Chapter 3. The Geometry of n Dimensions ∗ Vector Spaces
4. Lines and Line Segments
The term “line” shall always mean a line extending indefinitely (never a line segment, which is only a part of a line).
¯ a ¯b ~ u ¯ x ¯ 0 Figure 17
To obtain all points of a straight line in E2 or E3, we take a “vector” ~u = −ab→ (joining two given points ¯a and ¯bon the line) and then, so to say, “stretch” it indefinitely in both direc- tions, i.e., multiply ~u by all possible scalarst ∈E1(positive, negative, and 0). Now, by definition,
~u =−ab→= ¯b−a¯ =−→0b−−0a→
(see Figure 17). The “position vector” −0x→ of any point ¯x on the line ab is, geometrically, the sum of −0a→ and −ax:→ −0x→ = 0a−→+−ax. Here the vector→ −ax→ is a scalar multiple of −ab→=~u; specifically, −ax→=t~u, where
t= | −→ ax| |~u| or t=− |−ax→| |~u|
according to whether the vectors−ax→and~uhave the same or opposite directions. Thus we have
¯
x=−0x→=0a−→+−ax→= ¯a+t~u.
Conversely, every point of that form (for any t ∈E1) lies on the line ab. Thus
the line ab in E2 or E3 is exactly the set of all points x¯ of the form ¯
x= ¯a+t~u= ¯a+t(¯b−a),¯ t∈E1. By varying t, we obtain all points of ab.
It is natural to accept this as adefinition of a line in En. Definition 1.
Theline passing through two given points ¯a, ¯b∈En(¯a 6= ¯b) (equivalently, the line passing through ¯a in the direction of a vector ~u=−ab→= ¯b−¯a) is
the set of all points ¯x∈En of the form
¯
x= ¯a+t~u= ¯a+t(¯b−¯a),
where t is a variable which takes on all real values (we call it a real parameter). In symbols,
Line ab ={x¯∈En |x¯= ¯a+t~u for some t ∈E1}; ~u= ¯b−¯a=−ab→6=~0. (1) Briefly, we call it “the line ¯x = ¯a +t~u” or “the line ¯x = ¯a +t(¯b− ¯a)”; instead, we may write ¯x = (1−t)¯a+t¯b (rearranging brackets). The formula ¯
x = ¯a+t~u (respectively, ¯x = ¯a+t(¯b−¯a)) is called the equation of the line (more precisely, itsparametric equation). In the first case, we say that the line is given by a point ¯a and a direction ~u; in the second case, it is determined by two of its points, ¯a and ¯b. In terms of the coordinates of ¯x, ¯a and ~u (or ¯b), the parametric equation is equivalent to n simultaneous equations (called the parametric coordinate equations of the line):
xk=ak+tuk =ak+t(bk−ak), k = 1,2, . . . , n. (2)
It is a great advantage of the vector notation that one vector equation replaces n coordinate equations.
Now, since the vector ~u (used to form the line) is anyway being multiplied by arbitrary scalars t, it is clear that the line (1) will not gain or lose any of its points if ~u is replaced by some scalar multiple c~u (c 6= 0). In particular, we may replace ~u by its unit ~u/|~u| (taking c = 1/|~u|). Thus we may always assume (if desirable) that ~u is a unit vector itself. In this case the equation ¯
x = ¯a+t~u (and the equations (2)) are said to be normal. To normalize an equation of a line means to replace ~u by ~u/|~u|. Since c may also be negative, the line (1) does not change if we replace ~u by −~u; thus the direction of a line is not uniquely determined: we always have two choices of the unit vector ~u. If, however, a particular ~u is prescribed in advance, we speak of a directed line. The coordinates of the direction vector ~u (or any of its scalar multiples c~u) are called a set of n direction numbers for the line (1); of course, there are infinitely many such sets corresponding to different values of c. In particular, the direction cosines of ~u (i.e., the components of the unit vector ~u/|~u|) are called a set of direction cosines of the line. (There are precisely two such sets, namely the direction cosines of ~u and those of −~u.)
In addition to changing the vector ~u, we may also alter the parameter t. Indeed, since t is anyway supposed to take on all real values, nothing will change if we replace it by some other variable expressionθ which likewise runs over all real values, e.g., by θ = 1−t. Thus, every line has infinitely many parametric equations, depending on the choice of the parameter. We can also entirely eliminate the parameter from equations (2) by rewriting them as follows
§4. Lines and Line Segments 147
(assuming that bk−ak6= 0), and then dropping “t” on the right, if desirable:
x1−a1 b1−a1 = x2−a2 b2−a2 =· · ·= xn−an bn−an =t. (3)
One can write the equations in that form even if some of the denominators vanish. It is then understood that the corresponding numerators are to be equated to 0, e.g., xk −ak = 0, and this equation replaces the (senseless)
equation involving the fraction with the vanishing denominator. Note that the xk in (3) and (2) are variables.
Dropping t in (3), we are left with n −1 equations between n fractions involving only the (fixed) coordinates of ¯a and ¯band the (variable) coordinates of ¯x. A point ¯xthen belongs to the line ab if and only if its coordinates satisfy these n−1 equations (called the nonparametric equations of a line through two given points). If, instead, the line is given in terms of one point ¯a and a direction vector~u = ¯b−a, then, replacing¯ bk−ak by uk, we get
x1−a1 u1 = x2−a2 u2 =· · ·= xn−an uk . (4)
Here theuk form a set of direction numbers. Normalizing (i.e., dividing the
uk by |~u|= p
u2
1+u22+· · ·+u2n), we get a set of direction cosines of ab.
If~uand~v are the direction vectors of two lines, we also callh~u, ~vi(as defined in §3) the angle between the two lines. This angle is uniquely determined if the lines are directed; otherwise, by changing the sign of ~u or~v, one can also change the sign of cosh~u, ~vi. (Verify this!) Thus one obtainstwo angles, α and π−α. Two lines are said to be perpendicular or orthogonal if ~u⊥~v, i.e., if
~u·~v = n
X
k=1
ukvk = 0.
They are said to be parallel if one of~u and~v is a scalar multiple of the other, say~u=c~v; in this case, we also say that the vectors ~uand~v arecollinear (see
Definition 3 in §2).
Note 1. More precisely, we say that ~u and ~v are vector-collinear to mean that ~u = c~v or ~v = c~u. On the other hand, it is customary to say that three
points a, ¯¯ b, ¯c are collinear iff they lie on one and the same line (a different notion!).
If, in the parametric equation ¯x= ¯a+t~u, or ¯
x= ¯a+t(¯b−¯a) = (1−t)¯a+t¯b,
we lett vary not over all ofE1 but only over some subset ofE1, then we obtain only a part of the line ab. In particular, by letting t vary over some interval
in E1, we obtain what is called a line segment in En. (We reserve the name “interval” for another kind of sets, to be defined in §7. In E1, both kinds of
sets coincide with ordinary intervals.) Exactly as in E1, we have four types of
such line segments. We define them below.
Definition 2.
Given two points ¯a and ¯b in En, we define the open line segment from ¯a
to ¯b, denoted L(¯a,¯b), as the set of all points ¯x ∈En of the form ¯
x= ¯a+t(¯b−¯a) = (1−t)¯a+t¯b,
where t varies over the interval (0,1)⊂E1, i.e., 0< t <1. In symbols,
L(a, b) ={x¯∈En |x¯= ¯a+t(¯b−a) for some¯ t ∈(0,1)}.
This is also briefly written asL(a, b) ={¯a+t(¯b−¯a)|0< t <1}, i.e., “the set of all points ¯a+t(¯b−¯a) for 0 < t <1.”
Similarly, theclosed line segment L[¯a,¯b] is L[¯a,¯b] ={¯a+t(¯b−¯a)|0≤t≤1}; the half-open line segment is
L(¯a,¯b] ={¯a+t(¯b−¯a)|0< t≤1}, and the half-closed one is
L[¯a,¯b) ={¯a+t(¯b−¯a)|0≤t <1}.
In all cases, ¯a and ¯b are called the endpoints of the line segment, and
|¯b−¯a| is called its length.
Note 2. (i) The line segments are also defined in case ¯a = ¯b (“degenerate case”). (ii) Settingt = 0 ort= 1, we obtain the endpoints ¯aand ¯b, respectively. The other points are obtained as t varies between 0 and 1.
Examples.
Take three points in E3: ¯a = (0, −1, 2), ¯b = (1, 1, 1), ¯c = (3, 1, −1).
Then the line ab has the parametric equation ¯x = ¯a+t(¯b−¯a); or, in coordinates,
x1 = 0 +t(1−0) =t, x1 =−1 + 2t, x3 = 2−t;
or, writing (x, y, z) for (x1, x2, x3),
x=t, y =−1 + 2t, z = 2−t. Eliminating t (as in formula (3)), we obtain
x 1 = y+ 1 2 = z−2 −1 ; or, normalizing, x 1/√6 = y+ 1 2/√6 = z−2 −1/√6,
§4. Lines and Line Segments 149
where (1, 2, −1) is a set of direction numbers (coordinates of the vector ~u=−→ab = ¯b−¯a), while 1 √ 6, 2 √ 6, −1 √ 6
is a set of direction cosines (coordinates of the unit vector ~u/|~u|). A set of direction numbers for the line bc is obtained from the vector~v =−→bc = ¯
c−¯b = (2, 0, −2); the direction cosines are (2/√8, 0, −2/√8). Using formula (4), we obtain the coordinate equations in the symbolic form (not normalized) x−1 2 = y−1 0 = z−1 −2 ; i.e., x−1 2 = z−1 −2 and y−1 = 0. The angle between −ab→ and −→bc is given by
cosh~u, ~vi= ~u·~v |~u| |~v| = 4 √ 48 = 1 √ 3.
∗Note 3. Any line, ¯x = ¯a + t~u, in En is an isomorphic copy of E1, in the sense of §14 of Chapter 2. Indeed, let us define a mapping f on E1 by setting (∀t ∈ E1) f(t) = ¯a+t~u (with ¯a and ~u fixed) and let L denote the given line. Clearly, as t varies over E1, f(t) varies over L; thus f is a map of E1 onto L. This map is also easily proved to be one-to-one, and it becomes an ordered-field–isomorphism if operations and inequalities inL are defined as follows: Let ¯x=f(t), ¯x′=f(t′); then, by definition,
x+x′ =f(t+t′), x¯x¯′ =f(tt′), and ¯x <x¯′ iff t < t′ (cf. Problem 10 below).
Problems on Lines, Angles, and Directions in
E
n1. Prove in detailCorollary 1(b) andCorollary 3of§3. Also show that the angleh~u, ~vidoes not change if~uand~v are multiplied by some scalars of the same sign. What if the scalars are of different signs?
2. Prove geometrically (in E3) that the dot product ~v ·~u, where ~u is a
unit vector, is the orthogonal (directed) projection of~v on the directed line ¯x = ¯a +t~u (where ¯a is arbitrary but fixed). Define analogously projections of vectors on directed lines in En.
3. Find the mutual angles between the vectors ~u, ~v, and w~ specified in
Problem 1 of §1 (do cases (a)–(d) separately). Also normalize these vectors and find their direction cosines. Verify by actual computation, in at least one case, that Formula (b) of Corollary 1 in §3 holds. Are any two of the vectors perpendicular?
4. Let ~u, ~v ∈E3, and let
~
w= (u2v3−u3v2, u3v1−u1v3, u1v2−u2v1).
Show that ~u⊥w~ and ~v ⊥w.~
Note: The vector w~ so defined is called the cross product of ~u and ~v and is denoted by~u×~v or symbolically by the “determinant”
~ı ~ ~k u1 u2 u3 v1 v2 v3 ,
where~ı,~,~k are the basic unit vectors in E3. Show that
~u×~v =−(~v×~u) and that in general
(~u×~v)×~x6=~u×(~v×~x).
(Give a counterexample!) Also prove that two lines ¯x = ¯a +t~u and ¯
x= ¯b+t~v in E3 are parallel iff ~u×~v = 0. (Note that cross products are
defined only in E3.)
5. Find a vector (unit) inE3, with positive coordinates, which forms equal angles with the axes (i.e., with the basic unit vectors). Solve a similar problem in E4.
6. Given three points in E4: ¯a = (0, 0, −1, 2), ¯b = (2, 4,−3, −1), ¯c =
(5, 4, 2, 0). Find the angles of the triangle ¯a¯b¯cand the equations of its sides, in nonparametric form. Normalize the equations. For each side give a set of direction numbers and direction cosines.
6’. Let ¯bbe any point on the line ¯x= ¯a+t~u. Show that this line coincides with the line ¯x = ¯b+θ~u.
[Hint: Let ¯b= ¯a+t0~u. Find θ.]
7. A globe (solid sphere) in En, with center ¯p and radius ǫ > 0, is by definition the set
{x¯∈En |ρ(¯x,p)¯ < ǫ},
denoted Gp¯(ǫ). Show that if ¯a,¯b ∈ Gp¯(ǫ), then also L[¯a,¯b] ⊆ Gp¯(ǫ).
Prove the same property (calledconvexity) also for the closed globe Gp¯(ǫ) ={x¯∈En |ρ(¯x,p)¯ ≤ǫ}.
Disprove it for the nonsolid sphere
Sp¯(ǫ) ={x¯∈En|ρ(¯x,p) =¯ ǫ}.
§4. Lines and Line Segments 151
8. In Problem 6 find the nonparametric equations of the lines through each vertex parallel to the opposite side of the triangle ¯a¯bc. Find also the¯ points of intersection of these three lines.
9. Prove that if a vector ~v in En is perpendicular to each of the n basic unit vectors, i.e.,~v·~ek = 0,k = 1,2, . . . , n, then necessarily~v =~0. Infer
that if~v·~x= 0 for all x, then¯ ~v =~0.
10. Prove that the map f defined inNote 3 of §4 is one-to-one.
[Hint: Show thatt6=t′
=⇒ |f(t)−f(t′
)|=ρ(f(t), f(t′
))6= 0.]
Next, verify that the lineLis an ordered field, with zero elementf(0) = a and unity f(1), under operations and ordering as defined in Note 3, and that f(t+t′) = f(t) +f(t′) and f(tt′) = f(t)·f(t′), by definition. ∗(Hence infer that f is an isomorphism between the fields E1 and L.)
11. (i) Given a point ¯p ∈ En and a line ¯x = ¯a +t~u (|~u| = 1), find the
orthogonal projection of ¯p on the line, i.e., a point ¯x0 = ¯a+t0~u such that −→x0p⊥~u.
[Hint: By Problem 2,t0= (¯p−¯a)·~u; verify that (¯p−x¯0)·~u= 0 if ¯x0= ¯a+t0~u.]
(ii) Show that
ρ(¯p,x¯0) =|p¯−x¯0|= q
|p¯−¯a|2−t2
0 =|p¯−¯a| |sinα|, where α=h~u, p¯−a¯i.
[Hint: Use the formulas
|p¯−x¯0|2= (¯p−x¯0)·(¯p−x¯0)
and
|sinα|=p1−cos2α.]
(iii) Noting that ¯a is an arbitrary point on the line, infer that ρ(¯p,x¯0) is the least distance from ¯p to a point ¯a on the line.
12. Find the three altitudes of the triangle ¯a¯b¯c of Problem 6. (Use Prob- lem 11.)
13. Given two nonparallel lines in En: ¯x = ¯a+t~u and ¯y = ¯b+θ~v, where t, θ are real parameters and |~u| =|~v|= 1. Find two points ¯x and ¯y on these lines such that (¯x−y)¯ ⊥~u and simultaneously (¯x−y)¯ ⊥~v. Infer from Problem 11 that, for these points, ρ(¯x,y) is the shortest distance¯ between a point on one line and a point on the other line.
[Hint: We have to satisfy the simultaneous equations in two unknowns:
(¯x−y¯)·~u= 0 and (¯x−y¯)·~v= 0.
Substitute ¯x= ¯a+t~uand ¯y= ¯b+θ~v, and transform the two equations into (¯a−¯b)·~u+t−θ(~u·~v) = 0 and (¯a−¯b)·~v−θ+t(~u·~v) = 0.