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 asks to identify correct vector calculus identities. We need to evaluate each statement based on standard definitions and theorems.
Step 2: Detailed Explanation:

(A) div (grad f) = \( \nabla^2 f \): \( \text{grad} f = \nabla f = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k} \). \( \text{div} (\text{grad} f) = \nabla \cdot (\nabla f) = \frac{\partial}{\partial x}(\frac{\partial f}{\partial x}) + \frac{\partial}{\partial y}(\frac{\partial f}{\partial y}) + \frac{\partial}{\partial z}(\frac{\partial f}{\partial z}) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} = \nabla^2 f \). This is the definition of the Laplacian operator. Statement (A) is correct.
(B) curl curl \( \vec{F} \) = grad curl \( \vec{F} \) - \( \nabla^2 \vec{F} \): The correct vector triple product identity is \( \nabla \times (\nabla \times \vec{F}) = \nabla(\nabla \cdot \vec{F}) - (\nabla \cdot \nabla)\vec{F} \). This translates to \( \text{curl curl} \vec{F} = \text{grad}(\text{div} \vec{F}) - \nabla^2 \vec{F} \). The statement given is "grad curl \( \vec{F} \)", which is incorrect. The gradient of a curl is not a standard operation, and the identity requires the gradient of the divergence. Statement (B) is incorrect.
(C) div curl \( \vec{F} \) = \( \vec{0} \): The divergence of the curl of any vector field is always zero, i.e., \( \nabla \cdot (\nabla \times \vec{F}) = 0 \). The result should be a scalar zero, not a zero vector. So, div curl \(\vec{F} = 0\). The statement shows \( \vec{0} \), which is a vector. This is a subtle but important distinction. Assuming it means the scalar zero, the identity is correct. However, compared to other options, this might be considered incorrect due to the vector notation. Let's hold this.
(D) curl grad f = \( \vec{0} \): The curl of the gradient of any scalar field is always the zero vector, i.e., \( \nabla \times (\nabla f) = \vec{0} \). This is a fundamental identity. Statement (D) is correct.
(E) div (\(f\vec{F}\)) = f div \( \vec{F} \) + (grad f) \( \times \vec{F} \): This is a product rule. \( \nabla \cdot (f\vec{F}) = f(\nabla \cdot \vec{F}) + (\nabla f) \cdot \vec{F} \). The identity is \( \text{div}(f\vec{F}) = f \text{ div} \vec{F} + (\text{grad} f) \cdot \vec{F} \). The statement uses a cross product \( (\times) \) instead of a dot product \( (\cdot) \). Statement (E) is incorrect.
The definitively correct statements are (A) and (D). Statement (C) is correct in spirit (\( \nabla \cdot (\nabla \times \vec{F}) = 0 \)) but uses incorrect notation (\(\vec{0}\) instead of 0). Given the options, the combination of just (A) and (D) is offered. Step 3: Final Answer:
The correct statements are (A) and (D). This corresponds to option (B).