Chapter 2 3–Space: lines and planessterk/Bouwkunde/2db60-chap2.pdf24 3–Space: lines and planes...

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 vectors

2.1.1 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 usedto describe points in 3-space by triples of numbers: the (signed) distances to planes formedby two of these axes. We call such a set of perpendicular axes through a given point acartesian coordinate system. Each of the axes is called a coordinate axis and the point ofintersection 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 axesare 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 screwpositioned along the z–axis following this movement would move in the direction of thepositive z–axis. Every two of the three axes span a plane: the y, z–plane, the x, z–planeand the x, y–plane, respectively. The signed distances u, v and w of a point P in 3–spaceto each of these planes in the given order are the coordinates of P ; we usually put thethree together as follows: (u, v, w) and still call this the coordinates of P . Notations forpoints that occur often: (x, y, z), (x1, x2, x3), (a, b, c), (a1, a2, a3), (u, v, w), etc. For a pointP with coordinates (x, y, z) we also write P = (x, y, z).

21

22 3–Space: lines and planes

x

z

y4

P=(2,3,4)

x

y

z

Figure 2.1: Left: The third coordinate of P = (2, 3, 4) is the signed distance of P tothe x, y–plane; so if the point were below the x, y–plane, the third coordinate would havebeen negative. Right: Cartesian coordinate systems are taken to be right–handed: a screwpositioned along the z–axis moves along the positive z–axis if turned in the direction fromx–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 hascoordinates (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 ourdisposal. A vector is a quantity with both magnitude and direction. A well–known example

A

B C

D

E

F

x

y

z

p=OP

O

AB

Figure 2.2: The arrow with tail at A and head at B represents a vector (left). The twoarrows 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 boldfacesymbol, like p (right).

2.1 3–space and vectors 23

of a vector occurring in physics is velocity. A vector is usually represented by an arrow in thesense 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, AB denotes an arrow with tail at A andhead at B. It points from A to B. If two arrows have the same length and the samedirection, we say that they determine the same vector. If AB and CD have the samelength and direction, we agree to write AB = CD (equal as vectors). In practice, weoften use a representative arrow to talk about a certain vector.

• Vectors in 3–space (represented by arrows) with tail at the origin O = (0, 0, 0) andhead at a point P = (x, y, z) play a special role. The corresponding vector is usuallyrepresented by the corresponding lower case boldface symbol, p in this case. Wealso write p = (x, y, z), which should be read as: the vector determined by the tail(0, 0, 0) and head (x, y, z). If asked to draw p, we usually draw this representativearrow.

• To assign coordinates/numbers to any vector, say the vectorAB, whereA = (a1, a2, a3)and B = (b1, b2, b3), we note that this vector can also be represented by the arrowwith tail at the origin and head at (b1 − a1, b2 − a2, b3 − a3). And so we writeAB = (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 = (1, 0, 0) usually represented by an arrow with tail at (0, 0, 0) and head at(1, 0, 0), the standard basis vector of length 1 in the direction of the positivex–axis.

(*) e2 = (0, 1, 0) usually represented by an arrow with tail at (0, 0, 0) and head at(0, 1, 0), the standard basis vector of length 1 in the direction of the positivey–axis.

(*) e3 = (0, 0, 1) usually represented by an arrow with tail at (0, 0, 0) and head at(0, 0, 1), the standard basis vector of length 1 in the direction of the positivez–axis.

So keep in mind: vectors are often, but not always In the following, vectors are usually(but not always) usually refer to directed segments with their tail in the origin, unless thecontext suggests otherwise.

Also note that in writing on paper or on the blackboard the use of boldface symbolsis not convenient. So we also use notations like x = (x1, x2, x3), where we underline thesymbol 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 contextof geometry.


• Scalar multiplication and vector addition. Vectors can be added and multipliedby scalars in the following way:

x+ y = (x1 + y1, x2 + y2, x3 + y3) addition, sum of vectorsλx = (λx1, λx2, λx3) scalar multiplication.

So the first coordinate of the sum of two vectors is the sum of the first coordinatesof the two vectors, etc. By combining these operations more complicated vectors canbe built from two or more vectors. For instance, x− y is the sum of x and (−1) · y;the vector 2x + 3y is obtained by multiplying x by 2 and y by 3 and adding theresults.

The operations addition and scalar multiplication have a clear geometric meaning.To add two vectors u and v, here is what happens geometrically. Move v, say, so

u

v

u+v

u 2u

Figure 2.3: Vector addition (left) and scalar multiplication (right).

that its tail coincides with the head of u. The new vector whose tail is at the tail ofu and whose head is at the head of v represents the sum. A scalar multiple of u isobtained by changing the length of the vector u by the scalar factor, say λ. If λ > 0then the direction remains the same, but if λ < 0, then the direction of the vector isreversed.

• The zero vector. A vector with length 0 is called the zero vector. It is usuallydenoted by 0. In coordinates: 0 = (0, 0, 0). If we multiply any vector p by 0, weget 0, i.e., 0 · p = 0. Adding 0 to any other vector q yields q again. In symbols:0+ q = q. Also, q− q = 0 fo any vector q.

• Using the standard basis vectors. In terms of the standard basis vectors, anyvector x = (x1, x2, x3) can be written as a combination of them in a straightforwardway: x1 e1+x2 e2+x1 e1. This is not the moment to understand in which situationsthis is useful.

• Length. The length |p| of a vector p = (a, b, c) is

√a2 + b2 + c2.

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

O

x

y

z

P=(a,b,c)

Q=(a,b,0)(a,0,0)

Figure 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 + b2 + c2.

• 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 = x1y1 + x2y2 + x3y3.

This product does not have an immediate geometric interpretation, but will show upindirectly in various geometrically relevant situations, such as the length of a vectorand the angle between two vectors:

For instance, the squared length of x equals the inner product of x with itself:

|x|2 = x2

1+ x2

2+ x2

3= x1x1 + x2x2 + x3x3 = x • x.

• 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 PQ. We define the distance between twovectors as the distance between their endpoints, or, equivalently, the length of thedifference. 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.


• Perpendicular vectors. Using the dot product we can express when two vectorsare perpendicular. It turns out (no proof here) that this is the case precisely if theirinner product equals 0:

x • y = 0 ⇔ x and y perpendicular.

• Angle between vectors. If a and b are nonzero vectors, then the angle α betweenthem is usually computed via its cosine. From the Cosine Law the following relationinvolving the dot product can be shown to hold:

cosα =a • b|a| · |b| .

To determine the angle, two steps are therefore required: first determine the cosineof 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 cosα = 0 in thiscase.

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 equals0:: (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 determinecosα:

cosα =(1, 1, 0) • (1, 2, 1)|(1, 1, 0)| · |(1, 2, 1)| =

1 · 1 + 1 · 2 + 0 · 1√11 + 12 ·

√12 + 22 + 12

=3√

2 ·√6=

√3 ·

√3√

4 ·√3=

√3

2.

Hence the angle is 30◦ or π/6 radians.

2.2 Describing lines and planes

2.2.1 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 linepasses and to give the line’s direction, say given by the nonzero vector v. In terms ofvectors, start with a vector p0 with end point P0 and add an arbitrary scalar multiple λvof v to it:

p0 + λv.

2.2 Describing lines and planes 27

As λ ranges from −∞ to ∞, all vectors with endpoints on the line can be obtained. Notsurprisingly, the vector v is called a direction vector of the line. Of course, any nonzeromultiple 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

+ λ

av

a v

Figure 2.5: Parametric description of a line involves a support vector p0 and a nonzerodirection vector v. The line can be seen as ‘resting’ on p0. Any vector on the line isobtained 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, weget (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 tospecify the plane’s direction by giving the direction of a line which is perpendicularto the plane. The direction of that line can be given by a single nonzero vector (thelength 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.


• Here is the example.Suppose the point P0 = (2, 1, 3) is on the plane U , and let p0 = OP0 be the vectorpointing to P0. Suppose moreover that n = (1, 2, 2) is perpendicular to U . Our taskis to catch those vectors x = (x, y, z) whose endpoints are in the plane. Now such anendpoint is in U if the direction from P0 to this endpoint is perpendicular to n. Thisdirection is represented by the vector x − p0 = (x − 2, y − 1, z − 3). So we requirethat 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 thesame equation. For example, 6x+ 12y + 12z = 60.

– The coefficients of x, y, z in the equation x+ 2y+ 2z = 10 form a vector whichis (a multiple of) the vector n = (1, 2, 2) perpendicular to the plane we startedwith. 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.

xy

zn

x

p0

P0

P

Figure 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 wesuppose to be perpendicular to the plane we wish to describe. The plane throughP and perpendicular to n consists of all the points P = (x, y, z) such that P0P isperpendicular to n, i.e., (x− x0, y− y0, z − z0) is perpendicular to (a, b, c). In vectorform:

(a, b, c) • (x− x0, y − y0, z − z0) = 0,


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 singlesymbol, 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 tothe 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 everycoefficient 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 coefficientsby a number so that the equation looks more pleasant. For instance,

3

2x1 −

7

5x2 +

2

3x3 =

9

2

looks better in the form 45x1− 42x2+20x3 = 135 (where the first mentioned equation hasbeen multiplied through by 30).

What we just said is one instance of the fact that we are free to change the appearance ofan equation according to our specific needs, as long as we do not violate any mathematicalrule. 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).

• x2 =2

3x1 +

5

3x3 −

10

3(the second variable has been ‘isolated’ on one side).

• 2(x1 − 1)− 3(x2 + 1) + 5(x3 − 1) = 0 shows clearly that (1,−1, 1) is on the plane.


2.2.5 Vector parametric description of a plane

A second way of representing a plane is by a vector parametric equation, similar to thedescription 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 instancethe plane U given by x1 + 2x2 − 3x3 = 0. Take two vectors in the plane which are notmultiples 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 bytaking suitable multiples of the vectors and then adding them, see Figure (2.7).

a

bx

p

a

b

Figure 2.7: Left–hand side: If U is a plane through the origin and a and b are two vectorswhich are not along the same line, then any vector x in the plane can be obtained by addingsuitable multiples of a and b. Right–hand side: the ingredients of a vector parametricdescription 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 thevectors/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. Theway to find a vector parametric equation of a plane, starting from an equation, is tosolve the equation (and don’t forget: there are infinitely many solutions; it’s a plane afterall). 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 asx1 = −2x2 + 3x3 + 4, then we clearly see, that for any given values of x2 and x3, there isexactly one value of x1 such that the triple (x1, x2, x3) is on the plane. So assign the valueλ to x2 and µ to x3. Then x1 = 4− 2λ + 3µ. We gain more insight in the solutions if we


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 usuallycalled 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 descriptionof 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 maylook quite different. See the following example. Equations of a plane, however, showless variation: if we only consider equations of the form ax + by + cz = d, then twosuch 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 ofthe 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 andz = 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 perpendicularto (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). Thenthe 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. Thisillustrates the fact that vector parametric descriptions of planes are far from beingunique. Incidentally, in picking two vectors perpendicular to (1, 1, 1), usually twovectors are chosen which look relatively simple, like (1,−1, 0) and (0, 1,−1). Ourchoice of vectors (2, 3,−5) and (1, 1,−2) is correct, but may be more cumbersomein computations. The advantage of the method in a) is that it produces relativelysimple vectors.


c) If you follow the method as explained in a) but solve for y rather than for x, thenthe resulting parametric equation is again somewhat different: start by rewriting theequation 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 parametricdescriptions. 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 beendiscussed above. Here, we discuss how to transform vector parametric descriptions intoequations. 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 haveto 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 isperpendicular 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 = 0a+ b+ 3c = 0

for a, b and c. To solve this system of two equations, first eliminate a from the secondequation 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, ourequation 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 instancethe support vector (2, 1, 3):

(−5) · 2 + 2 · 1 + 1 · 3 = d.

In conclusion, an equation for U is −5x+ 2y + z = −5.

2.3 The relative position of lines and planes: intersections 33

2.3 The relative position of lines and planes: inter-

sections

2.3.1 Intersecting a line and a plane

The intersection of a line and a plane usually consists of exactly one point. If the line andthe plane happen to be parallel, the intersection may be empty or consist of the wholeline. How does this work out in actual computations? Here is an example. Suppose theplane 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 λfor which the corresponding point on ℓ belongs to U as well, i.e., satisfies the equation forU . 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 theparametric 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 youfind that line of intersection explicitly. Suppose U and V are two planes, with equationsx1 + 2x2 − x3 = 4 and 2x1 + x2 − 5x3 = 2, respectively. We have to solve both equationssimultaneously, i.e., find all (x1, x2, x3) which satisfy both equations. To manipulate theequations efficiently, we write them as follows:

x1 + 2x2 − x3 = 42x1 + 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 = 4x2 + x3 = 2.

Now get rid of x2 in the first equation by subtracting the new second equation from thefirst:

x1 + − 3x3 = 0x2 + 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.


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: angles

2.4.1 The angle between two lines

It is fairly straightforward to define the angle between two lines ℓ and m, although onedetail 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 anglebetween 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◦ and90◦).

To summarize: If a and b are vectors in the directions of the lines ℓ and m, resp., then theangle α between ℓ and m is computed from

cosα =|a • b||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 linem 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

cosα =|(1, 0, 1) • (1, 1, 2)||(1, 0, 1)| · |(1, 1, 2)| =

3√2 ·

√6=

√3 ·

√3√

2 ·√2 ·

√3=

√3

2.

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 weregiven 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 Ushould be precisely. The reason is that the plane contains so many directions. So, whichdirection of the plane should we compare with the direction of the line? Suppose ℓ and Umeet in a point P . The plane U contains a whole family of lines through P . Any such linemakes an angle with ℓ, and the angle with such a line in the plane varies as this line variesin the family. (There is one case where the angle does not change, do you see which casethis is?)

l

P

U

l

m

P

Uβ

Figure 2.9: To find the angle between a line and a plane, it is convenient to introduce aline perpendicular to the plane as in the right–hand picture. If β is the angle between thesetwo 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 characterizedby the direction of a line perpendicular to the plane. So, let us take a line m through Pwhich is perpendicular to the plane U . Of course, we know what the angle between ℓ andm is. Now this angle is of course not really what we want, but 90◦ minus this angle turnsout to be the appropriate angle.

Summarizing: to compute the angle α between the line ℓ with direction a and the planeU with vector n perpendicular to U , solve α from

sinα =|a • n||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 need90◦ minus this angle.) Here, we are using cos(90◦ − α) = sinα.

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)


from the parametric description and we need a vector perpendicular to the plane U , forexample, 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

sinα =|a • n||a| · |n| =

3√2√6=

3√2√2√3=

√3

2.

Therefore the angle is 60◦.

(b) Or we compute the angle β between a and n from

cos β =|a • n||a| · |n| =

√3

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 adefinite angle with each other, but it takes some thinking to realize that an appropriate wayto make precise what this angle is, is to look at the angle between two vectors perpendicularto the two planes (cf. the case of a line and a plane). The only detail you have to be carefulabout is that you may have to replace one of the vectors by its opposite vector in order toguarantee that the angle is acute (at most 90◦), or introduce an absolute value just like wedid in the case of a line and plane.

To summarize: To compute the angle between the planes U and V , take a nonzerovector u perpendicular to U and a nonzero vector v perpendicular to V . The angle αbetween U and V is then computed from

cosα =|u • v||u| · |v| .

Figure 2.10: To find the angle between two planes, determine the angle between two vectorsperpendicular to the two planes, respectively.

In particular, two planes are perpendicular if the two vectors perpendicular to theplanes are perpendicular.

2.5 The relative position of points, lines and planes: distances 37

2.4.6 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 . Bothvectors are taken from the coefficients of the two equations. Now solve α from

cosα =|(1, 0,−1) • (1,−1,−2)||(1, 0,−1)| · |(1,−1,−2)| =

3√2 ·

√6=

√3

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 Pwhich is perpendicular to ℓ. Now W meets U along a line m, and W meets V along a linen. Finally, compute the angle between the lines m and n.

W

α

α

m n

P

Figure 2.11: An alternative approach to computing the angle between two intersecting planesU and V . The picture shows the cross section with the plane W perpendicular to the line ofintersection ℓ of U and V . The intersection of U and W is denoted by m; the intersectionof V and W is denoted by n.

To see why this approach leads to the same answer, just realize that normal vectors toU and V with their tails at P lie in W . This is illustrated in Fig. (2.11). This approachusually leads to more complicated computations.

2.5 The relative position of points, lines and planes:

distances

2.5.1 Computing the distance between two points is straightforward, but the computation ofdistances between points and lines, points and planes, etc., is more subtle. In this sectionwe 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


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 thepoint Q such that PQ is perpendicular to ℓ is the point we are looking for.

P

l

RQ

Figure 2.12: The shortest distance between a point P and a point on the line ℓ occurs forthe point Q where PQ is perpendicular to the line ℓ. For instance, applying Pythagoras’theorem to the triangle PQR, shows that |PQ| < |PR|.

Let us work this out in a specific example, where P is the point (7, 2,−3) and ℓ is theline 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 vectorQP is then (−5 + λ,−3− λ, 4 + 2λ).

• The next step is to solve λ from the condition that PQ is perpendicular to thedirection 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 andthe point Q on ℓ closest to P is therefore (substitute for example in the parametricequation 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 betweenP 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. Bythe distance between a point P and a plane U we mean the shortest possible distancebetween P and any of the points of U . But how do we find such a point in the plane? It isgeometrically obvious that we can locate such a point by moving from P in the directionperpendicular to U until we meet U , i.e., we need the line through P perpendicular to theplane. 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 theequation x− 2y + 2z = 7.

2.5 The relative position of points, lines and planes: distances 39

P

U Q

distance

locate Q

Figure 2.13: To find the distance between P and the plane U , first determine the intersectionof U and the line through P perpendicular to U . Then calculate the distance between Pand the point of intersection.

• First we look for the line through P and perpendicular to U . A vector perpendicularto U is easily extracted from its equation: (1,−2, 2). So the line through P withdirection 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 as25 + 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 andQ:

√

(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? Thenthere 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 PQ isperpendicular 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 multiplesof each other, you take a point on one line and compute the distance to the other line, justlike before.

In case the direction vectors are not multiples of one another, the computation of thedistance turns out to be more complicated. Suppose ℓ and m are two such lines. Among allpoints P on ℓ and all points Q on m we need to find points for which the distance |PQ| is


minimal. As before, this condition turns out to be related to right angles: we have to findP on ℓ and Q onm in such a way that PQ is perpendicular to both ℓ andm. The procedureis best illustrated with an example. Suppose ℓ is given by (1, 3, 3)+λ(2, 0, 1) and m is given

l

m

Figure 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 equationsin 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 andµ = 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.

2.6 Geometric operations: translating lines and planes 41

2.6 Geometric operations: translating lines and planes

2.6.1 In designing shapes, you may want to move or rotate certain elements of your design, saya window or a wall. The question is how you do this mathematically. Such mathematicalcomputations are at the basis of computer software implementations. In this section, wewill 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 directionover a given distance. We can represent this direction and distance by a vector t. Anyvector (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 effectof a translation on lines.

If a line ℓ is given by p+ λv, then a translation over t produces again a line, now withparametric equation (p+t)+λv. For example, if ℓ is given by (1, 0, 6)+λ(1, 1, 1), then aftertranslation 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 youtranslate 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, ourdiscussion splits into two parts.

• In the case where the plane is described by a vector parametric description, thesituation 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 thatthe description of the plane is implicit. If we translate a vector (x, y, z) of U then theresult 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 thisvector back, so subtract (2, 5,−1), then the result should be on U , i.e., satisfy theequation of U . So we substitute (u, v, w) − (2, 5,−1) = (u − 2, v − 5, w + 1) in theequation 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 distancebetween the planes is not equal to the length of t, since t is not perpendicular to U .


If you realize that translating a plane produces a parallel plane, another strategysuggests 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 andsubstituting 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 = dto 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 appearancehave been built during the last few years. From a distance they look somewhat like a fleetof ships from historic times, which is probably the reason for the district’s name: Armada’sPaleiskwartier.

Now some of the buildings are identical in shape (or at least at first glance), so theycould be considered as translates of one given central building. Apart from details, wecan consider this problem as a problem of translating planes (to be fair: the shape of thebuildings 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-directionis assumed to point upward) in some cartesian coordinate system. Suppose we want totranslate 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) inU1, 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 similarway. 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 shouldhave the form x− 3y = ... and b) (2, 3, 0) is on U1, etc.)

2.7 Geometric operations: rotating lines and planes

2.7.1 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’sinverse: if you apply the two rotations after each other, the total effect will be that nothinghas happened). So rotating a single point isn’t that difficult. But suppose we want to rotatea whole plane. Let us concentrate on the first rotation where any point (x, y, z) is rotatedto (−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)

2.7 Geometric operations: rotating lines and planes 43

x

(x,y)

y

(−y,x)90o

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) 7→ (−y, x, z)). This provides us with the vector parametric equationof 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 equationx + 2y + z = 5. After rotation we get a new plane, which we call V . If we take apoint (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). Thispoint (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 normalvector (1, 2, 1). This results in the vector (−2, 1, 1) and this should be the normalvector 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 coefficient5 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 supposewe rotate V again over 90◦ in the same direction to obtain a plane W . If (x, y, z) is a pointof 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 ofU . Can you explain the minus signs by directly considering a rotation over 180◦?


2.7.3 Fans of planes

If two planes U and V are not parallel, they intersect along a line, say ℓ. By rotating oneof the planes around ℓ, we generate a whole family of planes, the fan of planes on ℓ. Ofcourse, all these planes contain ℓ.

The question is: what is the equation of a member of such a fan, given the equationsof U and V ? To answer this question, we take two specific planes, say U : x− y − z = −1and 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 pointof ℓ satisfies both equations, but then every point of ℓ also satisfies a combination of thetwo 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 theleft–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 examplewe compute a member of the fan which is perpendicular to U . The equation of a generalmember 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 weobtain 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 thisequation and therefore to the same plane.)

2.8 Geometric operations: reflecting lines and planes

2.8.1 Mirror images

Sometimes you do not even notice it at first sight when two (types of) buildings havebeen 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

2.8 Geometric operations: reflecting lines and planes 45

changed: a right–handed screw changes into a left–handed screw. One immediate propertyof a reflection is that taking the mirror image of the mirror image of a point or vectorbrings you back to the original point or vector.

In this section we restrict our discussion to reflections where the mirror is a coordinateplane 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

x

z

(−x,z) (x,z)

Figure 2.16: In picturing the reflection in the y, z–plane, we have left out the y–axis for thesake of simplicity and pictured the resulting 2–d version of reflection in the z–axis. Thevector (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). Theimage of a line, say ℓ : (2, 3,−4)+λ(1, 2, 3), is obtained by reflecting each point of the lineindividually: 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 ifthe plane is given by an equation, we have to work backwards again, just like we did fortranslations and rotations, or exploit a normal vector of the plane. Suppose the plane Uhas equation 2x + y − 3z = 7. If (u, v, w) is a vector on the mirror image U ′ of U , thenreflecting the vector yields the vector (−u, v, w) in U , so (−u, v, w) satisfies the equationof U :

2(−u) + v − 3w = 7.

Therefore, U ′ has equation −2u+ v− 3w = 7. If the mirror is not a coordinate plane, likethe y, z–plane, things usually become a lot harder to describe. In later chapters we will gointo this in more detail.

In the case at hand, the ‘normal vector approach’ runs as follows. Reflect the normalvector (2, 1,−3) to obtain (−2, 1,−3), so the equation of U ′ is of the form −2x+y−3z = dfor 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


the vector (x, y, z) is easily seen to be (y, x, z) (just change the first two coordinates). Fromthis description in coordinates it is also clear that reflecting twice in succession brings youback to where you started:

(x, y, z)reflect7−→ (y, x, z)

reflect7−→ (x, y, z).

2.8.3 Example. Two towers of a building are designed as mirror images. Both towers havetriangular horizontal cross sections. In a suitable coordinate system, the walls of one ofthese 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 thevector (u, v, w) is (v, u, w) and is on the plane V : y − x = −1, so u − v = −1. Thereforethe equation of the mirror image V ′ of the plane V : y − x = −1 is x − y = −1 (so theroles 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 = 3y − x = −1 x− y = −1y = −1 x = −1

2.8.4 More complicated reflections

A straightforward observation on reflections in a plane U is the following. If you connecta 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 translatesinto the following strategy for computing mirror images:

a) Start at P and move along the line through P which is perpendicular to U , so firstset 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 theparameter, say λ0, the line intersects the plane U . So p + λ0v is ‘halfway’ betweenP and its mirror image.

c) But then p+ 2λ0v 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 planex+y+z = 3. The idea is to start moving from P in the direction (1, 1, 1) (perpendicular toU). Then locate on this line the second point which has the same distance to U . Let ℓ bethe line (2, 3, 4)+ λ(1, 1, 1). First find out where this line meets the plane U , so substitutein 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

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 thedifference (−2,−1, 0)− (2, 3, 4) = (−4,−4,−4) is perpendicular to U .

2.9 Tesselations of planes

2.9.1 Sometimes the outside of a flat wall is given a special pattern. For instance, the plane iscovered with rectangles, triangles or other shapes with a certain degree of regularity orsymmetry. 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 onthe left and a triangular pattern on the right. In the case of a full plane, the patterns aresupposed to extend infinitely.

of symmetry that can occur in such subdivisions of a plane or in producing decorations ofa plane. This section only touches upon the subject. Deeper investigations require moreadvanced 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 ofsymmetry. 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 intoitself. For instance, if you translate over 2a, the whole pattern is moved two steps to

a

b

Figure 2.18: Translational symmetry.


the right. The translated pattern coincides precisely with the original pattern. But a

translation over, say,1

2·a creates a pattern which does not coincide with the original

pattern.

b) Rotational symmetry. The patterns have various centers of symmetry. A rotationover a specific angle transforms the pattern into itself.

center

r

r

Figure 2.19: Rotational symmetry. All black dots are centers of symmetry, at least if youimagine 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 intothemselves by reflecting them in suitable lines. In the picture a few of such lines areindicated.

lines of symmetry

Figure 2.20: Mirror symmetry or reflectional symmetry. If you think of the pattern asextending indefinitely, reflecting in any of the indicated lines transforms the picture intoitself.

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, scientistshave 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): ifthe pattern has translational symmetry over the vectors a and b and rotational symmetry

2.9 Tesselations of planes 49

over 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 cosα and sinα. 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 = m1a + m2b

Rb = n1a + n2b.

Now the occurrence of the integers m1, m2, n1 and n2 forces the numbers cosα and sinαto be relatively simple. In turn, that means that there are only a few possibilities for α.


2.10 Exercises

1 Pictures of operations on vectors

a) Make a drawing of vector subtraction: draw two vectors x and y and indicate howx− 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).

b) Evaluate (x, y, z) • (2, 1, 3). Also evaluate (1√14

(2x+ y + 3z)(2, 1, 3)) • (2, 1, 3).

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 forevery vector x the vectors x and (x • a) · a have the same inner product with a. Usepart c) to conclude that x − (x • a) · a and a are perpendicular. Show in a picturehow this relates to the orthogonal projection of x on the line λ · a.

2.10 Exercises 51

5 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 relatedto ‘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 toboth 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 ℓcan also be found as the answer to a minimization problem: find the vector on ℓ which isclosest to O.

a) If ℓ is the line (−1,−3, 5) + λ(2, 4,−2), then describe the distance of 0 = (0, 0, 0) toa vector on ℓ as a function of λ.

b) Can you use the square of this function just as well for our problem? Find theminimum of the function you have chosen to work with. Which vector on ℓ lies atthis distance from the origin?

c) Let’s call the vector you have found p. Compute the dot product p • (2, 4,−2) andexplain 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.


c) Geometrically, it is clear that the answer to b) should be a plane. Use your geometricintuition to find directly, without using a) or b), a vector perpendicular to the planeand 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 bothterms to the left–hand side, don’t expand, and use a2 − b2 = (a + b)(a − b) in asuitable way.]

10 Describing a plane through three given points

It is quite obvious that through three distinct points which are not on a line preciselyone plane passes. This exercise is about finding the plane’s parametric description and itsequation. 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 isnot on ℓ.

b) Use PQ and PR as direction vectors and set up a parametric description of the planeU 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) andpasses 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.

2.10 Exercises 53

13 Computing intersections

a) Find the intersection of the line (0, 2, 3) + λ(1,−1, 1) and the plane with equationx+ 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 producesa point Q. Finally, compute the distance between P and Q.

a) Explain why PQ 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 withvector parametric equation (3, 2, 4) + λ(2, 1, 2). So first derive the equation of theplane 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 andintersect with ℓ and m. Then compute the distance between the resulting points of inter-section.

a) Explain why this approach is wrong in general.


b) Comment on the following approach: Take a plane containing ℓ and rotate it aroundℓ 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 theplanes.

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 thedistance 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 equationx+ 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 planex+ y + z = a, and then computing the distance between the origin and the point ofintersection.

b) For which values of a will the plane x+ y + z = a meet the sphere x2 + y2 + z2 = 1?

c) For which value(s) of a will the intersection of the plane and the sphere be a circlewhich 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) 7→ (−y, x, z). Find the equation ofthe 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 planescontaining the y–axis, the other fan consists of the planes containing the line x = 5, z = 0.

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 = 0and z = 1. Now this construction is rotated over 45◦ around the y–axis counterclockwise.First we try to describe this rotation explicitly.

a) Explain why e1 = (1, 0, 0) is transformed into1√2(1, 0, 1) and e3 = (0, 0, 1) is trans-

formed into1√2(−1, 0, 1). What happens to e2 = (0, 1, 0)?

b) Take any vector (x, y, z) and rewrite it as xe1 + ye2 + ze3. Now rotate (x, y, z) by

rotating its three components xe1, ye2 and ze3. Show that this leads to (1√2(x −

z), y,1√2(x+ z)).

c) Use the transformation in b), i.e.,

(x, y, z) 7→ (1√2(x− z), y,

1√2(x+ z)),

to find the bounding planes of the rotated cube.

(Examples of such houses, the so–calles ‘kubushuizen’ with similar rotated positions havebeen built in Helmond.)

25 Inside a building a room is constructed with perpendicular sloping walls. These wallsshould 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 possibilitiesfor such walls, just two are needed).

b) What is the equation of the member of the fan which makes an angle of 45◦ with thex, y–plane?


l

Date post:	13-Jul-2020
Category:	Documents
Upload:	others
View:	4 times
Download:	0 times

Chapter 2 3–Space: lines and planessterk/Bouwkunde/2db60-chap2.pdf24 3–Space: lines and planes...

Documents