Answer
$x=\sqrt 2-2\sqrt 2t; y=\sqrt 2 +2\sqrt 2t; z=4$
Work Step by Step
The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$
Given: $f(x,y,z)=x^3+3x^2+y^3+4xy-z^3=0$
The equation of tangent line is $v=-2\sqrt 2i+2 \sqrt 2j+0k$
or, $v=\lt -2\sqrt 2,2\sqrt 2,0\gt$
Now, we have the parametric equations for $\nabla f(\sqrt 2, \sqrt 2, 4)=\lt -2\sqrt 2,2\sqrt 2,0\gt$ as follows:
$x=\sqrt 2-2\sqrt 2t; y=\sqrt 2 +2\sqrt 2t; z=4+0t=4$
Thus, $x=\sqrt 2-2\sqrt 2t; y=\sqrt 2 +2\sqrt 2t; z=4$
