We are asked to compute the divergence of the vector field \( \mathbf{a} \times (\mathbf{r} \times \mathbf{a}) \), where \( \mathbf{a} \) is a constant vector and \( \mathbf{r} \) is the position vector.
The vector triple product identity is given by:
\[ \mathbf{r} \times (\mathbf{a} \times \mathbf{b}) = (\mathbf{r} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{r} \cdot \mathbf{a}) \mathbf{b} \]
Substitute \( \mathbf{b} = \mathbf{a} \) into the identity to simplify the cross product \( \mathbf{r} \times (\mathbf{a} \times \mathbf{a}) \). Since the cross product of any vector with itself is zero, we get:
\[ \mathbf{r} \times (\mathbf{a} \times \mathbf{a}) = \mathbf{0} \]
Thus, we need to compute the divergence of the expression \( \mathbf{a} \times (\mathbf{r} \times \mathbf{a}) \). Using the vector identity for the divergence of a cross product:
\[ \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B}) \]
For a constant vector \( \mathbf{a} \), the divergence simplifies to:
\[ \text{div}[\mathbf{a} \times (\mathbf{r} \times \mathbf{a})] = 2a^2 \]
The correct answer is option (A), \( 2a^2 \).
Let a vector \[ \vec a=\sqrt{2}\,\hat i-\hat j+\lambda \hat k,\ \lambda>0, \] make an obtuse angle with the vector \[ \vec b=-\lambda^2\hat i+4\sqrt{2}\hat j+4\sqrt{2}\hat k \] and an angle \( \theta \), \(\frac{\pi}{6}<\theta<\frac{\pi}{2}\), with the positive \(z\)-axis. If the set of all possible values of \( \lambda \) is \((\alpha,\beta)-\{\gamma\}\), then\(\alpha+\beta+\gamma\) is equal to ____________.
Let $\vec a = 2\hat i + \hat j - 2\hat k$, $\vec b = \hat i + \hat j$ and $\vec c = \vec a \times \vec b$. Let $\vec d$ be a vector such that $|\vec d - \vec a| = \sqrt{11}$, $|\vec c \times \vec d| = 3$ and the angle between $\vec c$ and $\vec d$ is $\frac{\pi}{4}$. Then $\vec a \cdot \vec d$ is equal to