Show that the principal unit normal vector $\mathbf{N}(t)$ of the helix $$\begin{equation*} \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k} \end{equation*}$$

is orthogonal to the $z$-axis.

Solution We begin by finding $\mathbf{r^{\prime} }( t)$ and $\left\Vert \mathbf{r}^{\prime} ( t)\right \Vert.$ $$\begin{equation*} \mathbf{r}^{\prime} (t)=-\!\sin t\mathbf{i}+\cos t\mathbf{j}+\mathbf{k}\qquad \left\Vert \mathbf{r}^{\prime }(t)\right\Vert =\sqrt{ (-\!\sin t)^{2}+(\cos t)^{2}+1}=\sqrt{2} \end{equation*}$$

Then the unit tangent vector is $\mathbf{T}(t)=\dfrac{\mathbf{r}^{\prime }(t)}{||\mathbf{r}^{\prime }(t)||}=\dfrac{-\!\sin t\mathbf{i}+\cos t\mathbf{j}+\mathbf{k}}{\sqrt{2}}.$

Figure 14 $\mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t \mathbf{k}$

Since $$\begin{equation*} \mathbf{T}^{\prime} (t)=\dfrac{-\!\cos t\mathbf{i}-\sin t\mathbf{j}}{\sqrt{2} }\qquad \hbox{and}\qquad \left\Vert \mathbf{T}^{\prime }(t)\right\Vert =\sqrt{ \left( -\dfrac{\cos t}{\sqrt{2}}\right) ^{2}+\left( -\dfrac{\sin t}{\sqrt{2}} \right) ^{2}}=\dfrac{1}{\sqrt{2}} \end{equation*}$$

we have $$\begin{equation*} \mathbf{N}(t)=\dfrac{\mathbf{T}^{\prime} (t)}{\left\Vert \mathbf{T}^{\prime} (t)\right\Vert }=-\!\cos t\mathbf{i}-\sin t\mathbf{j} \end{equation*}$$

Since the direction of the $z$-axis is $\mathbf{k}$, it follows that $\mathbf{N}(t)\,{\cdot}\, \mathbf{k}=0$ for all $t.$ So, $\mathbf{N}(t)$ is always orthogonal to the $z$-axis, as shown in Figure 14.