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