The vector function expressed by $F=a_x\left(5y-k_{1}z\right)+a_y\left(3z+k_{2}x\right)+a_z\left(k_{3}y-4x\right)$ Represents a conservative field, where ax, ay, az are unit vectors along x, y and z directions, respectively. The values of constant k₁, k₂, k₃ are given by:

The correct option is k₁ = 4, k₂ = 5, k₃ = 3.

Step-by-Step Explanation:

A vector field $F=F_x{a}_x+F_y{a}_y+F_z{a}_z$ is said to be conservative if its curl is identically zero, i.e.,

$\nabla <!--\; \nabla \; -->\times F=0$

From the given vector field expression, we extract the individual scalar components:

$F_x=5y-k_{1}z$

$F_y=3z+k_{2}x$

$F_z=k_{3}y-4x$

Now, let us calculate each component of the curl of $F$ and set them equal to zero:

1. X-component of the curl:

$\frac{\partial <!--\; \partial \; -->F_z}{\partial <!--\; \partial \; -->y}-\frac{\partial <!--\; \partial \; -->F_y}{\partial <!--\; \partial \; -->z}=0$

Substituting the partial derivatives:
$k_3-3=0$
⇒ k₃ = 3

2. Y-component of the curl:

$\frac{\partial <!--\; \partial \; -->F_x}{\partial <!--\; \partial \; -->z}-\frac{\partial <!--\; \partial \; -->F_z}{\partial <!--\; \partial \; -->x}=0$

Substituting the partial derivatives:
$-k_1-(-4)=0$
⇒ $-k_1+4=0$
⇒ k₁ = 4

3. Z-component of the curl:

$\frac{\partial <!--\; \partial \; -->F_y}{\partial <!--\; \partial \; -->x}-\frac{\partial <!--\; \partial \; -->F_x}{\partial <!--\; \partial \; -->y}=0$

Substituting the partial derivatives:
$k_2-5=0$
⇒ k₂ = 5

Therefore, the values of the constants for the field to be conservative are k₁ = 4, k₂ = 5, k₃ = 3.