(1)
v1v2
= (x1i+y1j+z1k)(x2i+y2j+z2k)
= -(x2 + y2 + z2)+(y1z2-z1y2)i + (z1x2 - x1z2)j +(x1y2-y1x2)k
= -(v1 ·v2) + (v1 ×v2).
Thus the product of pure quaternions combines the scalar product and the vector product of the equivalent vectors.

(2) For any quaternion v=x i + y j + z k where |v| = Ö(x2 + y2 + z2) = 1 we have
v2
=-(v1 ·v1) + (v1 ×v1)
= -(x2 + y2 + z2) + 0
= -1
.
For all values of q and f there are quaternions v = cosqcosfi + cosqsinfj + sinqk and by elementary trig:
cos2 qcos2 f+ sin2 qcos2 f+ sin2 f
= (cos2 q+ sin2 q)cos2 f+ sin2 f
= cos2 f+sin2 f
=1
so there are infinitely many quaternions whose square is -1.

(3) (i)As u0 is a point on the plane, by the rules of quaternion multiplication:

u0n = -u0·n + u0 ×n = u0 ×n

n u0 = - n ·u0 + n ×u0 = n ×u0.

Hence u0n = - n u0 so F(u0) = n u0 n = n(-n u0) = - n2u0 = u0.

(ii) Here we shall use n2 = -1:
F(u0+t n) = F(u0) + tF(n) = u0 + t n n n = u0 -t n.