Tangent, Normal and Binormal Vectors — Question 9

Question 9 – Solution

Strategy. Normalizing a nonzero vector multiplies it by a positive scalar, which does not change whether its dot product with $\mathbf a$ is zero.

Step 1: Tangent direction $$\mathbf r'(t)=\left\langle -2\sin t,2\cos t,1\right\rangle, \qquad \|\mathbf r'(t)\|=\sqrt 5.$$ Thus $\mathbf T\cdot\mathbf a=0$ exactly when $\mathbf r'\cdot\mathbf a=0$.

Step 2: Solve $$\begin{align*} \mathbf r'(t)\cdot\mathbf a &=-2\sin t+2\cos t=0\\ &\Longrightarrow\quad \sin t=\cos t\\ &\Longrightarrow\quad \tan t=1. \end{align*}$$ On $[0,2\pi)$, $$\boxed{t=\frac\pi 4,\ \frac{5\pi}4}.$$ Neither solution is lost by using tangent because $\cos t\ne 0$ at both values.

Step 3: Points $$\mathbf r\left(\frac\pi 4\right) =\boxed{\left\langle \sqrt 2,\sqrt 2,\tfrac\pi 4\right\rangle},$$ and $$\mathbf r\left(\frac{5\pi}4\right) =\boxed{\left\langle -\sqrt 2,-\sqrt 2,\tfrac{5\pi}4\right\rangle}.$$ At either parameter, substituting into $-2\sin t+2\cos t$ gives zero, verifying perpendicularity.