https://eater.net/quaternions/video/intro
计算f(i)的坐标
q = a+bi+cj+dk
q*i*q^(-1)= (-b + ai + dj - ck)(a-bi-cj-dk) （单位四元数的逆等价于共轭）
= (-ab + (a^2)i + adj - ack) +
[(b^2)*i + ab -bdji +bcki ] +
[bcj + -acij +cd +(c^2)kj] +
[bdk - adik - (d^2)jk - cd]
= (-ab + (a^2)i + adj - ack) +
[(b^2)*i + ab + bdk +bcj ] +
[bcj + -ack +cd +(c^2)(-i)] +
[bdk - ad(-j) - (d^2)i - cd]
= (-ab + (a^2)i + adj - ack) +
[ab + (b^2)*i +bcj + bdk ] +
[cd - (c^2)i + bcj -ack ] +
[- cd - (d^2)i + adj + bdk]
= 0 + (a^2 + b^2 - c^2 - d^2)i + (2ad+ 2bc)j + (2bd - 2ac)k
a=0.92 d= 0.38 b=c =0
= 0 + (0.8464 - 0.1444)i + (2*0.92*0.38)j +0k
= 0 + 0.702i + 0.6992j +0k
a=0.85 b=0.35 c=0.19 d=0.35
0 + (0.7225+0.1225-0.0361-0.1225)i +(0.595+0.133)j+(0.245 - 0.323)k =
0 + 0.6864i +0.728j + -0.078k