Solution to Exercise 3.84.
Whoever despises the high wisdom of mathematics nourishes himself on delusion.
—Leonardo da Vinci (1452–1519)

Exercise 3.84. Find a product of reflections that maps X(–2, 5, 1), Y(–2, 7, 1), and Z(–5, 5, 1) to X'(4, 3, 1), Y'(6, 3, 1), and Z'(4, 0, 1).

 

Two Methods are given.

Method 1.

        By definition, the line of reflection is the perpendicular bisector of segment XX'. The midpoint of segment XX' is (1, 4, 1). We find line XX' by Proposition 3.1,

 

Evaluating the determinant, we find the line  XX' is [1, 3, –13]. A line perpendicular to line XX' has the form l[3, –1, l₃]. Since the midpoint is on the line l[3, –1, l₃],

 

That is, l₃ = 1. Hence, l[3, –1, 1] is the perpendicular bisector of segment XX'.
        We first find the point of intersection of line h and l, by applying Proposition 3.2.

 

Hence,  Thus, the point of intersection C is (–1/3, 0, 1). The measure of the angle between lines h and l is

 

Note:

 

and

 

Hence, by Proposition 3.15,

 

The image of X is X', Y(–2, 7, 1) is Y_l(26/5, 23/5, 1) and Z(–5, 5, 1) is Z_l(32/5, 6/5, 1).

        We repeat the above process using Y_l and Y' as in the proof of Theorem 3.27. The next line of reflection is the perpendicular bisector of segment Y_lY'. The midpoint of segment Y_lY' is (28/5, 19/5, 1). We find line Y_lY',

 

Evaluating the determinant, we find the line Y_lY' is [2, 1, –15]. A line perpendicular to line Y_lY' has the form m[1, –2, m₃]. Since the midpoint is on the line m[1, –2, m₃],

 

That is, m₃ = 2. Hence, m[1, –2, 2] is the perpendicular bisector of segment Y_lY'.
        We find the point of intersection of line h and m, by applying Proposition 3.2.

 

Hence,  Thus, the point of intersection C is (–2, 0, 1). The measure of the angle between lines h and m is

 

Note:

 

and

 

Hence, by Proposition 3.15,

 

The image of X' is X', Y_l(26/5, 23/5, 1) is Y'(6, 3, 1) and Z_l(32/5, 6/5, 1) is Z_m( 4, 6, 1).

        We repeat the above process using Z_m and Z' as in the proof of Theorem 3.27. The next line of reflection is the perpendicular bisector of segment Z_mZ'. The midpoint of segment Z_mZ' is (4, 3, 1). We find line Z_mZ',

 

Evaluating the determinant, we find the line Z_mZ' is [1, 0, –4]. A line perpendicular to line Z_mZ' has the form n[0, 1, n₃]. Since the midpoint is on the line n[0, 1, n₃],

 

That is, n₃ = –3. Hence, n[0, 1, –3] is the perpendicular bisector of segment Z_mZ'.

        Note that the lines h and n are parallel since the measure of the angle between lines h and n is

 

Thus, the vector (0, 3, 0) translates h to n. Hence, by Proposition 3.15,

 

The image of X' is X', Y' is Y', and Z_m( 4, 6, 1) is Z'(4, 0, 1).

        Hence, the product R_nR_mR_l  defined by

 

maps X to X', Y to Y', and Z to Z'.

 

Method 2.

        By definition, the line of reflection is the perpendicular bisector of segment XX'. The midpoint of segment XX' is (1, 4, 1). We find line XX' by Proposition 3.1,

 

Evaluating the determinant, we find the line  XX' is [1, 3, –13]. A line perpendicular to line XX' has the form l[3, –1, l₃]. Since the midpoint is on the line l[3, –1, l₃],

 

That is, l₃ = 1. Hence, l[3, –1, 1] is the perpendicular bisector of segment XX'. We apply Proposition 3.6 to find the matrix of a direct isometry T that maps h[0, 1, 0] to l[3, –1, 1]. There is a nonzero real number k such that klT = h.

 

 

implies

 

Hence,    and b = 3a + 1. For a matrix of a direct isometry that maps line h to line l, let a = 0 and b = 1,

 

Hence, R_l is defined by

 

 

The image of X is X', Y(–2, 7, 1) is Y_l(26/5, 23/5, 1) and Z(–5, 5, 1) is Z_l(32/5, 6/5, 1).

        We repeat the above process using Y_l and Y' as in the proof of Theorem 3.27. The next line of reflection is the perpendicular bisector of segment Y_lY'. The midpoint of segment Y_lY' is (28/5, 19/5, 1). We find line Y_lY',

 

Evaluating the determinant, we find the line Y_lY' is [2, 1, –15]. A line perpendicular to line Y_lY' has the form m[1, –2, m₃]. Since the midpoint is on the line m[1, –2, m₃],

 

That is, m₃ = 2. Hence, m[1, –2, 2] is the perpendicular bisector of segment Y_lY'. We find the matrix of a direct isometry T that maps h[0, 1, 0] to m[1, –2, 2]. There is a nonzero real number k such that kmT = h.

 

 

implies

 

Hence,    and a = 2b – 2. For a matrix of a direct isometry that maps line h to line m, let a = 0 and b = 1,

 

Hence, R_m is defined by

 

 

The image of X' is X', Y_l(26/5, 23/5, 1) is Y'(6, 3, 1) and Z_l(32/5, 6/5, 1) is Z_m( 4, 6, 1).

        We repeat the above process using Z_m and Z' as in the proof of Theorem 3.27. The next line of reflection is the perpendicular bisector of segment Z_mZ'. The midpoint of segment Z_mZ' is (4, 3, 1). We find line Z_mZ',

 

Evaluating the determinant, we find the line Z_mZ' is [1, 0, –4]. A line perpendicular to line Z_mZ' has the form n[0, 1, n₃]. Since the midpoint is on the line n[0, 1, n₃],

 

That is, n₃ = –3. Hence, n[0, 1, –3] is the perpendicular bisector of segment Z_mZ'. We find the matrix of a direct isometry T that maps h[0, 1, 0] to n[0, 1, –3]. There is a nonzero real number k such that knT = h.

 

 

implies

 

Hence, k = 1,   and b = 3. For a matrix of a direct isometry that maps line h to line n, let a = 0,

 

Hence, R_n is defined by

 

 

The image of X' is X', Y' is Y', and Z_m( 4, 6, 1) is Z'(4, 0, 1).

        Hence, the product R_nR_mR_l  defined by

 

maps X to X', Y to Y', and Z to Z'.