Tangent, Normal and Binormal Vectors — Question 6

Question 6 – Solution

Strategy. The cross product $\mathbf r'\times\mathbf r''$ gives the binormal direction without differentiating a normalized vector.

Step 1: Tangent $$\|\mathbf r'(t_0)\|=\sqrt{1+4+4}=3, \qquad \boxed{\mathbf T(t_0)=\frac 13\left\langle 1,2,2\right\rangle}.$$

Step 2: Binormal direction $$\begin{align*} \mathbf r'\times\mathbf r'' &=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&2&2\\2&0&1\end{vmatrix} =\left\langle 2,3,-4\right\rangle. \end{align*}$$ Its length is $\sqrt{4+9+16}=\sqrt{29}$, so $$\boxed{\mathbf B(t_0)=\frac 1{\sqrt{29}}\left\langle 2,3,-4\right\rangle}.$$

Step 3: Principal normal In a right-handed frame, $\mathbf B\times\mathbf T=\mathbf N$. Thus $$\begin{align*} \mathbf N(t_0) &=\frac 1{3\sqrt{29}}\left\langle 2,3,-4\right\rangle\times\left\langle 1,2,2\right\rangle\\ &=\boxed{\frac 1{3\sqrt{29}}\left\langle 14,-8,1\right\rangle}. \end{align*}$$ Its squared numerator length is $196+64+1=261=9(29)$, so it has unit length. Direct dot products with $\mathbf T$ and $\mathbf B$ are zero, verifying the frame.