Question

Let \( \vec{F} \) be the vector valued function and f be a scalar function. Let \( \nabla = \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \) then,
(A) div (grad f) = \( \nabla^2 f \)
(B) curl curl \( \vec{F} \) = grad curl \( \vec{F} \) - \( \nabla^2 \vec{F} \)
(C) div curl \( \vec{F} \) = \( \vec{0} \)
(D) curl grad f = \( \vec{0} \)
(E) div (\(f\vec{F}\)) = f div \( \vec{F} \) + (grad f) \( \times \vec{F} \)
Choose the correct answer from the options given below:

• A. (A), (B) and (C) only
• B. (A) and (D) only
• C. (A), (B) and (D) only
• D. (B), (D) and (E) only

Hint:

Memorize the two fundamental "zero" identities:
The curl of a gradient is always the zero vector: \( \nabla \times (\nabla f) = \vec{0} \).
The divergence of a curl is always zero: \( \nabla \cdot (\nabla \times \vec{F}) = 0 \).
Also, remember the "vector BAC-CAB rule" analogue: \( \nabla \times (\nabla \times \vec{F}) = \nabla(\nabla \cdot \vec{F}) - (\nabla \cdot \nabla)\vec{F} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question requires identifying valid vector calculus identities. Each statement must be assessed against established definitions and theorems.
Step 2: Detailed Explanation:

(A) div (grad f) = \( abla^2 f \): The gradient of a scalar function f is \( \text{grad} f = abla f = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k} \). The divergence of this gradient is \( \text{div} (\text{grad} f) = abla \cdot (abla f) = \frac{\partial}{\partial x}(\frac{\partial f}{\partial x}) + \frac{\partial}{\partial y}(\frac{\partial f}{\partial y}) + \frac{\partial}{\partial z}(\frac{\partial f}{\partial z}) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} \), which is the Laplacian, \( abla^2 f \). Statement (A) is correct.
(B) curl curl \( \vec{F} \) = grad curl \( \vec{F} \) - \( abla^2 \vec{F} \): The correct vector identity for the curl of a curl is \( abla \times (abla \times \vec{F}) = abla(abla \cdot \vec{F}) - (abla \cdot abla)\vec{F} \), which translates to \( \text{curl curl} \vec{F} = \text{grad}(\text{div} \vec{F}) - abla^2 \vec{F} \). The statement uses "grad curl \( \vec{F} \)", which is not a standard operation and deviates from the correct identity. Statement (B) is incorrect.
(C) div curl \( \vec{F} \) = \( \vec{0} \): The divergence of the curl of any vector field is always the scalar zero: \( abla \cdot (abla \times \vec{F}) = 0 \). The statement uses \( \vec{0} \), which denotes the zero vector. While the magnitude is zero, the result is a scalar, not a vector. If interpreted as the scalar zero, the identity holds. However, the notation is imprecise.
(D) curl grad f = \( \vec{0} \): The curl of the gradient of any scalar field is identically the zero vector: \( abla \times (abla f) = \vec{0} \). This is a fundamental identity. Statement (D) is correct.
(E) div (\(f\vec{F}\)) = f div \( \vec{F} \) + (grad f) \( \times \vec{F} \): The product rule for divergence states: \( abla \cdot (f\vec{F}) = f(abla \cdot \vec{F}) + (abla f) \cdot \vec{F} \). This means \( \text{div}(f\vec{F}) = f \text{ div} \vec{F} + (\text{grad} f) \cdot \vec{F} \). The statement incorrectly uses a cross product \( (\times) \) instead of the dot product \( (\cdot) \). Statement (E) is incorrect.
The definitively correct statements are (A) and (D). Statement (C) is conceptually correct but uses vector notation for a scalar result. Considering standard vector calculus identities, (A) and (D) are the unequivocally correct statements. Step 3: Final Answer
The correct statements are (A) and (D). This corresponds to option (B).