Finding the Angle Between Velocity and Acceleration Vectors

[BGriffin]

Question : r(t) is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0.

r(t)=(3t+1)i + (sqrt(3t))j + (t²)k

I found the velocity and acceleration vectors. They are ;

v(t)= (3)i + (sqrt(3)/(2sqrt(t)))j + (2t)k

a(t)= (0)i + (-sqrt(3)/(4t^3/2))j + (2)k

I don't know how to find the angle between them. Can anybody help me?

 

[pka]

Question : r(t) is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0. r(t)=(3t+1)i + (sqrt(3t))j + (t²)k
I don't know how to find the angle between them.

\(\displaystyle \arccos\left(\dfrac{v(0)\cdot a(0)}{\|v(0)\|\|a(0)\|}\right)\)

 

[BGriffin]

So

v(0) = 3i
a(0) = 2k

then a dot b = a1b1 + a2b2 + a3b3 which means a dot b = 0 ?

arccos(0) = 90 = pi/2

is this right?