## University Calculus: Early Transcendentals (3rd Edition)

$2x+2y+z=6$ and $x=1+2t; y=1+2t; z=2+t$
The vector equation can be calculated as: $\nabla f(r_0) \cdot (r-r_0)=0$ The equation of the tangent for $\nabla f( 1,1,2)=\lt 2,2,1 \gt$ is: $2(x-1)+2(y-1)+1(z-2)=0$ or, $2x-2+2y-2+z-2 =0 \implies 2x+2y+z=6$ Now, the parametric equations can be written as: $r-r_0+\nabla f(r_0) t$ for $\nabla f( 1,1,2)=\lt 2,2,1 \gt$: Thus, $x=1+2t; y=1+2t; z=2+t$ Hence, $2x+2y+z=6$ and $x=1+2t; y=1+2t; z=2+t$