Question

Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}, \, x, y, z \in \mathbb{R}. \) Which of the following is FALSE?

• A. \( \nabla(\vec{a} \cdot \vec{r}) = \vec{a} \)
• B. \( \nabla \cdot (\vec{a} \times \vec{r}) = 0 \)
• C. \( \nabla \times (\vec{a} \times \vec{r}) = \vec{a} \)
• D. \( \nabla \cdot ((\vec{a} \cdot \vec{r})\vec{r}) = 4(\vec{a} \cdot \vec{r}) \)

Hint:

When using vector identities, always apply the standard relations: \( \nabla \cdot (\vec{a} \times \vec{r}) = 0 \), \( \nabla \times (\vec{a} \times \vec{r}) = -2\vec{a} \).

Answer 1

The Correct Option is C

The problem involves vector calculus operations. Let's analyze each option to identify the false statement.

1. Given \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\), we start with the gradient of the dot product:
   • Apply the formula: \(\nabla(\vec{a} \cdot \vec{r}) = \nabla(x + y + z)\), as the dot product \(\vec{a} \cdot \vec{r} = x + y + z\).
   • The gradient is: \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\), which is consistent with the given statement: \(\nabla(\vec{a} \cdot \vec{r}) = \vec{a}\). Hence, this statement is true.
2. For the divergence of the cross product:
   • The cross product is: \(\vec{a} \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = (z-y)\hat{i} + (x-z)\hat{j} + (y-x)\hat{k}\).
   • The divergence is: \(\nabla \cdot (\vec{a} \times \vec{r}) = \frac{\partial}{\partial x}(z-y) + \frac{\partial}{\partial y}(x-z) + \frac{\partial}{\partial z}(y-x) = 0\).
   • This matches the given statement: \(\nabla \cdot (\vec{a} \times \vec{r}) = 0\), so it is true.
3. For the curl of the cross product, using the vector triple product identity:
   • Apply: \(\nabla \times (\vec{a} \times \vec{r}) = (\vec{r} \cdot \nabla)\vec{a} - (\vec{a} \cdot \nabla)\vec{r} + \vec{a}(\nabla \cdot \vec{r}) - \vec{r}(\nabla \cdot \vec{a})\).
   • Here, \((\nabla \cdot \vec{a}) = 0\) and \((\nabla \cdot \vec{r}) = 3\).
   • The expression simplifies to: \(\nabla \times (\vec{a} \times \vec{r}) = \vec{a}\) reduces to: \(- (\vec{a} \cdot \nabla)\vec{r}\), which does not simplify only to \(\vec{a}\).
   • This is inconsistent with the option given, confirming it's false.
4. For the divergence of a scalar multiplied vector:
   • Compute: \(\nabla \cdot ((\vec{a} \cdot \vec{r})\vec{r}) = \nabla \cdot ((x+y+z)(x\hat{i} + y\hat{j} + z\hat{k}))\).
   • Applying the product rule: \((\nabla \cdot \vec{F}) = (\nabla f \cdot \vec{r}) + (f(\nabla \cdot \vec{r}))\) where \(f = (\vec{a} \cdot \vec{r})\) and \(\vec{F} = f\vec{r}\) gives: \(4(x+y+z) = 4(\vec{a} \cdot \vec{r})\).
   • This matches \(\nabla \cdot ((\vec{a} \cdot \vec{r})\vec{r}) = 4(\vec{a} \cdot \vec{r})\), verifying it's true.

Thus, the false statement is \(\nabla \times (\vec{a} \times \vec{r}) = \vec{a}\).