Tangent, Normal and Binormal Vectors — Question 2

Question 2 – Solution

Strategy. The helix has constant speed, which makes each normalization especially transparent.

Step 1: Unit tangent $$\mathbf r'(t)=\left\langle -3\sin t,3\cos t,4\right\rangle,\qquad \|\mathbf r'(t)\|=\sqrt{9\sin^2t+9\cos^2t+16}=5.$$ Thus $$\boxed{\mathbf T(t)=\left\langle -\tfrac 35\sin t,\tfrac 35\cos t,\tfrac 45\right\rangle}.$$

Step 2: Principal unit normal $$\mathbf T'(t)=\left\langle -\tfrac 35\cos t,-\tfrac 35\sin t,0\right\rangle, \qquad \|\mathbf T'(t)\|=\frac 35.$$ Therefore $$\boxed{\mathbf N(t)=\left\langle -\cos t,-\sin t,0\right\rangle}.$$

Step 3: Binormal $$\begin{align*} \mathbf B(t)&=\mathbf T(t)\times\mathbf N(t)\\ &=\boxed{\left\langle \tfrac 45\sin t,-\tfrac 45\cos t,\tfrac 35\right\rangle}. \end{align*}$$ Its norm is $1$, and direct dot products give $\mathbf T\cdot\mathbf N=\mathbf T\cdot\mathbf B=\mathbf N\cdot\mathbf B=0$. The defining cross product confirms the right-handed orientation.