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# Lines and planes in space

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### Lines and planes in space

1. 1. Chapter 2 3–Space: lines and planes In this chapter we • discuss how to describe points, lines and planes in 3–space. • introduce the language of vectors. • discuss various matters concerning the relative position of lines and planes: in par- ticular, intersections and angles. • show how to translate and rotate lines and planes in relatively simple cases. 2.1 3–space and vectors2.1.1 Coordinatizing 3–space The physical world around us is called three–dimensional because through any point pre- cisely three mutually perpendicular axes (and no more) can pass. Such axes can be used to describe points in 3-space by triples of numbers: the (signed) distances to planes formed by two of these axes. We call such a set of perpendicular axes through a given point a cartesian coordinate system. Each of the axes is called a coordinate axis and the point of intersection of the three axes is called the origin. Every point of a coordinate axis cor- responds to a real number. As you know, often, but not necessarily always, these axes are called x–axis, y–axis and z–axis. By convention, we agree that the coordinate sys- tem be right–handed : if you turn the positive x–axis to the positive y–axis, then a screw positioned along the z–axis following this movement would move in the direction of the positive z–axis. Every two of the three axes span a plane: the y, z–plane, the x, z–plane and the x, y–plane, respectively. The signed distances u, v and w of a point P in 3–space to each of these planes in the given order are the coordinates of P ; we usually put the three together as follows: (u, v, w) and still call this the coordinates of P . Notations for points that occur often: (x, y, z), (x1 , x2 , x3 ), (a, b, c), (a1 , a2 , a3 ), (u, v, w), etc. For a point P with coordinates (x, y, z) we also write P = (x, y, z). 21
2. 2. 22 3–Space: lines and planes z z P=(2,3,4) y 4 y x x Figure 2.1: Left: The third coordinate of P = (2, 3, 4) is the signed distance of P to the x, y–plane; so if the point were below the x, y–plane, the third coordinate would have been negative. Right: Cartesian coordinate systems are taken to be right–handed: a screw positioned along the z–axis moves along the positive z–axis if turned in the direction from x–axis to y–axis. Most discussions in this chapter will focus on 3–space. But with some adaptations (usually simpliﬁcations) the material makes sense in 2–space.2.1.2 Example. Given the point P with coordinates (x, y, z), the point Q with coordinates (x, y, 0) is the vertical projection of P on the x, y–plane. The distance between P and Q is |z| (note the absolute value). Similarly, the (horizontal) projection on the y, z–plane has coordinates (0, y, z). Finally, the horizontal projection on the x, z–plane has coordinates (x, 0, z).2.1.3 Vectors: introduction In dealing with geometric issues, it is convenient to have the language of vectors at our disposal. A vector is a quantity with both magnitude and direction. A well–known example z E p=OP F B C O y A AB D x Figure 2.2: The arrow with tail at A and head at B represents a vector (left). The two arrows CD and EF have the same length and direction, and so represent the same vector (middle). A vector with tail in the origin is usually represented by a lower case boldface symbol, like p (right).
3. 3. 2.1 3–space and vectors 23 of a vector occurring in physics is velocity. A vector is usually represented by an arrow in the sense of a directed line segment; the length of the segment corresponds to the magnitude. For us the most important aspects about the notion of a vector are summarized below. • For points A and B in 2–space or 3–space, the vector AB is an arrow with tail at A and head at B. (It points from A to B.) Only its magnitude (length) and its direction are of importance, not the positioning of the arrow in the following sense: if two vectors have the same length and the same direction, we consider them equal. The representing arrows need not coincide at all (but one can be translated so that it coincides with the other). • The vectors in 3–space with tail at the origin O = (0, 0, 0) and head at a point P = (x, y, z) play a special role. Such a vector OP is usually represented by the corresponding lower case boldface symbol, say p. If we want to emphasize the vector aspect, we write p = (x, y, z) rather than P = (x, y, z). The very scrupulous reader may ﬁnd this abuse of notation a bit distressing. In practice, it almost never goes wrong. • To assign coordinates/numbers to any vector, say the vector AB with A = (a1 , a2 , a3 ) and B = (b1 , b2 , b3 ), we note that this vector can also be represented by the arrow with tail at the origin and head at (b1 − a1 , b2 − a2 , b3 − a3 ). And so we write AB = (b1 − a1 , b2 − a2 , b3 − a3 ). • In 3–space with a cartesian coordinate system, we single out three special vectors, the so–called standard basis vectors. – e1 with tail at (0, 0, 0) and head at (1, 0, 0) (the standard basis vector of length 1 along the positive x–axis). – e2 with tail at (0, 0, 0) and head at (0, 1, 0) (the standard basis vector of length 1 along the positive y–axis). – e3 with tail at (0, 0, 0) and head at (0, 0, 1) (the standard basis vector of length 1 along the positive z–axis). In the following, vectors usually refer to directed segments with their tail in the origin, unless the context suggests otherwise. Also note that in writing on paper or on the blackboard the use of boldface symbols is not convenient. So we also use notations like x = (x1 , x2 , x3 ), where we underline the symbol for a vector.2.1.4 The ﬁrst arithmetic with vectors We can do a kind of arithmetic with vectors that turns out to make sense in the context of geometry.
5. 5. 2.1 3–space and vectors 25 Of course, this is the same as the distance from the origin to the point with coordi- nates (a, b, c). The squared length is therefore |p|2 = a2 + b2 + c2 z P=(a,b,c) O y (a,0,0) Q=(a,b,0) xFigure 2.4: The length of a vector with end√point at (a, b, c) is computed by repeatedapplication of Pythagoras’ theorem: |OQ| = a2 + b2 and |OP | = |OQ|2 + |QP |2 =√ a2 + b 2 + c 2 . • Dot or inner product. The dot product or inner product of two vectors x = (x1 , x2 , x3 ) and y = (y1 , y2 , y3 ) is deﬁned as x • y = x 1 y1 + x2 y2 + x3 y3 . This product does not have an immediate geometric interpretation, but will show up indirectly in various geometrically relevant situations, such as the length of a vector and the angle between two vectors: For instance, the squared length of x equals the inner product of x with itself: |x|2 = x2 + x2 + x2 = x1 x1 + x2 x2 + x3 x3 = x • x. 1 2 3 • Distance. The distance between the points P = (x, y, z) and Q = (a, b, c) is (x − a)2 + (y − b)2 + (z − c)2 ; it equals the length of the line segment P Q. We deﬁne the distance between two vectors as the distance between their endpoints, or, equivalently, the length of the diﬀerence. The distance between p = (x, y, z) and q = (a, b, c) is |p − q| = |(x − a, y − b, z − c)| = (x − a)2 + (y − b)2 + (z − c)2 .
6. 6. 26 3–Space: lines and planes • Perpendicular vectors. Using the dot product we can express when two vectors are perpendicular. It turns out (no proof here) that this is the case precisely if their inner product equals 0: x • y = 0 ⇔ x and y perpendicular. • Angle between vectors. If a and b are nonzero vectors, then the angle α between them is usually computed via its cosine. From the Cosine Law the following relation involving the dot product can be shown to hold: a•b cos α = . |a| · |b| To determine the angle, two steps are therefore required: ﬁrst determine the cosine of the angle using the formula above, then solve for the angle. In the case of perpen- dicular vectors, the angle between them is 90◦ or π/2 radians, since cos α = 0 in this case.2.1.5 Examples. Here are a few examples of the notions mentioned above. a) 2 · (3, 4, −1) − 3 · (2, 5, 1) = (0, −7, −5). √ √ b) The length of the vector p = (1, 2, 3) is |p| = 12 + 22 + 32 = 14. c) The inner product of the vectors (1, 2, 3) and (2, −1, 4) is (1, 2, 3) • (2, −1, 4) = 1 · 2 + 2 · (−1) + 3 · 4 = 12. d) The vectors (1, 1, 2) and (2, 2, −2) are perpendicular since their inner product equals 0:: (1, 1, 2) • (2, 2, −2) = 1 · 2 + 1 · 2 + 2 · (−2) = 0. e) To compute the angle α between the vectors (1, 1, 0) and (1, 2, 1) we ﬁrst determine cos α: √ √ √ (1, 1, 0) • (1, 2, 1) 1·1+1·2+0·1 3 3· 3 3 cos α = =√ √ =√ √ =√ √ = . |(1, 1, 0)| · |(1, 2, 1)| 11 + 12 · 12 + 22 + 12 2· 6 4· 3 2 Hence the angle is 30◦ or π/6 radians. 2.2 Describing lines and planes2.2.1 Describing a line by a vector parametric equation To specify a line ℓ in 3–space, it suﬃces to give a point, say P0 , through which the line passes and to give the line’s direction, say given by the nonzero vector v. In terms of vectors, start with a vector p0 with end point P0 and add an arbitrary scalar multiple λv of v to it: p0 + λv.
7. 7. 2.2 Describing lines and planes 27 As λ ranges from −∞ to ∞, all vectors with endpoints on the line can be obtained. Not surprisingly, the vector v is called a direction vector of the line. Of course, any nonzero multiple of v can be used as direction vector, i.e., p0 + µ(3v) describes the same line. Here is an example. The line ℓ passing through P0 = (2, 3, 5) and having direction vector a + λv a v Figure 2.5: Parametric description of a line involves a support vector p0 and a nonzero direction vector v. The line can be seen as ‘resting’ on p0 . Any vector on the line is obtained by adding a suitable multiple of v to p0 . x = (1, −1, 2) is (2, 3, 5) + λ(1, −1, 2). Any value of λ produces a speciﬁc vector or point on the line. For instance, for λ = 2, we get (2, 3, 5) + 2 · (1, −1, 2) = (4, 1, 9). An arbitrary point on the line can be described as (2 + λ, 3 − λ, 5 + 2λ). Two remarks are in place here. • Any vector on the line can be taken as the support vector. So, for example, (4, 1, 9) + µ(1, −1, 2) is the same line, since (4, 1, 9) is on the line. • Any two direction vectors diﬀer by a (nonzero) multiple. This means that, for exam- ple, (2, 3, 5) + ρ(−2, 2, −4) is the same line.2.2.2 Describing a plane in 3–space by an equation Suppose V is a plane in 3–space. If you want to explain to someone else which plane it is, it suﬃces to give the following information: • a point P0 through which the plane passes, and • its ‘direction’. Well, a plane contains many directions. Now a smart move is to specify the plane’s direction by giving the direction of a line which is perpendicular to the plane. The direction of that line can be given by a single nonzero vector (the length of the vector is irrelevant, only its direction matters). Let us translate this into mathematics, ﬁrst in an example and then in the general case.
8. 8. 28 3–Space: lines and planes • Here is the example. Suppose the point P0 = (2, 1, 3) is on the plane U , and let p0 = OP0 be the vector pointing to P0 . Suppose moreover that n = (1, 2, 2) is perpendicular to U . Our task is to catch those vectors x = (x, y, z) whose endpoints are in the plane. Now such an endpoint is in U if the direction from P0 to this endpoint is perpendicular to n. This direction is represented by the vector x − p0 = (x − 2, y − 1, z − 3). So we require that this vector is perpendicular to n = (1, 2, 2), i.e., (x − 2, y − 1, z − 3) • (1, 2, 2) = 0 or (x − 2) + 2 · (y − 1) + 2 · (z − 3) = 0. Of course, we can rewrite this equation as x + 2y + 2z = 10, but the form x − 2 + 2(y − 1) + 2(z − 3) = 0 shows clearly that (2, 1, 3) is on the plane. Please note that: – For any nonzero real number t, the equation tx + 2ty + 2tz = 10t represents the same equation. For example, 6x + 12y + 12z = 60. – The coeﬃcients of x, y, z in the equation x + 2y + 2z = 10 form a vector which is (a multiple of) the vector n = (1, 2, 2) perpendicular to the plane we started with. This is not a coincidence as we will see below. – The coordinates of P0 satisfy the equation: 2 + 2 · 1 + 2 · 3 = 10. The last two items provide an easy check on the correctness of the equation obtained. n z P0 p0 P x y xFigure 2.6: Given a point P0 in the plane and a vector n perpendicular to the plane, anyvector x satisfying (x − p0 ) • n = 0 has its endpoint on the plane. • Here is the general story. Let P0 = (x0 , y0 , z0 ) be a point and let n = (a, b, c) be a nonzero vector, which we suppose to be perpendicular to the plane we wish to describe. The plane through P and perpendicular to n consists of all the points P = (x, y, z) such that P0 P is perpendicular to n, i.e., (x − x0 , y − y0 , z − z0 ) is perpendicular to (a, b, c). In vector form: (a, b, c) • (x − x0 , y − y0 , z − z0 ) = 0,
9. 9. 2.2 Describing lines and planes 29 and in equation form (i.e., we have expanded the inner product): a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0. Of course, an equivalent form is ax + by + cz = ax0 + by0 + cz0 (the right–hand side is a constant). If we replace the right–hand side by a single symbol, the general form of the equation of a plane becomes: ax + by + cz = d. The vector (a, b, c) is a vector perpendicular to the plane.2.2.3 Deﬁnition. (The point–normal equation of a plane) The plane which passes through the point P0 = (x0 , y0 , z0 ) and which is perpendicular to the direction of the nonzero vector n = (a, b, c) has equation a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0. In vector form, the equation is usually written as: n • (x − p0 ) = 0, where the vector p0 corresponds to the point P0 .2.2.4 Remark. Suppose the plane U has equation 2x1 − 3x2 + 5x3 = 10. If you multiply every coeﬃcient by the same number, say 5, then the resulting equation, 10x1 −15x2 +25x3 = 50, describes U as well. In practice we usually (but not necessarily) multiply the coeﬃcients by a number so that the equation looks more pleasant. For instance, 3 7 2 9 x1 − x2 + x3 = 2 5 3 2 looks better in the form 45x1 − 42x2 + 20x3 = 135 (where the ﬁrst mentioned equation has been multiplied through by 30). What we just said is one instance of the fact that we are free to change the appearance of an equation according to our speciﬁc needs, as long as we do not violate any mathematical rule. Here are a few examples of the equation 2x1 − 3x2 + 5x3 = 10 in diﬀerent guises: • 2x1 − 3x2 + 5x3 − 10 = 0 (all terms on one side). 2 5 10 • x2 = x1 + x3 − (the second variable has been ‘isolated’ on one side). 3 3 3 • 2(x1 − 1) − 3(x2 + 1) + 5(x3 − 1) = 0 shows clearly that (1, −1, 1) is on the plane.
10. 10. 30 3–Space: lines and planes2.2.5 Vector parametric description of a plane A second way of representing a plane is by a vector parametric equation, similar to the description of a line. First we discuss an example, where we use an intuitive approach, then we turn to a general strategy. So here is the ﬁrst example. Suppose the plane passes through (0, 0, 0), like for instance the plane U given by x1 + 2x2 − 3x3 = 0. Take two vectors in the plane which are not multiples of each other (they point in really distinct directions), for example (2, −1, 0) and (3, 0, 1). Geometrically, it is quite clear that any vector in the plane can be obtained by taking suitable multiples of the vectors and then adding them, see Figure (2.7). b b a x a p Figure 2.7: Left–hand side: If U is a plane through the origin and a and b are two vectors which are not along the same line, then any vector x in the plane can be obtained by adding suitable multiples of a and b. Right–hand side: the ingredients of a vector parametric description of a plane: a vector p with endpoint in the plane and two direction vectors a and b. For varying λ and µ, the vector p + λa + µb runs through all vectors of the plane. So, any vector in the plane can be written as a so–called linear combination λ(2, −1, 0) + µ(3, 0, 1) of (2, −1, 0) and (3, 0, 1) for suitable λ and µ. Note that in this way we have described the vectors/points of the plane explicitly: every value of λ and µ produces a point in the plane. For example, for λ = 2 and µ = −3 we ﬁnd 2 · (2, −1, 0) − 3 · (3, 0, 1) = (−5, −2, −3). Next we turn to a general strategy for ﬁnding vector parametric descriptions. The way to ﬁnd a vector parametric equation of a plane, starting from an equation, is to solve the equation (and don’t forget: there are inﬁnitely many solutions; it’s a plane after all). Suppose V is the plane with equation x1 + 2x2 − 3x3 = 4; so V is parallel to U , since the vector (1, 2, −3) is perpendicular to both planes. If we rewrite this equation as x1 = −2x2 + 3x3 + 4, then we clearly see, that for any given values of x2 and x3 , there is exactly one value of x1 such that the triple (x1 , x2 , x3 ) is on the plane. So assign the value λ to x2 and µ to x3 . Then x1 = 4 − 2λ + 3µ. We gain more insight in the solutions if we
11. 11. 2.2 Describing lines and planes 31 rewrite this explicit description in vector notation as follows: (x1 , x2 , x3 ) = (4 − 2λ + 3µ, λ, µ) = (4, 0, 0) + λ(−2, 1, 0) + µ(3, 0, 1). In this case you see (again) that V is parallel to U (how?). The vector (4, 0, 0) is usually called a support vector , while (−2, 1, 0) and (3, 0, 1) are called direction vectors. To summarize: we have described two ways of ﬁnding a vector parametric description of a plane: • Solve an equation describing the plane and rewrite the solutions in the form p + λa + µb. • Find any vector p whose endpoint is in the plane and ﬁnd two vectors a and b representing two ‘independent’ directions in the plane. Then the endpoints of p + λa + µb for varying λ and µ run through all points of the plane. Please note that the same plane can be described by parametric descriptions which may look quite diﬀerent. See the following example. Equations of a plane, however, show less variation: if we only consider equations of the form ax + by + cz = d, then two such equations describe the same plane precisely when the coeﬃcients diﬀer by a common (nonzero) multiple.2.2.6 Example. To ﬁnd a vector parametric description of the plane U : x + y + z = 4, any of the following approaches can be taken. a) We solve for x, so we ﬁrst rewrite the equation as x = 4 − y − z. If we let y = λ and z = µ, then x = 4 − λ − µ. In vector form this becomes: (x, y, z) = (4 − λ − µ, λ, µ) = (4, 0, 0) + λ(−1, 1, 0) + µ(−1, 0, 1). Intuitively: the plane ‘rests’ on (4, 0, 0) and an arbitrary vector in the plane is de- scribed by adding any combination of the vectors (−1, 1, 0) and (−1, 0, 1) to (4, 0, 0). b) Alternatively, by looking carefully at the equation, pick any vector in U , say (1, 1, 2). We use this as the support vector. Now take any two vectors which are perpendicular to (1, 1, 1) and which are not multiples of one another (‘independent vectors’), i.e., pick two ‘independent’ solutions of u + v + w = 0, e.g., (2, 3, −5) and (1, 1, −2). Then the plane is described as (x, y, z) = (1, 1, 2) + λ(2, 3, −5) + µ(1, 1, −2). Note that this description is quite diﬀerent from the previous description. This illustrates the fact that vector parametric descriptions of planes are far from being unique. Incidentally, in picking two vectors perpendicular to (1, 1, 1), usually two vectors are chosen which look relatively simple, like (1, −1, 0) and (0, 1, −1). Our choice of vectors (2, 3, −5) and (1, 1, −2) is correct, but may be more cumbersome in computations. The advantage of the method in a) is that it produces relatively simple vectors.
12. 12. 32 3–Space: lines and planes c) If you follow the method as explained in a) but solve for y rather than for x, then the resulting parametric equation is again somewhat diﬀerent: start by rewriting the equation as y = 4 − x − z, then set x = λ and z = µ. Finally, you get (x, y, z) = (0, 4, 0) + λ(1, −1, 0) + µ(0, −1, 1).2.2.7 Transforming vector parametric descriptions into equations We have come across two ways of representing planes: by equations and by parametric descriptions. To go from the ﬁrst to the second comes down to solving an equation repre- senting a plane and rewriting the solutions in the appropriate vector form. This has been discussed above. Here, we discuss how to transform vector parametric descriptions into equations. This is best illustrated by an example. Suppose the plane U is given by (2, 1, 3) + λ(1, 2, 1) + µ(1, 1, 3). Our task is to ﬁnd an equation of the form ax + by + cz = d representing U , i.e., we have to ﬁnd a, b, c and d. This problem splits in two parts: • To ﬁnd a, b and c note that the vector (a, b, c) is perpendicular to the plane, so is perpendicular to both direction vectors (1, 2, 1) and (1, 1, 3). This means: (a, b, c) • (1, 2, 1) = 0, and (a, b, c) • (1, 1, 3) = 0. So we have to solve: a + 2b + c = 0 a + b + 3c = 0 for a, b and c. To solve this system of two equations, ﬁrst eliminate a from the second equation by subtracting the ﬁrst equation from the second: a + 2b + c = 0 − b + 2c = 0. To make the ﬁrst equation simpler, add twice the new second equation to the ﬁrst: a + 5c = 0 − b + 2c = 0. So a = −5c and b = 2c. So one vector perpendicular to (1, 2, 1) and (1, 1, 3) is, for example, the vector (a, b, c) = (−5, 2, 1) obtained by taking c = 1. Thus, our equation looks like −5x + 2y + z = d and it remains to ﬁnd d. • To ﬁnd d in −5x + 2y + z = d is simple: just substitute any vector of U , for instance the support vector (2, 1, 3): (−5) · 2 + 2 · 1 + 1 · 3 = d. In conclusion, an equation for U is −5x + 2y + z = −5.
13. 13. 2.3 The relative position of lines and planes: intersections 33 2.3 The relative position of lines and planes: inter- sections2.3.1 Intersecting a line and a plane The intersection of a line and a plane usually consists of exactly one point. If the line and the plane happen to be parallel, the intersection may be empty or consist of the whole line. How does this work out in actual computations? Here is an example. Suppose the plane U has equation 2x1 + 3x2 − 5x3 = 3, and the line ℓ has the parametric description (−1, 4, −3) + λ(1, −1, 2). Finding the intersection comes down to ﬁnding the value(s) of λ for which the corresponding point on ℓ belongs to U as well, i.e., satisﬁes the equation for U . So we substitute (−1 + λ, 4 − λ, −3 + 2λ) in the equation: 2(−1 + λ) + 3(4 − λ) − 5(−3 + 2λ) = 3. This reduces to 25 − 11λ = 3, so that λ = 2. Now substitute this value of λ in the parametric equation of ℓ to ﬁnd the point of intersection: (1, 2, 1).2.3.2 Intersecting two planes Geometrically it is clear that in general two planes meet along a line. This is how you ﬁnd that line of intersection explicitly. Suppose U and V are two planes, with equations x1 + 2x2 − x3 = 4 and 2x1 + x2 − 5x3 = 2, respectively. We have to solve both equations simultaneously, i.e., ﬁnd all (x1 , x2 , x3 ) which satisfy both equations. To manipulate the equations eﬃciently, we write them as follows: x1 + 2x2 − x3 = 4 2x1 + x2 − 5x3 = 2. Subtract the ﬁrst equation 2 times from the second (and leave the ﬁrst equation as it is): x1 + 2x2 − x3 = 4 −3x2 − 3x3 = −6. Divide the resulting second equation by −3 to obtain: x1 + 2x2 − x3 = 4 x2 + x3 = 2. Now get rid of x2 in the ﬁrst equation by subtracting the new second equation from the ﬁrst: x1 + − 3x3 = 0 x2 + x3 = 2. In this stage, both x1 and x2 can be expressed in terms of x3 : to see this more clearly, rewrite as follows: x1 = 3x3 x2 = −x3 + 2.
14. 14. 34 3–Space: lines and planes Assign an arbitrary value λ to x3 , then we get (x1 , x2 , x3 ) = (3λ, −λ + 2, λ) = (0, 2, 0) + λ(3, −1, 1), the parametric description of a line with direction vector (3, −1, 1). 2.4 The relative position of lines and planes: angles2.4.1 The angle between two lines It is fairly straightforward to deﬁne the angle between two lines ℓ and m, although one detail has to be taken care of. a) Take a vector a in the direction of ℓ and a vector b in the direction of m; Figure 2.8: To ﬁnd the angle between two lines, choose vectors in the direction of the lines. b) Then compute the angle between a and b using the inner product; if the angle between a and b is obtuse, then replace a by −a and compute the angle between −a and b. In other words, make sure you end up with an acute angle (between 0◦ and 90◦ ). To summarize: If a and b are vectors in the directions of the lines ℓ and m, resp., then the angle α between ℓ and m is computed from |a • b| cos α = . |a| · |b| By using the absolute value in the numerator, we ensure that the fraction is non–negative. Consequently, we get an acute angle. √2.4.2 Example. The line ℓ is given by (2, 7, −1)+λ(1, 0, 1) and the line m is given by (2, π, −1)+ µ(1, 1, 2). To compute the angle α between ℓ and m we take a = (1, 0, 1) and b = (1, 1, 2) (the other vectors occurring in the parametric descriptions are irrelevant), and solve α from √ √ √ |(1, 0, 1) • (1, 1, 2)| 3 3· 3 3 cos α = =√ √ =√ √ √ = . |(1, 0, 1)| · |(1, 1, 2)| 2· 6 2· 2· 3 2
15. 15. 2.4 The relative position of lines and planes: angles 35 Therefore, the angle is 60◦ . Note that in this example, the absolute value signs in |(1, 0, 1)• (1, 1, 2)| were not necessary. This would have been diﬀerent if, for example, the line ℓ were given by (2, 7, −1) + λ(−1, 0, −1).2.4.3 The angle between a line and a plane You quickly realize that it is not so clear what the angle between a line ℓ and a plane U should be precisely. The reason is that the plane contains so many directions. So, which direction of the plane should we compare with the direction of the line? Suppose ℓ and U meet in a point P . The plane U contains a whole family of lines through P . Any such line makes an angle with ℓ, and the angle with such a line in the plane varies as this line varies in the family. (There is one case where the angle does not change, do you see which case this is?) m U U β l l P P Figure 2.9: To ﬁnd the angle between a line and a plane, it is convenient to introduce a line perpendicular to the plane as in the right–hand picture. If β is the angle between these two lines, then the angle between the line and the plane is deﬁned to be 90◦ − β. But there is a way out if you realize that the direction of a plane is also characterized by the direction of a line perpendicular to the plane. So, let us take a line m through P which is perpendicular to the plane U . Of course, we know what the angle between ℓ and m is. Now this angle is of course not really what we want, but 90◦ minus this angle turns out to be the appropriate angle. Summarizing: to compute the angle α between the line ℓ with direction a and the plane U with vector n perpendicular to U , solve α from |a • n| sin α = , |a| · |n| and make sure to take α in the range between 0◦ and 90◦ . (Note the occurrence of sin: with cos we would ﬁnd the angle between ℓ and the line perpendicular to U , but we need 90◦ minus this angle.) Here, we are using cos(90◦ − α) = sin α.2.4.4 Example. To compute the angle between the line ℓ with parametric equation (2, 0, 3) + √ λ(0, 1, 1) and the plane U with equation x1 +x2 +2x3 = 13, we need the vector a = (0, 1, 1)
16. 16. 36 3–Space: lines and planes from the parametric description and we need a vector perpendicular to the plane U , for example, n = (1, 1, 2) (taken from the coeﬃcients in the plane’s equation). Now we can proceed in two ways: (a) Solve for α in √ |a • n| 3 3 3 sin α = =√ √ =√ √ √ = . |a| · |n| 2 6 2 2 3 2 Therefore the angle is 60◦ . (b) Or we compute the angle β between a and n from √ |a • n| 3 cos β = = , |a| · |n| 2 so that β = 30◦ . The angle between ℓ and U is then 90◦ − 30◦ = 60◦ .2.4.5 The angle between two planes Suppose U and V are two planes which meet along the line ℓ. Of course they seem to make a deﬁnite angle with each other, but it takes some thinking to realize that an appropriate way to make precise what this angle is, is to look at the angle between two vectors perpendicular to the two planes (cf. the case of a line and a plane). The only detail you have to be careful about is that you may have to replace one of the vectors by its opposite vector in order to guarantee that the angle is acute (at most 90◦ ), or introduce an absolute value just like we did in the case of a line and plane. To summarize: To compute the angle between the planes U and V , take a nonzero vector u perpendicular to U and a nonzero vector v perpendicular to V . The angle α between U and V is then computed from |u • v| cos α = . |u| · |v| Figure 2.10: To ﬁnd the angle between two planes, determine the angle between two vectors perpendicular to the two planes, respectively. In particular, two planes are perpendicular if the two vectors perpendicular to the planes are perpendicular.
17. 17. 2.5 The relative position of points, lines and planes: distances 372.4.6 Example. Consider the planes U and V with equations x1 −x3 = 7 and x1 −x2 −2x3 = 25, respectively. To compute the angle between the two planes, consider the vector (1, 0, −1) which is perpendicular to U and the vector (1, −1, −2) which is perpendicular to V . Both vectors are taken from the coeﬃcients of the two equations. Now solve α from √ |(1, 0, −1) • (1, −1, −2)| 3 3 cos α = =√ √ = . |(1, 0, −1)| · |(1, −1, −2)| 2· 6 2 Therefore the angle is 30◦ .2.4.7 The angle between two planes: alternative approach Here is another approach to compute the angle between two intersecting planes U and V . Fix a point P on the line of intersection ℓ of U and V . Then take a plane W through P which is perpendicular to ℓ. Now W meets U along a line m, and W meets V along a line n. Finally, compute the angle between the lines m and n. W α P α m n Figure 2.11: An alternative approach to computing the angle between two intersecting planes U and V . The picture shows the cross section with the plane W perpendicular to the line of intersection ℓ of U and V . The intersection of U and W is denoted by m; the intersection of V and W is denoted by n. To see why this approach leads to the same answer, just realize that normal vectors to U and V with their tails at P lie in W . This is illustrated in Fig. (2.11). This approach usually leads to more complicated computations. 2.5 The relative position of points, lines and planes: distances2.5.1 Computing the distance between two points is straightforward, but the computation of distances between points and lines, points and planes, etc., is more subtle. In this section we are not aiming for explicit formulas, but for strategies to compute distances.2.5.2 The distance between a point and a line Suppose P is a point and ℓ is a line. The distance between P and a point on ℓ varies as this
18. 18. 38 3–Space: lines and planes point varies through ℓ. So the question is: for which point on ℓ is this distance minimal? A bit of experimentation with the Theorem of Pythagoras (see Fig. (2.12)) shows that the point Q such that P Q is perpendicular to ℓ is the point we are looking for. P l Q R Figure 2.12: The shortest distance between a point P and a point on the line ℓ occurs for the point Q where P Q is perpendicular to the line ℓ. For instance, applying Pythagoras’ theorem to the triangle P QR, shows that |P Q| < |P R|. Let us work this out in a speciﬁc example, where P is the point (7, 2, −3) and ℓ is the line with parametric description (2, −1, 1) + λ(1, −1, 2). • First, an arbitrary point Q on ℓ is described by (2 + λ, −1 − λ, 1 + 2λ). The vector QP is then (−5 + λ, −3 − λ, 4 + 2λ). • The next step is to solve λ from the condition that P Q is perpendicular to the direction vector (1, −1, 2) of ℓ, i.e., solve (−5 + λ, −3 − λ, 4 + 2λ) • (1, −1, 2) = 0. This comes down to (−5+λ)−(−3−λ)+2(4+2λ) = 0 or 6+6λ = 0. So λ = −1 and the point Q on ℓ closest to P is therefore (substitute for example in the parametric equation of ℓ) (2, −1, 1) − (1, −1, 2) = (1, 0, −1). • Finally, the distance between P and the line ℓ is calculated as the distance between P and Q: √ √ (7 − 1)2 + (2 − 0)2 + (−3 − −1)2 = 44 = 2 11.2.5.3 The distance between a point and a plane The distance between a point P and a point Q in the plane U varies as Q varies. By the distance between a point P and a plane U we mean the shortest possible distance between P and any of the points of U . But how do we ﬁnd such a point in the plane? It is geometrically obvious that we can locate such a point by moving from P in the direction perpendicular to U until we meet U , i.e., we need the line through P perpendicular to the plane. Just as in the previous case (2.5.2), this is based on Pythagoras’ theorem. Let us use this strategy when P is the point (5, −4, 6) and U is the plane given by the equation x − 2y + 2z = 7.
19. 19. 2.5 The relative position of points, lines and planes: distances 39 P locate Q distance U Q Figure 2.13: To ﬁnd the distance between P and the plane U , ﬁrst determine the intersection of U and the line through P perpendicular to U . Then calculate the distance between P and the point of intersection. • First we look for the line through P and perpendicular to U . A vector perpendicular to U is easily extracted from its equation: (1, −2, 2). So the line through P with direction vector (1, −2, 2) is described by (5, −4, 6) + λ(1, −2, 2). • The next step is to intersect this line with U . We substitute (5, −4, 6) + λ(1, −2, 2) in the equation: (5 + λ) − 2(−4 − 2λ) + 2(6 + 2λ) = 7. This is easily rewritten as 25 + 9λ = 7, so that λ = −2. For λ = −2, we ﬁnd the point Q = (5, −4, 6) − 2(1, −2, 2) = (3, 0, 2). So Q = (3, 0, 2) is the point in U closest to P . • Finally, the distance between P and U is calculated as the distance between P and √ Q: (5 − 3)2 + (−4 − 0)2 + (6 − 2)2 = 36 = 6. There is at least one but: what if the plane is given by a vector parametric equation? Then there are a couple of strategies available: • One strategy is to ﬁnd an equation for the plane and then proceed as before. • An alternative is to ﬁnd a point Q in the plane such that the direction of P Q is perpendicular to every direction in U .2.5.4 The distance between two non–intersecting lines If two distinct lines in 3–space do not intersect (in one point), then it makes sense to con- sider their distance. In case the lines are parallel, i.e., their direction vectors are multiples of each other, you take a point on one line and compute the distance to the other line, just like before. In case the direction vectors are not multiples of one another, the computation of the distance turns out to be more complicated. Suppose ℓ and m are two such lines. Among all points P on ℓ and all points Q on m we need to ﬁnd points for which the distance |P Q| is
20. 20. 40 3–Space: lines and planesminimal. As before, this condition turns out to be related to right angles: we have to ﬁndP on ℓ and Q on m in such a way that P Q is perpendicular to both ℓ and m. The procedureis best illustrated with an example. Suppose ℓ is given by (1, 3, 3)+λ(2, 0, 1) and m is given l mFigure 2.14: To ﬁnd the distance between two non–intersecting lines ℓ and m, ﬁnd vectorsp on ℓ and q on m such that q − p is perpendicular to both ℓ and m. The distance is thelength |q − p|of q − p.by (1, 6, −3) + µ(0, 1, 1). The lines are certainly not parallel since the direction vectors arenot multiples of one another. They may intersect, but then our distance computation willsimply yield 0. So begin by taking a vector p = (1, 3, 3) + λ(2, 0, 1) = (1 + 2λ, 3, 3 + λ) onℓ and taking q = (1, 6, −3) + µ(0, 1, 1) = (1, 6 + µ, −3 + µ) on m. Here are the steps tofollow: • First, we impose the condition that q − p be perpendicular to (2, 0, 1) and (0, 1, 1) (the direction vectors of ℓ and m, respectively), i.e., (q − p) • (2, 0, 1) = 0 and (q − p) • (0, 1, 1) = 0. Since q − p = (−2λ, 3 + µ, −6 + µ − λ), these conditions translate into two equations in the variables λ and µ: −4λ + (−6 + µ − λ) = 0 (3 + µ) + (−6 + µ − λ) = 0. Rearranging leads to −5λ + µ = 6 −λ + 2µ = 3. An easy calculation shows that this system has exactly one solution: λ = −1 and µ = 1. • Second, corresponding to this solution we ﬁnd the vector p = (−1, 3, 2) and q = (1, 7, −2). The distance between the lines is therefore √ (1 − −1)2 + (7 − 3)2 + (−2 − 2)2 = 36 = 6.
21. 21. 2.6 Geometric operations: translating lines and planes 41 2.6 Geometric operations: translating lines and planes2.6.1 In designing shapes, you may want to move or rotate certain elements of your design, say a window or a wall. The question is how you do this mathematically. Such mathematical computations are at the basis of computer software implementations. In this section, we will brieﬂy discuss a few aspects of translations.2.6.2 Translating lines A fairly simple geometric operation is translation: we move an object in a given direction over a given distance. We can represent this direction and distance by a vector t. Any vector (x, y, z) is moved to another vector, given by (x, y, z) + t. If t = (2, 5, −1), then (x, y, z) is translated to (x + 2, y + 5, z − 1). No problem here. Now let us turn to the eﬀect of a translation on lines. If a line ℓ is given by p + λv, then a translation over t produces again a line, now with parametric equation (p+t)+λv. For example, if ℓ is given by (1, 0, 6)+λ(1, 1, 1), then after translation over t = (2, 5, −1) we obtain the line with support vector (1, 0, 6) + (2, 5, −1) = (3, 5, 5) and direction vector (1, 1, 1), i.e., the line (3, 5, 5) + λ(1, 1, 1). Of course, if you translate a line then the resulting line is parallel with the line you started with.2.6.3 Translating planes Next, we investigate what happens to a plane when we translate that plane over a vector, say t = (2, 5, −1) as before. Since we have two standard ways of representing a plane, our discussion splits into two parts. • In the case where the plane is described by a vector parametric description, the situation is similar to the case of a line. Suppose the plane U is described by (1, −1, 0) + λ(1, 1, 0) + µ(0, 4, 1), then the eﬀect of translating over t = (2, 5, −1) is: (2, 5, −1) + (1, −1, 0) + λ(1, 1, 0) + µ(0, 4, 1) or (3, 4, −1) + λ(1, 1, 0) + µ(0, 4, 1). • Suppose the plane U is given by its equation x − y + 4z = 2. The trouble now is that the description of the plane is implicit. If we translate a vector (x, y, z) of U then the result is (2, 5, −1) + (x, y, z), but then what? Instead, it is better to work backwards: we begin with a vector from the translated plane, say (u, v, w). If we translate this vector back, so subtract (2, 5, −1), then the result should be on U , i.e., satisfy the equation of U . So we substitute (u, v, w) − (2, 5, −1) = (u − 2, v − 5, w + 1) in the equation of U : (u − 2) − (v − 5) + 4(w + 1) = 2. This reduces to u − v + 4w = −5. So the plane U : x − y + 4z = 2 is translated to the plane V : x − y + 4z = −5. Of course, as the equations show, these planes are parallel. Note that the distance between the planes is not equal to the length of t, since t is not perpendicular to U .
22. 22. 42 3–Space: lines and planes If you realize that translating a plane produces a parallel plane, another strategy suggests itself: the translated plane must have an equation of the form x − y + 4z = d for some d. Now, d can be determined by translating a single point of U and substituting the result. For instance, (2, 0, 0) is in U . Translating the point gives (2, 0, 0) + (2, 5, −1) = (4, 5, −1). Substitute (4, 5, −1) in the equation x − y + 4z = d to ﬁnd that d = −5.2.6.4 Example. (Armada’s Paleiskwartier, Den Bosch) Close to the railway station in Den Bosch, a number of buildings with a special appearance have been built during the last few years. From a distance they look somewhat like a ﬂeet of ships from historic times, which is probably the reason for the district’s name: Armada’s Paleiskwartier. Now some of the buildings are identical in shape (or at least at ﬁrst glance), so they could be considered as translates of one given central building. Apart from details, we can consider this problem as a problem of translating planes (to be fair: the shape of the buildings is not ﬂat on all sides, so our discussion is limited to ﬂat parts only). Suppose one wall is given by (part of) the plane U : x − 3y = 0 (the z-direction is assumed to point upward) in some cartesian coordinate system. Suppose we want to translate this plane over (2, 3, 0), (3, 2, 0) and (5, 5, 0), resulting in the planes U1 , U2 and U3 . To ﬁnd the equation of the ﬁrst translate, we work backwards again: start with (u, v, w) in U1 , then (u, v, w) − (2, 3, 0) is in U , so (u − 2, v − 3, w) should satisfy (u − 2) − 3(v − 3) = 0. A bit of simpliﬁcation yields U1 : u − 3v = −7. The other translates are found in a similar way. The result is: U1 : u − 3v = −7, U2 : u − 3v = −3, U3 : u − 3v = −10. (In this case, a simpler strategy works faster, since you know that a) all equations should have the form x − 3y = ... and b) (2, 3, 0) is on U1 , etc.) 2.7 Geometric operations: rotating lines and planes2.7.1 Rotation around a coordinate axis If you rotate the point (x, y, z) around the z–axis over 90◦ , the result is either (−y, x, z) or (y, −x, z) depending on the direction of orientation (the two rotations are each other’s inverse: if you apply the two rotations after each other, the total eﬀect will be that nothing has happened). So rotating a single point isn’t that diﬃcult. But suppose we want to rotate a whole plane. Let us concentrate on the ﬁrst rotation where any point (x, y, z) is rotated to (−y, x, z). What would be the equation or parametric description of the new plane? Let us consider these two descriptions separately. • Let U be the plane given by the parametric description (0, 2, 1) + λ(2, −1, 0) + µ(1, 2, −5). So an arbitrary vector in U looks like (2λ + µ, 2 − λ + 2µ, 1 − 5µ). If we rotate this vector we obtain the vector (−2 + λ − 2µ, 2λ + µ, 1 − 5µ)
23. 23. 2.7 Geometric operations: rotating lines and planes 43 y (−y,x) 90o (x,y) x Figure 2.15: A rotation around the z–axis doesn’t aﬀect the z–coordinate; the ﬁgure con- centrates on what happens to the x and y coordinates. (apply (x, y, z) → (−y, x, z)). This provides us with the vector parametric equation of the rotated plane: (−2, 0, 1) + λ(1, 2, 0) + µ(−2, 1, −5). • Now suppose we want to rotate the plane U , but U is given through its equation x + 2y + z = 5. After rotation we get a new plane, which we call V . If we take a point (x, y, z) of V and rotate it back , then we should get a point of U . Now, rotating (x, y, z) back produces the point (y, −x, z) (since the two rotations are inverses). This point (y, −x, z) should satisfy the equation of U , so y + 2(−x) + z = 5. Rewriting, we get the equation −2x + y + z = 5. • A clever approach to rotating the plane U : x + 2y + z = 5 is to rotate its normal vector (1, 2, 1). This results in the vector (−2, 1, 1) and this should be the normal vector of the rotated plane V . So V has equation −2x + y + z = d for some d. Now, the point (0, 0, 5) belongs to U and remains ﬁxed when we apply the rotation. Substituting (0, 0, 5) in −2x + y + z = d gives d = 5. Therefore, the equation of V is −2x + y + z = 5. (There are more ways to argue that the right–hand side coeﬃcient 5 of the equation of U does not change for the rotation at hand.)2.7.2 Example. If we rotate the plane U with equation x + 2y + z = 5 around the z–axis in the ‘positive’ direction, we obtain the plane V with equation −2x + y + z = 5. Now suppose we rotate V again over 90◦ in the same direction to obtain a plane W . If (x, y, z) is a point of W , then rotating it backwards we ﬁnd the point (y, −x, z), a point that belongs to V , so should satisfy the equation of V . This leads to −2y + (−x) + z = 5, or −x − 2y + z = 5. This equation looks pretty much like the equation x + 2y + z = 5 of U . Can you explain the minus signs by directly considering a rotation over 180◦ ?
24. 24. 44 3–Space: lines and planes2.7.3 Fans of planes If two planes U and V are not parallel, they intersect along a line, say ℓ. By rotating one of the planes around ℓ, we generate a whole family of planes, the fan of planes on ℓ. Of course, all these planes contain ℓ. The question is: what is the equation of a member of such a fan, given the equations of U and V ? To answer this question, we take two speciﬁc planes, say U : x − y − z = −1 and V : 3x − y + 3z = 9. These planes turn out to intersect along ℓ: (1, 0, 2) + λ(2, 3, −1) (this follows from a computation similar to the one explained on p. 33). Now every point of ℓ satisﬁes both equations, but then every point of ℓ also satisﬁes a combination of the two equations, like ‘2 times the ﬁrst equation + 3 times the second equation’ , or 2(x − y − z) + 3(3x − y + 3z) = 2 · (−1) + 3 × 9, which simpliﬁes to 11x − 5y + 7z = 25. Likewise, any combination a(x − y − z) + b(3x − y + 3z) = −a + 9b (with a and b not both equal to 0) is an equation of a plane that contains ℓ. Sometimes, it is more convenient in this setting to bring all terms of the equation of a plane to the left–hand side: x − y − z + 1 = 0 and 3x − y + 3z − 9 = 0 describe the two planes and a(x − y − z + 1) + b(3x − y + 3z − 9) = 0 is, for ﬁxed a and b, the equation of a speciﬁc member of the fan.2.7.4 Example. The planes U : x + y = 1 and V : x + 2y = 3 generate a fan. In this example we compute a member of the fan which is perpendicular to U . The equation of a general member of the fan is a(x + y − 1) + b(x + 2y − 3) = 0, or (a + b)x + (a + 2b)y − a − 3b = 0. The plane with equation (a + b)x + (a + 2b)y − a − 3b = 0 is perpendicular to U if (a + b, a + 2b, 0) • (1, 1, 0) = 0. This reduces to 2a + 3b = 0. For a = 3 and b = −2 we obtain the plane with equation x − y + 3 = 0. (Other non–zero values of a and b which satisfy 2a + 3b = 0 lead to a multiple of this equation and therefore to the same plane.) 2.8 Geometric operations: reﬂecting lines and planes2.8.1 Mirror images Sometimes you do not even notice it at ﬁrst sight when two (types of) buildings have been constructed as mirror images of one another: they look so much the same. Sizes, distances and angles are preserved under reﬂections, but somehow the orientation has
25. 25. 2.8 Geometric operations: reﬂecting lines and planes 45 changed: a right–handed screw changes into a left–handed screw. One immediate property of a reﬂection is that taking the mirror image of the mirror image of a point or vector brings you back to the original point or vector. In this section we restrict our discussion to reﬂections where the mirror is a coordinate plane or another easy to describe plane. In particular, the mirrors we consider are ﬂat. To begin with, we consider the reﬂection in the y, z–plane. In this case, the mirror image z (−x,z) (x,z) x Figure 2.16: In picturing the reﬂection in the y, z–plane, we have left out the y–axis for the sake of simplicity and pictured the resulting 2–d version of reﬂection in the z–axis. The vector (x, z) is reﬂected into the vector (−x, z). The two chairs are mirror images. of any point (x, y, z) (or vector) is (−x, y, z) (so only the x–coordinate has changed). The image of a line, say ℓ : (2, 3, −4) + λ(1, 2, 3), is obtained by reﬂecting each point of the line individually: so the image of ℓ is ℓ′ : (−2, 3, −4) + λ(−1, 2, 3). To be precise, every vector (2+λ, 3+2λ, −4+3λ) of ℓ is mapped to (−2−λ, 3+2λ, −4+3λ) = (−2, 3, −4)+λ(−1, 2, 3). Similarly, the image of a plane given in parametric form is easily obtained. But if the plane is given by an equation, we have to work backwards again, just like we did for translations and rotations, or exploit a normal vector of the plane. Suppose the plane U has equation 2x + y − 3z = 7. If (u, v, w) is a vector on the mirror image U ′ of U , then reﬂecting the vector yields the vector (−u, v, w) in U , so (−u, v, w) satisﬁes the equation of U : 2(−u) + v − 3w = 7. Therefore, U ′ has equation −2u + v − 3w = 7. If the mirror is not a coordinate plane, like the y, z–plane, things usually become a lot harder to describe. In later chapters we will go into this in more detail. In the case at hand, the ‘normal vector approach’ runs as follows. Reﬂect the normal vector (2, 1, −3) to obtain (−2, 1, −3), so the equation of U ′ is of the form −2x + y − 3z = d for some d. To determine d, pick a point on U , say (2, 3, 0) and reﬂect it: (−2, 3, 0). Substitute in −2x + y − 3z = d and conclude that d = 7. The equation of U ′ is therefore −2x + y − 3z = 7.2.8.2 More reﬂections The case where the mirror is the plane x = y is still easy to handle. The mirror image of
26. 26. 46 3–Space: lines and planes the vector (x, y, z) is easily seen to be (y, x, z) (just change the ﬁrst two coordinates). From this description in coordinates it is also clear that reﬂecting twice in succession brings you back to where you started: reﬂect reﬂect (x, y, z) −→ (y, x, z) −→ (x, y, z).2.8.3 Example. Two towers of a building are designed as mirror images. Both towers have triangular horizontal cross sections. In a suitable coordinate system, the walls of one of these towers are (parts of) the three planes U : x = 3, V : y − x = −1 and W : y = −1. The mirror is the plane x = y. To ﬁnd the mirror image of the plane V : y − x = −1, start with a vector (u, v, w) on the mirror image V ′ of the plane. The mirror image of the vector (u, v, w) is (v, u, w) and is on the plane V : y − x = −1, so u − v = −1. Therefore the equation of the mirror image V ′ of the plane V : y − x = −1 is x − y = −1 (so the roles of x and y have been interchanged). Here is an overview of the equations of the planes and their mirror images. planes mirror images x=3 y=3 y − x = −1 x − y = −1 y = −1 x = −12.8.4 More complicated reﬂections A straightforward observation on reﬂections in a plane U is the following. If you connect a point P and its mirror image P ′ , then this segment is perpendicular to the plane U . Moreover, the distance from P to U is equal to the distance from P ′ to U . This translates into the following strategy for computing mirror images: a) Start at P and move along the line through P which is perpendicular to U , so ﬁrst set up the parametric equation of this line, say p + λv. b) Compute the intersection of the line with the plane U . For a speciﬁc value of the parameter, say λ0 , the line intersects the plane U . So p + λ0 v is ‘halfway’ between P and its mirror image. c) But then p + 2λ0 v is the mirror image (note the factor 2). By way of example let us compute the mirror image P ′ of the point P = (2, 3, 4) in the plane x+y +z = 3. The idea is to start moving from P in the direction (1, 1, 1) (perpendicular to U ). Then locate on this line the second point which has the same distance to U . Let ℓ be the line (2, 3, 4) + λ(1, 1, 1). First ﬁnd out where this line meets the plane U , so substitute in x + y + z = 3: (2 + λ) + (3 + λ) + (4 + λ) = 3. This equation reduces easily to 9 + 3λ = 3 so that λ = −2. So if you add −2 · (1, 1, 1) to (2, 3, 4) you end up in the plane. Therefore, if you go twice as far you will get the mirror
27. 27. 2.9 Tesselations of planes 47 image. In conclusion, the mirror image of P is obtained by adding −4 · (1, 1, 1) to (2, 3, 4), i.e., P ′ has coordinates (2, 3, 4) − 4 · (1, 1, 1) = (−2, −1, 0). To check the answer: see if ((2, 3, 4) + (−2, −1, 0))/2 is on the plane and check if the diﬀerence (−2, −1, 0) − (2, 3, 4) = (−4, −4, −4) is perpendicular to U . 2.9 Tesselations of planes2.9.1 Sometimes the outside of a ﬂat wall is given a special pattern. For instance, the plane is covered with rectangles, triangles or other shapes with a certain degree of regularity or symmetry. Over the years quite a bit of attention has gone into understanding the kinds Figure 2.17: Two examples of regular subdivisions of a plane: a rectangular pattern on the left and a triangular pattern on the right. In the case of a full plane, the patterns are supposed to extend inﬁnitely. of symmetry that can occur in such subdivisions of a plane or in producing decorations of a plane. This section only touches upon the subject. Deeper investigations require more advanced mathematical tools. See also [2] for more extensive discussions. The patterns shown in Figs (2.17), (2.18), (2.19) and (2.20) admit various kinds of symmetry. Here is a brief overview of the kinds of symmetry that are usually distinguished. a) Translational symmetry Translating the pattern over the vectors a, b or an inte- gral combination m · a + n · b (with m and n integers) transforms the pattern into itself. For instance, if you translate over 2a, the whole pattern is moved two steps to b a Figure 2.18: Translational symmetry.
28. 28. 48 3–Space: lines and planes the right. The translated pattern coincides precisely with the original pattern. But a 1 translation over, say, · a creates a pattern which does not coincide with the original 2 pattern. b) Rotational symmetry. The patterns have various centers of symmetry. A rotation over a speciﬁc angle transforms the pattern into itself. r r center Figure 2.19: Rotational symmetry. All black dots are centers of symmetry, at least if you imagine the pattern to extend indeﬁnitely. On the right the eﬀect of a rotation over 180◦ on one of the rectangles is illustrated. c) Mirror symmetry or reﬂectional symmetry. Patterns may be transformed into themselves by reﬂecting them in suitable lines. In the picture a few of such lines are indicated. lines of symmetry Figure 2.20: Mirror symmetry or reﬂectional symmetry. If you think of the pattern as extending indeﬁnitely, reﬂecting in any of the indicated lines transforms the picture into itself.2.9.2 Strong restrictions on rotational symmetry A pattern as discussed above may have more than one form of symmetry, for instance, both translational and rotational symmetry. As regards rotational symmetry, scientists have been able to prove that the only rotations that can occur are rotations over 180◦ , 120◦ , 90◦ and 60◦ . In essence this follows from the following considerations (no details): if the pattern has translational symmetry over the vectors a and b and rotational symmetry
29. 29. 2.9 Tesselations of planes 49over the angle α, then ﬁrst compute the vectors Ra and Rb obtained by rotating a andb over the angle α; this will involve expressions containing cos α and sin α. The next stepis to realize that these two rotated vectors should themselves be expressible as an integralcombination of a and b, i.e., there should be integers m1 , m2 , n1 and n2 such that Ra = m1 a + m2 b Rb = n1 a + n2 b.Now the occurrence of the integers m1 , m2 , n1 and n2 forces the numbers cos α and sin αto be relatively simple. In turn, that means that there are only a few possibilities for α.
30. 30. 50 3–Space: lines and planes 2.10 Exercises1 Pictures of operations on vectors a) Make a drawing of vector subtraction: draw two vectors x and y and indicate how x − y arises. b) Sketch the addition of 2x and 3y, given the vectors x and y.2 Vector arithmetic In this exercise you’ll be practicing elementary computational skills regarding vectors. a) Compute (2, 3, 5) − 2 · (1, 1, 1) + 4 · (1, −1, 3). b) Find the length of the vector (2, −2, 1). c) Find the distance between the vectors (1, 1, 1) and (1, 4, −4). d) Determine the angle between the vectors (1, 1, 2) and (1, 1, −1). e) Find a so that the vector (1, 2, a) is perpendicular to the vector (3, −1, −1). f) Find a so that the angle between the vectors (1, 2, 2) and (1, a, 0) is 45◦ .3 Find examples yourself a) Give three examples of vectors of length 1 with at least two non–zero coordinates. b) Give an example of two non–zero vectors whose inner product is 0. c) Find three vectors of length 13. How many vectors of length −13 are there? d) Give an example of two vectors whose inner product is negative.4 On the dot product or inner product a) Determine (1, 3, −1) • (2, −1, 1). 1 b) Evaluate (x, y, z) • (2, 1, 3). Also evaluate ( √ (2x + y + 3z)(2, 1, 3)) • (2, 1, 3). 14 c) Verify by writing out the appropriate expressions: if x•a = y•a, then (x−y)•a = 0. In that case x − y is perpendicular to a. Also verify that (λx) • y = λ(x • y) (here, λ is a real number). d) Suppose a is a vector of length 1, so |a| = 1 or, equivalently, a • a = 1. Show that for every vector x the vectors x and (x • a) · a have the same inner product with a. Use part c) to conclude that x − (x • a) · a and a are perpendicular. Show in a picture how this relates to the orthogonal projection of x on the line λ · a.
31. 31. 2.10 Exercises 515 Vector parametric description of lines In describing lines there is some ﬂexibility. This exercise focuses on this ﬂexibility. a) Explain why (1, 2, 1) + λ(2, 2, −1) and (1, 2, 1) + µ(−2, −2, 1) describe the same line. Now suppose you use the ﬁrst description and a fellow student uses the second one. If you want to talk about the same vector on the line, how would ‘your’ λ be related to ‘his/her’ µ? b) Find out if (3, 4, 0) is on the line (1, 2, 1) + λ(2, 2, −1). Then decide if (3, 4, 0) + λ(2, 2, −1) and (1, 2, 1) + µ(−2, −2, 1) describe the same line. c) Produce two more vector parametric descriptions of the line (1, 2, 1) + λ(2, 2, −1) (with diﬀerent support and direction vectors) and show why your answer is correct.6 Sharpening your geometric intuition Given a point P and two distinct lines ℓ and m. a) How many planes are there through P which are perpendicular to ℓ? b) How many lines are there through P and perpendicular to ℓ? c) How many lines are there (not necessarily through P ), which are perpendicular to both ℓ and m?7 Distances: a link with calculus Suppose ℓ is a line which does not pass through the origin. Then the distance of O to ℓ can also be found as the answer to a minimization problem: ﬁnd the vector on ℓ which is closest to O. a) If ℓ is the line (−1, −3, 5) + λ(2, 4, −2), then describe the distance of 0 = (0, 0, 0) to a vector on ℓ as a function of λ. b) Can you use the square of this function just as well for our problem? Find the minimum of the function you have chosen to work with. Which vector on ℓ lies at this distance from the origin? c) Let’s call the vector you have found p. Compute the dot product p • (2, 4, −2) and explain your answer.8 Points at equal distance to two given points Given the point P = (1, 2, 3) in 3–space. a) Determine the distance between (x, y, z) and P and the distance between (x, y, z) and the origin O. b) Starting from the two expressions in a) determine the equation of the points in 3– space with equal distance to O and P . Simplify your answer as far as possible.
32. 32. 52 3–Space: lines and planes c) Geometrically, it is clear that the answer to b) should be a plane. Use your geometric intuition to ﬁnd directly, without using a) or b), a vector perpendicular to the plane and a point in the plane. Construct the equation from these data. 9 Pitfalls when rewriting equations The equation 2x1 − 3x2 + 5x3 = 10 can be rewritten as, e.g., 3x2 = 2x1 + 5x3 − 10. a) What is wrong with rewriting the equation as (3x2 )2 = (2x1 + 5x3 − 10)2 ? b) Explain why this new equation actually represents two planes. [Hint: bring both terms to the left–hand side, don’t expand, and use a2 − b2 = (a + b)(a − b) in a suitable way.]10 Describing a plane through three given points It is quite obvious that through three distinct points which are not on a line precisely one plane passes. This exercise is about ﬁnding the plane’s parametric description and its equation. Here are the three points: P = (1, 2, −1), Q = (2, 0, 3) and R = (2, 1, 0). a) Set up the parametric equation for the line ℓ through P and Q and show that R is not on ℓ. b) Use P Q and P R as direction vectors and set up a parametric description of the plane U through P , Q and R. c) Starting from the result in b) ﬁnd an equation for U .11 Describing planes a) Find an equation of the plane U which is perpendicular to the vector (1, −1, 1) and passes through the point (2, −1, 1). b) Find a parametric equation of the plane U . c) Find an equation of the plane V which is parallel to U but passes through the origin (0, 0, 0). Also provide a parametric equation of this plane. d) Find an equation of the plane V which is parallel to U but passes through the point (3, 3, 3). Parametric equation?12 Transforming a parametric description of a plane into an equation The plane U has parametric description (1, 1, −1) + λ(1, 2, 0) + µ(0, 2, 3). a) Determine a vector perpendicular to the plane. b) Set up an equation for the plane.
33. 33. 2.10 Exercises 5313 Computing intersections a) Find the intersection of the line (0, 2, 3) + λ(1, −1, 1) and the plane with equation x + y + z = 7. b) Find the intersection of the planes 2x + 3y − z = 4 and 4x + 5y + 3z = 9. c) Find the intersection of the planes 2x + 3y − z = 4 and (1, 1, 0) + λ(5, −4, 0) + µ(0, 3, −5).14 Computing angles a) Find the angle between the lines (3, −2, 5) + λ(−1, −1, 0) and (2, 2, −1) + µ(1, 2, 1). b) Determine the angle between the plane 2x + 3y − z = 5 and the line (1, 0, 4) + λ(2, 3, −1). c) Determine the angle between the plane x + y = 14 and the line (0, 2, −3) + λ(0, 1, 1). d) Determine the (approximate) angle between the planes x+y+z = 3 and x+2y−z = 2.15 The distance between a point and a line Compute the distance between the point (1, 1, 1) and the line (7, −2, 4) + λ(2, −1, −1).16 On the distance between a point and a line The following method is proposed to compute the distance between a point P and a line ℓ: Take the plane U through P which is perpendicular to ℓ. Intersect U with ℓ; this produces a point Q. Finally, compute the distance between P and Q. a) Explain why P Q is perpendicular to ℓ. Conclude that the suggested method is valid. b) Use this approach to compute the distance between P = (3, 1, 0) and the line with vector parametric equation (3, 2, 4) + λ(2, 1, 2). So ﬁrst derive the equation of the plane through P which is perpendicular to ℓ.17 The distance between two lines Compute the distance between the lines λ(1, −1, 1) and (0, −4, −2) + µ(2, 3, 2).18 On the distance between two lines To compute the distance between the non–intersecting lines ℓ and m with diﬀerent di- rections, the following method is proposed: First take any plane perpendicular to ℓ and intersect with ℓ and m. Then compute the distance between the resulting points of inter- section. a) Explain why this approach is wrong in general.
34. 34. 54 3–Space: lines and planes b) Comment on the following approach: Take a plane containing ℓ and rotate it around ℓ until it is perpendicular to m; in this last case the plane meets m in a point P . Now compute the distance between P and ℓ.19 The distance between a point and a plane Compute the distance between (2, 1, 2) and the plane x − y + 3z = −4.20 The distance between two parallel planes To compute the distance between two parallel planes, a vector is needed which is per- pendicular to both planes and which has its tail in one plane and its head in the other. Suppose U has equation x + 2y + 3z = 5 and V is given by x + 2y + 3z = 19. a) Find a vector v which is perpendicular to both planes. Find a vector p in one of the planes. b) Use the line p + λv to ﬁnd a vector in the other plane. c) Use the previous results to compute the distance between the planes. d) Computing the distance between two parallel planes is related to computing the distance between a point and a plane. Explain this relation.21 The relative position of a plane and a sphere In Chapter 1, Ex. 1.1.6, the distance between the origin and the plane with equation √ x + y + z = a was given as |a|/ 3. a) Compute this distance by ﬁrst intersecting the line (x, y, z) = λ(1, 1, 1) with the plane x + y + z = a, and then computing the distance between the origin and the point of intersection. b) For which values of a will the plane x + y + z = a meet the sphere x2 + y 2 + z 2 = 1? c) For which value(s) of a will the intersection of the plane and the sphere be a circle which touches the x, y–plane, but is otherwise contained in the upperhalf space z ≥ 0?22 Translating, rotating and reﬂecting planes Let U be the plane with equation 2x − 3y + 5z = 0. a) Translate U over the vector (1, −2, 1). What is the equation of the new plane? b) Rotate the plane U over 90◦ according to (x, y, z) → (−y, x, z). Find the equation of the image of U . c) Reﬂect U in the x, y–plane. What is the equation of the resulting surface?23 Two fans of planes In designing the roof of a house, two fans of planes are used. One fan consists of the planes containing the y–axis, the other fan consists of the planes containing the line x = 5, z = 0.
35. 35. 2.10 Exercises 55 a) Use two planes from each fan to describe the two fans using parameters. b) Find a plane U1 from the ﬁrst fan that makes an angle of 45◦ with the x, y–plane. Then ﬁnd a plane U2 from the second fan that make an angle of 90◦ with U1 . c) Find a plane V1 from the ﬁrst fan that makes an angle of 30◦ with the x, y–plane. Then ﬁnd a plane V2 from the second fan that makes an angle of 90◦ with V1 .24 Constructing a house shaped like a cube A building is shaped like a cube, say with bounding planes x = 0, x = 1, y = 0, y = 1, z = 0 and z = 1. Now this construction is rotated over 45◦ around the y–axis counterclockwise. First we try to describe this rotation explicitly. 1 a) Explain why e1 = (1, 0, 0) is transformed into √ (1, 0, 1) and e3 = (0, 0, 1) is trans- 2 1 formed into √ (−1, 0, 1). What happens to e2 = (0, 1, 0)? 2 b) Take any vector (x, y, z) and rewrite it as xe1 + ye2 + ze3 . Now rotate (x, y, z) by 1 rotating its three components xe1 , ye2 and ze3 . Show that this leads to ( √ (x − 2 1 z), y, √ (x + z)). 2 c) Use the transformation in b), i.e., 1 1 (x, y, z) → ( √ (x − z), y, √ (x + z)), 2 2 to ﬁnd the bounding planes of the rotated cube. (Examples of such houses, the so–calles ‘kubushuizen’ with similar rotated positions have been built in Helmond.)25 Inside a building a room is constructed with perpendicular sloping walls. These walls should contain the line ℓ with parametric description λ(1, 1, 1). a) Find the equations of two walls which contain ℓ (there are inﬁnitely many possibilities for such walls, just two are needed). b) What is the equation of the member of the fan which makes an angle of 45◦ with the x, y–plane?
36. 36. 56 3–Space: lines and planes