an angle of 60 from e2 ,at an angle of 120 from e3 , at an angle of 45 from e4Equation 5.11 follows from the Cauchy-Schwarz theoremThe above shows that ku vk2 (kuk kvk)2 . Thetheorem then follows by taking the square root of bothsides.Note that for any non-zero vector v there are two unitvectors in the same direction as v. One has the samesense as v, the other has the opposite sanse of v.ku vk kuk kvkProof.ku vk2 u · u 2u · v v · v(5.9) u · u 2 u · v v · v(5.10) u · u 2kukkvk v · v(5.11) (kuk kvk)2Direction AnglesThe dot product gives us a new way of looking at unitvectors. Any particular entry in a unit vector cannot belarger than 1 or less than -1. The entries in a unit vectorturn out to have a very simple geometric interpretation.2 3v13In R the angles between any vector v 4v2 5 andv3the x1 , x2 and x3 axes are called the direction anglesof v and are represented by α, β, γ respectively. Theseare just the angles between v and the unit vectors i, j,and k. So we havev·iv1 kvk kikkvkv·jv2 cos β kvk kjkkvkv·kv3 cos γ kvk kkkkvkcos α We can now write3322 3 2cos αkvk cos αv1v 4v2 5 4kvk cos β 5 kvk 4cos β 5cos γkvk cos γv3The values cos α, cos β, and cos γ are called the direction cosines of v.The last example can be generalized in the followingway (we leave the details as an exercise). Let v be anynon-zero vector in Rn . If we normalize v then the entriesin the normalized vector are just the cosines of the anglesbetween v and the vectors in the standard basis of Rn .
Problems9Problems1. For the following pairs of vectors u and v calculatekuk, kvk, u · v, and the angle between u and v.»a. u » –34b. u » – 30–2, v 1» –1, v 52 31c. u 415, v 12 314 2532 326 177d. u 64 25, v 12 346 376 74 251e. u » –» –ab,v baf. u i j k, v 2i j 2k2. Use the dot product to find the angles of the triangles with the following verticesa. A(1, 4), B(3, 2), C(6, 1)b. A(1, 0, 1), B(0, 2, 1), C(2, 1, 0)c. A(1, 1, 2, 2), B(1, 2, 2, 1), C(2, 2, 1, 1)3. Given the points A(1, 1), B(3, 1), and C(4, k) findthe values of k for which triangle ABC is a righttriangle.2 32 313454. Let u1 1 and u2 405. Show that any vector112 31in Span {u1 , u2 } is orthogonal to v 4 25. 1»–» –acos(θ) sin(θ)5. Let x and R . Use thebsin(θ)cos(θ)dot product to find the angle between x and Rx.» –» –126. Let u and v . For what value(s) of k3ka. do u and v have the same length?b. are u and v orthogonal?c. are u and v parallel?d. is the distance from u to v one unit?2 323124547. Let u 2 and v 1 k5. For what value(s)31 kof ka. do u and v have the same length?b. are u and v orthogonal?c. are u and v parallel?d. is the distance from u to v 3 units?»–»–cos θsin θ8. Let u and v . For what value(s)sin θcos θof θa. do u and v have the same length?b. are u and v orthogonal?c. are u and v parallel?9. Normalize the following vectors and find the direction angles in each case:2 31a. 2v 3 b. 2v 3 6271 36 7c. v 6374 2 54 1 5445 245a. Suppose v is a vector in R2 and you knowthat this vector forms an angle of π/3 withi. Is this enough information to determinevector v? What are the possible values forthe angle between v and j?b. Suppose v is a vector in R3 which forms anangle of π/6 with i. What are the possiblevalues for the angle between v and k?c. Suppose v is a vector in R3 which forms anangle of π/6 with i and an angle of π/6 withj. What are the possible values for the anglebetween v and k?11. Use the dot product to prove that kx yk2 kx yk2 2 kxk2 kyk210.12. Show that if u v and u v have the same magnitude then u and v are orthogonal.13. It was pointed out in this section that uT v vT ufor any two vectors u and v in Rn . Is it also truethat uvT vuT ? Prove this or give a count
Vector Geometry In this chapter we will look more closely at certain ge-ometric aspects of vectors in Rn. We will first develop an intuitive understanding of some basic concepts by looking at vectors in R2 and R3 where visualization is easy, then we will extend these geometric intuitions to Rn for any n.
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12 VECTOR GEOMETRY 12.1 VectorsinthePlane Preliminary Questions 1. Answer true or false. Every nonzero vector is: (a) equivalent to a vector based at the origin. (b) equivalent to a unit vector based at the origin. (c) parallel to a vector based at the origin. (d) parallel to a unit vector based at the origin. solution (a) This statement is true. Translating the vector so that it is based on .
