Answer
We explain why ${\bf{v}} = \nabla {F_P} \times \nabla {G_P}$ is a direction vector for the tangent line to $C$ at $P$.
Work Step by Step
By assumption, the intersection of the two surfaces $F\left( {x,y,z} \right) = 0$ and $G\left( {x,y,z} \right) = 0$ is a curve $C$. Let $P$ be a point on $C$.
By Theorem 5, the gradient $\nabla {F_P}$ of $F\left( {x,y,z} \right)$ is a vector normal to the tangent plane to the surface $F\left( {x,y,z} \right) = 0$ at $P$. Similarly, the gradient $\nabla {G_P}$ of $G\left( {x,y,z} \right)$ is a vector normal to the tangent plane to the surface $G\left( {x,y,z} \right) = 0$ at $P$.
Next, we evaluate the cross product such that ${\bf{v}} = \nabla {F_P} \times \nabla {G_P}$. By the property of cross product, ${\bf{v}}$ is perpendicular to both $\nabla {F_P}$ and $\nabla {G_P}$. It follows that ${\bf{v}}$ must reside in the tangent plane to both surfaces. Since $P$ is a point on $C$, ${\bf{v}}$ is parallel to the tangent line. Hence ${\bf{v}}$ is a direction vector for the tangent line to $C$ at $P$.
