## Thomas' Calculus 13th Edition

$x=1; y=1+2t; z=1-2t$
The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ Given: $f(x,y,z)=x+y^2+z-4$ The equation of tangent line $v=0 i +2j-2k$ is given as follows: Now, we have the parametric equations for $\nabla f(1,1,1)=\lt 1,2,2 \gt$ are: $x=1; y=1+2t; z=1-2t$