Answer
a) $2y+3z=7$
b) $x=1,y=-1+4t; z=3+6t$
Work Step by Step
a. The vector equation is given as: $r(x,y,z)=r_0+t \nabla f(r_0)$
The equation of tangent line is given as:$\nabla f(1,-1,3)=\lt 0,4,6 \gt$
Now, $(0)(x-1)+(4)(y+1)+(6)(z-3)=0 \implies 4y+6z=14$
Thus, $2y+3z=7$
b. The vector equation is given as: $r(x,y,z)=r_0+t \nabla f(r_0)$
we have the parametric equations as follows:
$x=1+0t=1,y=-1+4t; z=3+6t$
Thus, $x=1,y=-1+4t; z=3+6t$
