Solution to Exercise 3.82.
Intuition is the conception of an attentive mind, so clear, so
distinct, and so effortless that we cannot doubt what we have so conceived.
—
Rene Descartes (1596–1650)
Exercise 3.82. Find a matrix of the reflection that maps the point
X(3, 8, 1) to Y(5, 1, 1) and find the image of Z(12, 7, 1).
By definition, the line of reflection is the perpendicular bisector of segment XY. The midpoint of segment XY is (4, 9/2, 1). We find line XY by Proposition 3.1,
Evaluating the determinant, we find the line XY is [7, 2, –37]. A line perpendicular to line XY has the form [2, –7, l3]. Since the midpoint is on the line [2, –7, l3],
That is, l3 = 47/2. Hence, l[4, –14, 47] is the perpendicular bisector of segment XY.
The remainder of the problem is similar to Exercises 3.80 and 3.81 (two methods are given).
Method 1.
We first find the point of intersection of h[0, 1, 0] and line l by applying Proposition 3.2.
Hence, Thus, the point of intersection C is
The measure of the angle between lines h and l is
Note:
and
Hence, by Proposition 3.15,
The image of Z(12, 7, 1) is Z'[642/53, 350/53, 1] since
Method 2.
We apply Proposition 3.6 to find the
matrix of a direct isometry T that
maps h[0, 1, 0] to l[4, –14, 47]. There is a nonzero real
number k such that klT = h.
implies
Hence,
and
For a matrix of a direct isometry that maps line h to line l, let a = 0 and b = 47/14,
Hence, Rl is defined by
The image of Z(12, 7, 1) is Z'[642/53, 350/53, 1] since
