Finding the Angle Between a Tangent Vector to a Helix and the Direction \(\mathbf{k}\)

Show that the acute angle between the tangent vector to the helix \[ \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}\quad 0\,{\leq}\,t\,{\leq}\,2\pi \]

and the direction \(\mathbf{k}\) is \(\dfrac{\pi }{4}\) radian.

Solution A tangent vector at any point on the helix is given by \begin{equation*} \mathbf{r}^{\prime} (t)=-\sin t\mathbf{i}+\cos t\mathbf{j}+\mathbf{k} \end{equation*}

Then \(\left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;= \sqrt{(-\sin t)^2 + \cos^2 t +1} = \sqrt{1+1} = \sqrt{2}.\)

The cosine of the acute angle \(\theta \) between \(\mathbf{r}^{\prime} (t)\) and \(\mathbf{k}\) is \begin{eqnarray*} &&\cos \theta =\frac{\mathbf{r}^{\prime} (t)\,{\cdot}\, \mathbf{k}}{\left\Vert \mathbf{r} ^{\prime} (t)\right\Vert \left\Vert \mathbf{k}\right\Vert }=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\qquad \color{#0066A7}{\hbox{$\mathbf{r}^{\prime} (t)\,{\cdot}\, \mathbf{k} = 1$}}\\ \end{eqnarray*}

So, \(\theta =\dfrac{\pi }{4}\) radian. See Figure 10.