Answer
$x=1-t,y=t,z=1-t$
Work Step by Step
Given: $r(t)= e^{-t}\cos t i+e^{-t}\sin t+e^{-t} k $
In order to find tangent vector $r'(t)$ we will have to take the derivative of $r(t)$ with respect to $t$.
Thus, $r'(t)=e^{-t}(-\cos t- \sin t) i+e^{-t}(-\sin t+ \cos t)-e^{-t} k $
$r'(0)=e^{-0}(-\cos(0)- \sin(0)) i+e^{-0}(-\sin (0)+ \cos (0))-e^{-(0)} k =-i+j-k$
Equation of tangent line passing through the points $(1,0,1)$ is:
$r(t)=\lt 1-t,t,1-t \gt$
Therefore, the parametric equations of the line are:
$x=1-t,y=t,z=1-t$
