Chapter 1 Euclidean Space - Rice University

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1Euclidean spaceChapter 1Euclidean spaceA. The basic vector spaceWe shall denote by R the field of real numbers. Then we shall use the Cartesian productRn R R . . . R of ordered n-tuples of real numbers (n factors). Typical notation forx Rn will bex (x1 , x2 , . . . , xn ).Here x is called a point or a vector , and x1 , x2 , . . . , xn are called the coordinates of x. Thenatural number n is called the dimension of the space. Often when speaking about Rn andits vectors, real numbers are called scalars.Special notations:R1R2R3xx (x1 , x2 ) or p (x, y)x (x1 , x2 , x3 ) or p (x, y, z).We like to draw pictures when n 1, 2, 3; e.g. the point ( 1, 3, 2) might be depicted as

2Chapter 1We define algebraic operations as follows: for x, y Rn and a R,x yax xx y (x1 y1 , x2 y2 , . . . , xn yn );(ax1 , ax2 , . . . , axn );( 1)x ( x1 , x2 , . . . , xn );x ( y) (x1 y1 , x2 y2 , . . . , xn yn ).We also define the origin (a/k/a the point zero)0 (0, 0, . . . , 0).(Notice that 0 on the left side is a vector, though we use the same notation as for the scalar0.)Then we have the easy facts:x y(x y) z0 xx x1x(ab)xa(x y)(a b)x0xa0 y x;x (y z);x;0;x;a(bx);ax ay;ax bx;0;0.in other words all the“usual” algebraic rulesare valid if they makesenseSchematic pictures can be very helpful. One nice example is concerned with the linedetermined by x and y (distinct points in Rn ). This line by definition is the set of all pointsof the form(1 t)x ty,Here’s the picture:where t R.

3Euclidean spaceThis picture really is more than just schematic, as the line is basically a 1-dimensionalobject, even though it is located as a subset of n-dimensional space. In addition, the closedline segment with end points x and y consists of all points as above, but with 0 t 1. Thissegment is shown above in heavier ink. We denote this segment by [x, y].We now see right away the wonderful interplay between algebra and geometry, somethingthat will occur frequently in this book. Namely, the points on the above line can be describedcompletely in terms of the algebraic formula given for the line. On the other hand, the line isof course a geometric object.It is very important to get comfortable with this sort of interplay. For instance, if wehappen to be discussing points in R5 , we probably have very little in our background thatgives us geometric insight to the nature of R5 . However, the algebra for a line in R5 is verysimple, and the geometry of a line is just like the geometry of R1 .Similarly, it is helpful to represent triangles with a picture in the plane of the page. Thusif we have three noncollinear points x, y, z in Rn , there is a unique plane which contains them.This plane lies in Rn of course, but restricting attention to it gives a picture that looks likean ordinary plane. The plane is the set of all points of the formp t1 x t2 y t3 z,where t1 t2 t3 1. Sometimes the scalars t1 , t2 , t3 are called barycentric coordinates ofthe point p.

4Chapter 1The point displayed inside this triangle is 13 (x y z), and is called the centroid of the triangle.PROBLEM 1–1. We need to examine the word collinear we have just used. In fact,prove that three points x, y, z in Rn lie on a line there exist scalars t1 , t2 , t3 , not allzero, such thatt1 t2 t3 0,t1 x t2 y t3 z 0.PROBLEM 1–2. Prove that x, y, 0 are collinear x is a scalar multiple of y or yis a scalar multiple of x.PROBLEM 1–3. Prove that if x, y, z in Rn are not collinear and if p belongs to theplane they determine, then the real numbers t1 , t2 , t3 such thatt1 t2 t3 1,t1 x t2 y t3 z p,are unique.

5Euclidean spacePROBLEM 1–4. In the triangle depicted above let L1 be the line determined by xand the midpoint 21 (y z), and L2 the line determined by y and the midpoint 21 (x z).Show that the intersection L1 L2 of these lines is the centroid. (This proves the theoremwhich states that the medians of a triangle are concurrent.)PROBLEM 1–5. Prove that the interior (excluding the sides) of the above triangle isdescribed by the conditions on the barycentric coordinatest1 t2 t3 1,t1 0, t2 0, and t3 0.As an example of our method of viewing triangles, think about an equilateral triangle. Ifwe imagine it conveniently placed in R2 , the coordinates of the vertices are bound to be rathercomplicated; for instance, here are two ways:x2x2(0, 3 )(0, 1)( 1, 0)(1, 0)x1x1( 3 /2, 1/2)( 3 /2, 1/2)But a really elegant positioning is available in R3 , if we simply place the vertices at (1, 0, 0),(0, 1, 0) and (0, 0, 1):

6Chapter 1(0, 1, 0)(1, 0, 0)(0, 0, 1)Now this looks much better if we view this triangle as it lies in the plane x1 x2 x3 1:We can’t draw coordinate axes in this plane, or even the origin, though we could imagine theorigin as sitting “behind” the centroid:origin

7Euclidean spacePROBLEM 1–6. Prove that the four points w, x, y, z in Rn are coplanar thereexist real numbers t1 , t2 , t3 , t4 , not all zero, such thatt1 t2 t3 t4 0,t1 w t2 x t3 y t4 z 0.As further evidence of the power of vector algebra in solving simple problems in geometry,we offerPROBLEM 1–7.Let us say that four distinct points w, x, y, z in Rn define aquadrilateral , whose sides are the segments [w, x], [x, y], [y, z], [z, w] in that order.zyzyxxwwProve that the four midpoints of the sides of a quadrilateral, taken in order, form thevertices of a parallelogram (which might be degenerate). In particular, these four pointsare coplanar. (Note that the original quadrilateral need not lie in a plane.) Express thecenter of this parallelogram in terms of the points w, x , y , and z.

8Chapter 1PROBLEM 1–8. We have seen in Problem 1–1 the idea of three points being collinear,and in Problem 1–6 the idea of four points being coplanar. These definitions can of coursebe generalized to an arbitrary number of points. In particular, give the correct analogousdefinition for two points to be (say) “copunctual” and then prove the easy result that twopoints are “copunctual” if and only if they are equal.PROBLEM 1–9. Give a careful proof that any three points in R1 are collinear; andalso that any four points in R2 are coplanar.It is certainly worth a comment that we expect three “random” points in R2 to be noncollinear. We would then say that the three points are in general position. Likewise, fourpoints in R3 are in general position if they are not coplanar.B. DistanceNow we are going to discuss the all-important notion of distance in Rn . We start with R2 ,where we have the advantage of really understanding and liking the Pythagorean theorem. Weshall freely accept and use facts we have learned from standard plane geometry. Thus we saythat the distance between x and y in R2 is#2axis x 2 y2 #1 axis x 1 y1 d(x, y) p(x1 y1 )2 (x2 y2 )2 .We can even proceed easily to R3 by applying Pythagoras to the right triangle with legs given

9Euclidean spaceby the two segments: (1) from (x1 , x2 , x3 ) to (y1 , y2 , x3 ) and (2) from (y1 , y2 , x3 ) to (y1 , y2 , y3 ).y( y 1, y2, x3 )xThe segment (1) determines the distancep(x1 y1 )2 (x2 y2 )2 ,and the segment (2) determines the distance x3 y3 .If we square these distances, add the results, and then take the square root, the distance wefind isp(x1 y1 )2 (x2 y2 )2 (x3 y3 )2 .Though we are hard pressed to draw a similar picture for R4 etc., we can easily imagine thesame procedure. For R4 we would consider the “horizontal” line segment from (x1 , x2 , x3 , x4 )to (y1 , y2 , y3 , x4 ), and then the “vertical” segment from (y1 , y2 , y3 , x4 ) to (y1 , y2 , y3 , y4 ). Thetwo corresponding distances arep(x1 y1 )2 (x2 y2 )2 (x3 y3 )2and x4 y4 ,respectively. (We have used our formula from the previous case of R3 .) We certainly want tothink of these two segments as the legs of a right triangle, so that the distance between x and

10Chapter 1y should come from Pythagoras by squaring the two numbers above, adding, and then takingthe square root:pd(x, y) (x1 y1 )2 (x2 y2 )2 (x3 y3 )2 (x4 y4 )2 .We don’t really need a picture to imagine this sort of construction for any Rn , so we areled toTHE PYTHAGOREAN DEFINITION. The distance between x and y in Rn isvu nuXd(x, y) t (xi yi )2 .i 1Clearly, d(x, y) d(y, x), d(x, x) 0, and d(x, y) 0 if x 6 y. We also say that d(x, y) isthe length of the line segment [x, y].This definition of course gives the “right”answer for n 2 and n 3. (It even works forpn 1, where it decrees that d(x, y) (x y)2 x y .)It will be especially convenient to have a special notation for the distance from a point tothe origin:DEFINITION. The norm of a point x in Rn is the numbervu nuXkxk d(x, 0) tx2i .i 1Thus we have equations likeR1R2R3R4::::k 3k 3k(3, 4)k 5 k(1, 2, 3)k 14k(1, 1, 1, 1)k 2.The idea of “norm” is important in many areas of mathematics. The particular definitionwe have given is sometimes given the name Euclidean norm.We have some easy properties:d(x, y)kaxk k xkk0kkxk kx yk; a kxk for a R;kxk;0;0 if x 6 0.

11Euclidean spaceHere’s a typical picture in R3 :Oxdx1x3x2kxk2 δ 2 x23δ 2 x21 x22We close this section with another “algebra-geometry” remark. We certainly are thinkingof distance geometrically, relying heavily on our R2 intuition. Yet we can calculate distancealgebraically, thanks to the formula for d(x, y) in terms of the coordinates of the points x andy in Rn .C. Right angleNow we turn to a discussion of orthogonality. We again take our clue from the Pythagoreantheorem, the square of the hypotenuse of a right triangle equals the sum of the squares of theother two sides. The key word we want to understand is right.Thus we want to examine a triangle in Rn (with n 2). Using a translation, we maypresume that the potential right angle is located at the origin. Thus we consider from thestart a triangle with vertices 0, x, y. As we know that these points lie in a plane, it makessense to think of them in a picture such as

12Chapter 1xOyWe are thus looking at the plane containing 0, x, y, even though these three points lie in Rn .We then say that the angle at 0 is a right angle if and only if the Pythagorean identityholds:d(x, y)2 d(x, 0)2 d(y, 0)2 .I.e., in terms of the Euclidean norm on Rn ,kx yk2 kxk2 kyk2 .(P)You probably have noticed that our reasoning has been somewhat circular in nature. For weused the Pythagorean idea to motivate the definition of distance in the first place, and now weare using distance to define right angle. There is an important subtlety at work here. Namely,in defining distance we worked with right angles in coordinate directions only, whereas nowwe are defining right angles in arbitrary (noncoordinate) directions. Thus we have achievedsomething quite significant in this definition.We feel justified in this definition because of the fact that it is based on our intuition fromthe Euclidean geometry of R2 . Our planes are located in Rn , but we want them to have thesame geometric properties as R2 .

13Euclidean spaceNow we perform a calculation based upon our use of algebra in this material:kx yk2 nX(xi yi )2i 1 nX(x2i 2xi yi yi2 )i 12 kxk 2nXxi yi kyk2 .i 1Thus the Pythagorean relation (P) becomesnXxi yi 0.i 1PIn summary, we have a right angle at 0 ni 1 xi yi 0.Based upon the sudden appearance of the above number, we now introduce an extremelyuseful bit of notation:DEFINITION. For any x, y Rn , the inner product of x and y, also known as the dotproduct, is the numberx y nXxi y i .I like to make this dot huge!i 1The above calculation thus says thatkx yk2 kxk2 2x y kyk2 .Just to make sure we have the definition down, we rephrase our definition of right angle:DEFINITION. If x, y Rn , then x and y are orthogonal (or perpendicular ) if x y 0.Inner product algebra is very easy and intuitive:x y(x y) z(ax) y0 xx x y x;x z y z;a (x y) ;0;kxk2 .

14Chapter 1The calculation we have performed can now be done completely formally:kx yk2 (x y) (x y) x x x y y x y y kxk2 2x y kyk2 .PROBLEM 1–10. As we have noted, the vector 0 is orthogonal to every vector. Showconversely that if x Rn is orthogonal to every vector in Rn , then x 0.PROBLEM 1–11. Given x, y Rn . Prove that x y x z y z for all z Rn .Here is an easy but astonishingly importantPROBLEM 1–12. Let x 6 0 and y be in Rn . As we know, the line determined by 0and x consists of all points of the form tx. Find the (unique) point on this line such thatthe vector y tx is orthogonal to x. Also calculate as elegantly as you can the distanceky txk.yxOtxPROBLEM 1–13. In the same situation, find the (unique) point on the line which isclosest to y. Comment?(Solutions: (y tx) x 0 y x tkxk2 0 t x y/kxk2 . Also we haveky txk2 kyk2 2tx y t2 kxk2 .

15Euclidean spaceFor the above value of t, we obtain2x y(x y)2x y kxk2kxk22(x y) kyk2 .kxk2ky txk2 kyk2 Thuspky txk kxk2 kyk2 (x y)2.kxkThis is the solution of 1–12. To do 1–13 write the formula above in the form· x y222ky txk kxk t 2t kyk22kxk¶2µ(x y)2x y2 kyk2 . kxk t kxk2kxk2The minimum occurs t x y/kxk2 . The comment is that the same point is the solutionof both problems.)Another way to handle Problem 1–13 is to use calculus to find the value of t which minimizesthe quadratic expression.You should trust your geometric intuition to cause you to believe strongly that the pointasked for in Problem 1–12 must be the same as that asked for in Problem 1–13.PROBLEM 1–14. This problem is a special case of a two-dimensional version of thepreceding two problems. Let n 2 and let M be the subset of Rn consisting of all pointsof the form x (x1 , x2 , 0, . . . , 0). (In other words, M is the x1 x2 plane.) Let y Rn .a. Find the unique x M such that y x is orthogonal to all points in M .b. Find the unique x M which is closest to y.BONUS. Since ky txk2 0, we conclude from the above algebra thatkxk2 kyk2 (x y)2 0. Furthermore, this can be equality y tx 0 y tx.Now that we have proved this, we state it as theSCHWARZ INEQUALITY. For any x, y Rn , x y kxk kyk.

16Chapter 1Furthermore, equality holds x 0 or y 0 or y is a scalar multiple of x.It is extremely useful to keep in mind schematic figures to illustrate the geometric significance of the sign of x y:ytxyOxy 0xOyxOx y 0txxx y 0The validity of these figures follows from the formula we have obtained for t, which impliesthat t and x y have the same sign.D. AnglesAmazingly, we can now use our understanding of right angle to define measurement ofangles in general. Suppose x and y are nonzero points in Rn which are not scalar multiplesof each other. In other words, x, y, and 0 are not collinear. We now scrutinize the plane inRn which contains x, y, and 0. Specifically, we examine the angle formed at 0 by the linesegments from 0 to x and from 0 to y, respectively.To measure this angle we use the most elementarydefinition of cosine from high school geometry. Wehave the three cases depicted at the end of Section C:yOx

17Euclidean spaceyyθθtxyOxOθxOtxIn all cases we let θ be the angle between the two segments we are considering. Thencos θ “adjacent side”tkxk .“hypotenuse”kykRemember that t x y/kxk2 . Therefore we have our desired formula for θ:yOθx0 θ πcos θ x y.kxk kykAt the risk of excessive repetition, this sketch of the geometric situation is absolutelyaccurate, as we are looking at a plane contained in Rn .Once again the interplay between algebra and geometry is displayed. For the definitionof the inner product x y is given as an algebraic expression in the coordinates of the twovectors, whereas now we see x y is intimately tied to the geometric idea of angle. What ismore, in case x and y are unit vectors (meaning that their norms equal 1), then cos θ x y.A nice picture for this is obtained by drawing a circle of radius 1, centered at 0, in the planex

18Chapter 1of 0, x, and y. Then x and y lie on this circle, and θ is the length of the shorter arc connectingx and y.arclengthθxθOySUMMARY. The inner product on Rn is a wonderful two-edged sword. First, x y is acompletely geometric quantity, as it equals the product of the lengths of the two factors andthe cosine of the angle between them. Second, it is easily computed algebraically in terms ofthe coordinates of the two factors.Any nonzero vector x produces a unit vector by means of the device of “dividing out” thenorm: x/kxk. Then the above formula can be rewrittencos θ xy .kxk kykREMARK. If the vertex of an angle is not the origin, then subtraction of points gives thecorrect formula:yzθxcos θ (x z) (y z).kx zk ky zk

19Euclidean spaceEXAMPLE. A triangle in R3 has vertices (1, 0, 0), (0, 2, 0), and (0, 0, 3). What are its threeangles? Solution:(1, 0, 0)AB(0, 2, 0)C(0, 0, 3)( 1, 2, 0) ( 1, 0, 3)1 .510504cos B ,659cos C .130cos A Approximately,A 81.87o ,B 60.26o ,C 37.87o .Notice how algebraic this is! And yet it produces valuable geometric information as to theshape of the given triangle.

20Chapter 1EXAMPLE. Compute the acute angle between the diagonals of a cube in R3 .θSolution: it is enough to arrange the cube so that its eight vertices are located at thepoints ( 1, 1, 1). Then the diagonals are the line segments from one vertex p to theopposite vertex p. These diagonals intersect at 0. The picture in the plane determined bytwo diagonals looks like this:

21Euclidean spacepqO-q-pWe are supposed to find the acute angle of their intersection, so we usecos θ p q.kpk kqk(If this turns out to be negative, as in the sketch, then we have π2 θ π and we useπ θ for the answer.) As p ( 1, 1, 1) and q ( 1, 1, 1), and p 6 q, we havekpk kqk 3 and p q 1. Thus1cos θ ,3givingθ arccos13( 70.5 ).PROBLEM 1–15.Consider two diagonals of faces of a cube which intersect at avertex of the cube. Compute the angle between them.

22Chapter 1PROBLEM 1–16. Given a regular tetrahedron (its four faces are equilateral triangles),locate its centroid. (You may define its centroid to be the average of its four vertices;in other words, the (vector) sum of the vertices divided by 4.) Then consider two linesegments from the centroid to two of the four vertices. Calculate the angle they form atthe center. (Here is displayed a particularly convenient location of a regular tetrahedron.)(1, 1, 1)( 1, 1, 1)( 1, 1, 1)(1, 1, 1)PROBLEM 1–17. Repeat the calculation of the receding problem but instead usea regular tetrahedron situated in R4 having vertices at the four unit coordinate vectors(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1).

23Euclidean spacePROBLEM 1–18.Two adjacent faces of a cube intersect in an edge and form adihedral angle, which is clearly π2 :dihedral angleSide view of cubeCalculate the dihedral angle formed by two faces of a regular tetrahedron.An important consequence of the Schwarz inequality is theTRIANGLE INEQUALITY. For any x, y Rn ,kx yk kxk kyk.PROOF. We simply compute as follows:kx yk2 (x y) (x y)kxk2 2x y kyk2kxk2 2 x y kyk2kxk2 2kxk kyk kyk2(kxk kyk)2 .(Schwarz inequality)QEDThe reason for the name “triangle inequality” can be seen in a picture:The shaded triangle has edges with lengths as shown, so the triangle inequality is thestatement that any edge of a triangle in Rn is less than the sum of the other two edges.

24Chapter 1y xx yy y O x xPROBLEM 1–19. Prove that the triangle inequality is an equality either x 0or y tx for some t 0. What does this mean geometrically?PROBLEM 1–20. Use the triangle inequality to prove that for any points x, y, z Rn ,d(x, y) d(x, z) d(z, y).And prove that equ

Euclidean space 3 This picture really is more than just schematic, as the line is basically a 1-dimensional object, even though it is located as a subset of n-dimensional space. In addition, the closed line segment with end points x and y consists of all points as above, but with 0 t 1. This segment is shown above in heavier ink.

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