AMU 2026 — Mathematics PYQ

1. Identify the Component Functions

The given vector field is:

$$\vec{F}=F_1\hat{i}+F_2\hat{j}+F_3\hat{k}$$

Comparing components, we have:

• $F_1=ax+y+a$

• $F_2=1$

• $F_3=-(x+y)=-x-y$

2. Compute $\text{Curl }\vec{F}$

The curl of a vector field is calculated using the determinant of a $3\times 3$ matrix:

$$\text{Curl }\vec{F}=\nabla \times \vec{F}=\left|\begin{array}{ccc}\hat{i}& amp;\hat{j}& amp;\hat{k}\\ \frac{\partial }{\partial x}& amp;\frac{\partial }{\partial y}& amp;\frac{\partial }{\partial z}\\ F_1& amp;F_2& amp;F_3\end{array}\right|$$

Expanding the determinant along the first row:

$$\text{Curl }\vec{F}=\hat{i}\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)-\hat{j}\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)+\hat{k}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)$$

Let's evaluate each bracket carefully using partial derivatives:

• $\hat{i}$ component:

  $$\frac{\partial }{\partial y}(-x-y)-\frac{\partial }{\partial z}(1)=-1-0=-1$$

• $\hat{j}$ component:

  $$-\left[\frac{\partial }{\partial x}(-x-y)-\frac{\partial }{\partial z}(ax+y+a)\right]=-[-1-0]=1$$

• $\hat{k}$ component:

  $$\frac{\partial }{\partial x}(1)-\frac{\partial }{\partial y}(ax+y+a)=0-1=-1$$

Assembling the parts gives:

$$\text{Curl }\vec{F}=-1\hat{i}+1\hat{j}-1\hat{k}$$

3. Apply the Condition $\vec{F}\cdot \text{Curl }\vec{F}=0$

Now, calculate the scalar dot product of the original vector field $\vec{F}$ and our freshly computed $\text{Curl }\vec{F}$:

$$\vec{F}\cdot \text{Curl }\vec{F}=(F_1)(-1)+(F_2)(1)+(F_3)(-1)=0$$

Substitute the component terms into the equation:

$$(ax+y+a)(-1)+(1)(1)+(-x-y)(-1)=0$$

Distribute the negative signs and simplify:

$$-ax-y-a+1+x+y=0$$

Notice that the $-y$ and $+y$ terms cancel each other out completely:

$$-ax-a+1+x=0$$

4. Solve for Constant '$a$'

Rearrange and group the remaining variables and constants:

$$(1-a)x+(1-a)=0$$

$$(1-a)(x+1)=0$$

For this equation to hold true universally across the entire vector field independent of the coordinates $(x,y,z)$, the coefficient involving '$a$' must equal zero:

$$1-a=0⟹a=1$$

Conclusion

The value of the constant '$a$' that satisfies the given condition is $1$.

This directly matches option (c).