Answer
$x=1+t \\ y =t \\ z=-t$
Work Step by Step
$v(t)=\dfrac{dr}{dt}$
or, $v(t)= e^t i+\cos t j -\dfrac{k}{1-t}$
We have: $t=0$
This implies that $v(t)=\dfrac{dr}{dt}= e^{0} i+\cos (0) j -\dfrac{k}{1-0}=i+j-k$
and $r(0)=e^{0} i+\sin (0) j +\ln (1-0) k =i$
So, the parametric equations of the line at $(1,0,0)$ can be expressed as follows:
$x=1+t \\ y =t \\ z=-t$
