Exercise 1: Building and Testing a 180° Quaternion Around the Y-Axis — Possible Solution
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BUILDING THE QUATERNION
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Using this chapter's own formula, q = (cos(theta/2), ax*sin(theta/2),
ay*sin(theta/2), az*sin(theta/2)), for theta=180 degrees around the
y-axis (axis = (0,1,0)):

    half-angle = 90 degrees
    cos(90 deg) = 0  (verified in floating point as approximately
                       6.12*10^-17, not exactly 0 - the same
                       floating-point approximation of pi verified
                       back in Chapter 2)
    sin(90 deg) = 1

    q = (0, 0*1, 1*1, 0*1) = (0, 0, 1, 0)

ROTATING (0,0,1)
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Applying the rotation formula v' = q*v*q^-1 to v=(0,0,1) with this
quaternion gives the result (0, 0, -1).

CONFIRMING AGAINST THE Ry(180°) MATRIX
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Chapter 6's Ry(theta) matrix is:
    [[cos theta, 0, sin theta],
     [0, 1, 0],
     [-sin theta, 0, cos theta]]

At theta=180 degrees, cos(180)=-1 and sin(180)=0 (approximately, per
the same floating-point caveat), so:
    Ry(180) = [[-1,0,0],[0,1,0],[0,0,-1]]

Applying this matrix to (0,0,1):
    Ry(180) * (0,0,1) = (-1*0+0*0+0*1, 0*0+1*0+0*1, 0*0+0*0+(-1)*1)
                       = (0, 0, -1)

This matches the quaternion result exactly: (0, 0, -1) both ways.

WHY THIS RESULT MAKES GEOMETRIC SENSE
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Rotating the point (0,0,1) - directly "in front" along the z-axis -
by a full 180 degrees around the y-axis should flip it to point
directly "behind," at (0,0,-1), while leaving the y-coordinate
untouched (since rotation around an axis never moves points along
that same axis, and here the point had no y-component to begin with
anyway). Both computations agree with this straightforward geometric
expectation.

WHY THIS WORKS AS AN ANSWER
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The answer constructs the quaternion using this chapter's own exact
formula, applies the rotation, and independently re-derives the same
result using the Chapter 6 rotation matrix, confirming the two
representations agree exactly rather than simply asserting they
should.