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10.19 For a rectangular 2вЂ™ X 3вЂ™ X 4вЂ™box, find the angle that the longest diagonal makes
with the 4вЂ™.side.
10.20 Use vector notation to prove that the diagonals of a rhombus are orthogonal to each
other. See Figure 10.23.
Figure
10.23
If Ilull = IIvII. (his quadrilateral is a rhombus.
10.21 Prove the following identities.
a) IIU + VII2 + IIU  Yl12 = 2llullвЂ™ + 2˜˜вЂњ˜˜*,
b) u v = a Ilu + вЂњII2 ;llu ˜11˜.
10.22 Prove that ifu and Y are orthogonal vectors, then IIu + ˜11вЂ™ = Ilull + [˜v˜˜˜. Explain
why this statement is called the general version of the Pythagorean Theorem.
10.23 The cross product is a commonlyused multiplication of vectors in R3, for which
the product of two vccto˜s in RвЂ™ is another vector in R3. It is defined as follows,
using the determinant notation introduced in Chapter 9:
Prow the following properties of the crossproduct:
cl) uxv= vxu,
b) u X v is perpendicular to u,
c) u X v is perpendicular to Y,
d) (ru) X v = r(u X v) = u x (IV),
i.) (u, + II?) x вЂќ = (u, x v) + (вЂњ2 x v),
f) I I вЂќ x вЂњ11˜ = 11вЂќ11˜1/вЂњ/12 (u v)вЂ˜.
(use item f and Theorem 10.3),
8) Ilu X VII = llullll˜ll sin B
h) uxu=O.
10.24 Show that the crossproduct can hc represented symbolically as
222 EVCLIDEAN SPACES [lOI
where e, = (1, 90). e2 = (0, 1, 0), and e3 = (40, 1). Treat the e;вЂ˜s as points or
symbols in the expansion of the determinant.
10.25 Use the cross product to find a vector perpendicular to both u and Y:
a) вЂќ = (LO, 1) v = (1, 1, 1B
h) вЂќ = (1,  1,2) Y = (0,5, 3).
10.26 Consider the parallelogram determined by the vectors u and Y in RвЂ™, as in Figure
10.24.
a) Show that the area of this parallelogram is 1111 X VII. [Hint: Express height h in
terms of u, Y, and 0.1
b) Find the area of the triangle in RвЂ™ whose vertices are (1, 1, 2), (0, 1, 3), and
(2 1, 0).
10.5 LINES
The fundamental objects of Euclidean geometry are points, lines, and planes.
These next two sections show how to describe lines and planes and their higher
dimensional analogues.
First. we will work with lines in RвЂ™. In high school algebra, we learn that
straight lines have an equation of the form
˜˜=mx,+b. (9)
The coefficient m is the slope of the line and the coefficient b is the yintercept. This
algebraic representation of the line is convenient for solving equations. However,
it is not the most useful equation for representing geometric objects. What is the
equation of line v in Figure 10.25? We cannot solve for x2 in terms of x1. More
important than the awkwardness of this special case is the need for an algebraic
representation which clearly expresses the geometry of the line. We will often find
a parametric reprrsmtarion of the line more useful.
110.51 223
LINES

Figure
Lines in RвЂ™. 10.25
A parametric representation of a point on a line uses a parameter f in the
coordinate expression of the point; more formally, a parametric representation
is an expression (x1 (I), x?(t)) with parameter f in RвЂ™. The point x = (xI,xz) is
on the line if and only if x = (x,(f*), x2(P)) for \ome value fX of the parameter
f. TV make matters concrete, you might think of f as representing time, and
the parameterization as describing the transversal of a path. The coordinates
(x,(f), x2(r)) dcscrihc the particular location which is reached at time 1.
A line is completely determined by two things: a point xg on the line and a
direction v in which to move from x0. Geometrically, to describe motion in the
direction v from the point x,,, we simply add scalar multiples of v to x,, as in Figure
10.26. The result is the parametric rcprcsentation
x(/) = x,, + tвЂќ. (IO)
Figure
10.26
ELKLlDEAN SPACES [lOi
224
Example 20.4 For example, line a in Figure 10.25 is the line which goes through
the point (4,Z) and moves directly to the northeastin the direction (1, 1). It
is described by the parameterization
X(9 = (RW>X2(4)
= (4,2) + t(1, 1)
= (4 + l 1, 2 + f. 1),
or x, =4+f.l (11)
xz=2+t.1. (12)
Figure 10.25 shows that (5, 3) and (1, 1) are on line a. The first point is reached
when t = 1, and the second when f = 3.
Note that the same line can be described by different parametric equations.
For example, we can also view line a in Figure 10.25 as the line through the
point (1, 1) in the direction (2,2). This yields the parameterization
(Xl(f), x*(t)) = (1, 1) + r(2,2) = (1 + 2t, 1 + 21)
With this parameterization, the line passes through (4,2) when f = 1.5 and
through (S,3) when I = 2.
Of course, the parameterization (10) works in all dimensions. For example,
the line in R3 through the point x,, = (2, 1,3) in the direction v = (4, 2,s) has
the parameterization
X(r) = (xl(t). X2(1)> x3(t))
= (2, 1, 3) + f(4, 2, 5)
= (2 + 4t, 1 ˜ 21, 3 + St).
Another way to determine a line is to identify two points on the line. Suppose
that x and y lie on a line LZ. Then, P can be viewed as the line which goes through
x and points in the direction y x. Thus, a parameterization for the line is
x(t) = x + r(y x)
= x + ry tx (13)
= ( 1 pr)x+ty.
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