Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 15

$x=1-2t; y=1; z=\dfrac{1}{2}+2t$

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