Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.5 - Tangential and Normal Components of Acceleration - Exercises 13.5 - Page 771: 2

$a=0 \ T + 0 \ N$

We calculate the velocity and acceleration as follows: $v(t)=\dfrac{dr}{dt}=3i+j-3k \implies |v(t)|=\sqrt {3^2+1^2+(-3)^2}=\sqrt {19}$ and $a(t)=\dfrac{d \ v(t)}{dt}= 0 $ $|a(t)|=\sqrt {0} \implies |a(t)|=0$ Now, $a_{N}=\sqrt {|a|^2 -a^2_{T}}=\sqrt {0^2-0^2}=0$ So, $a=a_T T+a_{N}=0 \ T + 0 \ N$