Question 9.3.3: At the point corresponding to t = π/2 on the circular helix ......
At the point corresponding to t=\pi / 2 on the circular helix in Example 2, find an equation of (a) the osculating plane, (b) the normal plane, and (c) the rectifying plane.
From \mathrm{r}(\pi / 2)=\langle 0,2,3 \pi / 2\rangle the point P in question is (0,2,3 \pi / 2).
(a) A normal vector to the osculating plane at P is
\mathrm{B}(\pi / 2)=\mathrm{T}(\pi / 2) \times \mathrm{N}(\pi / 2)=\frac{3}{\sqrt{13}} \mathrm{i}+\frac{2}{\sqrt{13}} \mathrm{k}.
To find an equation of a plane we do not require a unit normal, so in lieu of \mathrm{B}(\pi / 2) it is a bit simpler to use \langle 3,0,2\rangle. From (11) of Section 7.5 an equation of the osculating plane is
a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0 . (11)
3(x-0)+0(y-2)+2\left(z-\frac{3 \pi}{2}\right)=0 \quad \text { or } \quad 3 x+2 z=3 \pi.
(b) At the point P, the vector \mathrm{T}(\pi / 2)=\frac{1}{\sqrt{13}}\langle-2,0,3\rangle or \langle-2,0,3\rangle is normal to the plane containing \mathrm{N}(\pi / 2) and \mathrm{B}(\pi / 2). Hence an equation of the normal plane is
-2(x-0)+0(y-2)+3\left(z-\frac{3 \pi}{2}\right)=0 \quad \text { or } \quad-4 x+6 z=9 \pi.
(c) Finally, at the point P, the vector \mathrm{N}(\pi / 2)=\langle 0,-1,0\rangle is normal to the plane containing \mathrm{T}(\pi / 2) and \mathrm{B}(\pi / 2). An equation of the rectifying plane is
0(x-0)+(-1)(y-2)+0\left(z-\frac{3 \pi}{2}\right)=0 \quad \text { or } \quad y=2.
