Find Tangent Line to Path: x(t)=4costi-3sintj+5tk, t=\pi/3

[rgalvan2]

Find an equation for the line tangent to the given path at the indicated value for the parameter.

x(t)=4costi-3sintj+5tk, t=\pi/3

So what I did here was take x'(t) and then plugged in \pi/3 after that to get an equation containing i, j, and k.
x'(t)= -4sinti-3costj+5k
x'(\pi/3)=-4\sqrt{3}/2i-3/2j+5k
Am I doing this right? Calc III is confusing me so much. Any help would be appreciated. This is due Thursday. Thanks!

 

[Dick]

Looks all right so far.

 

[rgalvan2]

Alright sweet thanks a lot!