Question

Let \( f : \mathbb{R}^3 \to \mathbb{R} \) be a scalar field, \( \vec{v} : \mathbb{R}^3 \to \mathbb{R}^3 \) be a vector field and let \( \vec{a} \in \mathbb{R}^3 \) be a constant vector. If \( \vec{r} \) represents the position vector \( x\hat{i} + y\hat{j} + z\hat{k} \), then which one of the following is FALSE?

• A. \( \text{curl}(\vec{f} \vec{v}) = \text{grad(f)} \times \vec{v} + f \text{curl}(\vec{v}) \)
• B. \( \text{div}(\text{grad(f)}) = (\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + \frac{\partial^2 }{\partial z^2})f \)
• C. \( \text{curl}(\vec{a} \times \vec{r}) = 2|\vec{a}| \vec{r} \)
• D. \( \text{div} \left( \frac{\vec{r}}{|\vec{r}|^3} \right) = 0, \text{for} \vec{r} \neq \vec{0} \)

Hint:

Use vector calculus identities such as the product rule for curl and the divergence of the gradient to solve these types of questions.

Answer 1

The Correct Option is C

To evaluate which statement is FALSE among the given options, let's analyze each of them using vector calculus identities and properties.

1. Option 1: \(\text{curl}(\vec{f} \vec{v}) = \text{grad(f)} \times \vec{v} + f \text{curl}(\vec{v})\)
   This is a vector calculus identity that applies to the curl of the product of a scalar function \(f\) and a vector field \(\vec{v}\). It holds true as per the vector calculus rules. Thus, this statement is true.
2. Option 2: \(\text{div}(\text{grad(f)}) = \left(\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + \frac{\partial^2 }{\partial z^2}\right)f\)
   This expression represents the Laplacian of a scalar field \(f\) and is accurate. It evaluates to the divergence of the gradient, which matches the definition of the Laplacian. Hence, this statement is true.
3. Option 3: \(\text{curl}(\vec{a} \times \vec{r}) = 2|\vec{a}| \vec{r}\)
   For this identity, we apply the vector triple product identity: \(\text{curl}(\vec{a} \times \vec{r}) = \vec{a}(\nabla \cdot \vec{r}) - (\vec{a} \cdot \nabla)\vec{r}\). Since \(\nabla \cdot \vec{r} = 3\) for \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\), the result is not \(2|\vec{a}| \vec{r}\) but rather \(3\vec{a}\). This makes the statement false.
4. Option 4: \(\text{div} \left( \frac{\vec{r}}{|\vec{r}|^3} \right) = 0, \text{for } \vec{r} \neq \vec{0}\)
   This is the divergence of a vector field known as the inverse square law field. Its divergence is zero everywhere except at the origin where it is singular. Therefore, it is valid and true under the given condition \(\vec{r} \neq \vec{0}\).

Based on this analysis, the FALSE statement is clearly Option 3: \(\text{curl}(\vec{a} \times \vec{r}) = 2|\vec{a}| \vec{r}\).