Question

The value of \(a\), \(b\) and \(c\) such that \(\vec{F} = (3x - 4y + az)\hat{i} + (cx - 5y - 2z)\hat{j} + (x - by + 7z)\hat{k}\) is irrotational, are respectively:

• A. \(1,\ 2,\ -4\)
• B. \(-4,\ 2,\ 1\)
• C. \(2,\ 1,\ -4\)
• D. \(-4,\ 1,\ 2\)

Hint:

Irrotational means \(\nabla \times \vec{F} = 0\); equate the mixed partials in each curl component.

Answer 1

The Correct Option is A

Step 1: A vector field is irrotational when its curl vanishes, so we set \(\nabla \times \vec{F} = \vec{0}\). With \(F_x = 3x - 4y + az\), \(F_y = cx - 5y - 2z\), and \(F_z = x - by + 7z\), each component of the curl must be zero.

Step 2: The \(\hat{i}\) component of the curl is \(\dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z}\). Here \(\dfrac{\partial F_z}{\partial y} = -b\) and \(\dfrac{\partial F_y}{\partial z} = -2\), so \(-b - (-2) = 0\), giving \(b = 2\).

Step 3: The \(\hat{j}\) component is \(\dfrac{\partial F_x}{\partial z} - \dfrac{\partial F_z}{\partial x}\). Here \(\dfrac{\partial F_x}{\partial z} = a\) and \(\dfrac{\partial F_z}{\partial x} = 1\), so \(a - 1 = 0\), giving \(a = 1\).

Step 4: The \(\hat{k}\) component is \(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y}\). Here \(\dfrac{\partial F_y}{\partial x} = c\) and \(\dfrac{\partial F_x}{\partial y} = -4\), so \(c - (-4) = 0\), giving \(c = -4\).

Step 5: Collecting the results in the order \(a, b, c\): \[\boxed{a = 1,\quad b = 2,\quad c = -4}\]