Find the osculating plane and the curvature

[Bptrhp]

Homework Statement

Find the osculating plane and the curvature of the curve $r(t)=(a \cos(t)+b \sin(t), a \sin(t)+ b\cos(t), c \sin(2t)),t\in\mathbb{R}$

Relevant Equations

$B(t)=\dfrac{r'(t)\times r''(t)}{||r'(t)\times r''(t)||}$ and $\kappa=\dfrac{||r'(t)\times r''(t)||}{||r'(t)||^3}.$

I know the osculating plane is normal to the binormal vector $B(t)=(a,b,c)$. And since the point on which I am supposed to find the osculating plane is not given, I'm trying to find the osculating plane at an arbitrary point $P(x_0,y_0,z_0)$. So, if $R(x,y,z)$ is a point on the plane, the equation is given by:

$\langle (R-P),B(t)\rangle=0 \, \Rightarrow \,\langle(x-x_0,y-y_0,z-z_0),(a,b,c)\rangle=0\, \Rightarrow \, a(x-x_0)+b(y-y_0)+c(z-z_0)=0.$

The problem is, I tried to compute the binormal vector using the formula
\begin{align*}
B(t)=\dfrac{r'(t)\times r''(t)}{||r'(t)\times r''(t)||}
\end{align*}
but the result of the cross product between $r'(t)=(-a \sin(t)+b \cos(t),a \cos(t)-b \sin(t), 2c \cos(2t))$ and $r''(t)=(-a \cos(t)-b \sin(t), -a \sin(t)- b\cos(t), -4c \sin(2t))$ got very long.

For the curvature, I have a similar problem, since I'm trying to use the formula
\begin{align*}
\kappa=\dfrac{||r'(t)\times r''(t)||}{||r'(t)||^3}.
\end{align*}
I have no idea whether I'm on the right track. I appreciate any help!

 

[haruspex]

I found it a bit easier breaking up r as $(a,b,0)\cos(t)+(b,a,0)\sin(t)+(0,0,c)\sin(2t)$, and maintaining that through the differentiations. When it comes to the cross product, some combinations are easily discarded as zero.
Still pretty messy, though.