## Calculus (3rd Edition)

We have $$\|r(t)\|=\sqrt{9\cos^2t +25\sin^2t+16\cos^2t}=25.$$ So, $\|r(t)\|$ is constant. This means that $r(t)$ is orthogonal to $r'(t)$ since $\frac{d}{dt} \|r(t)\|^2=2r(t)\cdot r'(t)=0$. Moreover, $r'(t)=\lt -3\sin t, 5\cos t, -4\sin t\gt$ Hence we have $$r(t)\cdot r'(t)=-9\sin t \cos t +25 \sin t \cos t -16 \sin t \cos t=0$$ That is, $r(t)$ is orthogonal to $r'(t)$.