Answer
We use Theorem 1 to show that ${\bf{F}} = \left( {3,1,2} \right)$ is conservative.
In general, we prove that any constant vector field ${\bf{F}} = \left( {a,b,c} \right)$ is conservative.
Work Step by Step
We have the vector field: ${\bf{F}} = \left( {3,1,2} \right)$.
Suppose there is a function $f\left( {x,y,z} \right) = 3x + y + z$ such that
$\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right) = \left( {3,1,2} \right)$
By definition, $f$ is a potential function for ${\bf{F}} = \left( {3,1,2} \right)$ and ${\bf{F}} = \left( {3,1,2} \right)$ is conservative.
Similarly for a constant vector field ${\bf{F}} = \left( {a,b,c} \right)$.
Suppose there is a function $f\left( {x,y,z} \right) = ax + by + cz$ such that
$\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right) = \left( {a,b,c} \right)$
By definition, $f$ is a potential function for ${\bf{F}} = \left( {a,b,c} \right)$ and ${\bf{F}} = \left( {a,b,c} \right)$ is conservative.
