Answer
a) $x-3y-z=-1$
b) $x=1+t,y=1-3t; z=-1-t$
Work Step by Step
a) Since, we have the vector equation $r(x,y,z)=r_0+t \nabla f(r_0)$
The equation of the tangent line is: $\nabla f(1,1,-1)=\lt 1,-3,-1 \gt$
Thus, $1(x-1)-3(y-1)-1(z+1)=0$
or, $x-3y-z=-1$
b) Since, we have the vector equation $r=r_0+t \nabla f(r_0)$
Now, the parametric equations are:
$x=1+t,y=1-3t; z=-1-t$
