Answer
a) $x+y+z=3$ b) $x=1+2t,y=1+2t; z=1+2t$
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: $\nabla f(1,1,1)=\lt 2,2,2 \gt$
Thus, $2(x-1)+2(y-1)+2(z-1)=0$
or, $2x+2y+2z=6 \implies x+y+z=3$
b) Since, we have the vector equation $r=r_0+t \nabla f(r_0)$
Now, the parametric equations are: $\nabla f(1,1,1)=\lt 2,2,2 \gt$ are
$x=1+2t,y=1+2t; z=1+2t$
Hence, a) $x+y+z=3$ b) $x=1+2t,y=1+2t; z=1+2t$
