The problem involves two vectors a and b with magnitudes of 3, separated by an angle of $\theta = \frac{\pi}{3}$, both lying in the $xz$-plane. We need to calculate the cross product $\mathbf{a} \times \mathbf{b}$.

Solution:

1. The magnitude of the cross product is given by: $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin\theta$ Substituting $|\mathbf{a}| = |\mathbf{b}| = 3$ and $\theta = \frac{\pi}{3}$: $|\mathbf{a} \times \mathbf{b}| = 3 \cdot 3 \cdot \sin\left(\frac{\pi}{3}\right) = 9 \cdot \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{2}.$

2. Since both vectors lie in the $xz$-plane, their cross product will be perpendicular to this plane, pointing in the $y$-direction (using the right-hand rule). Thus: $\mathbf{a} \times \mathbf{b} = \frac{9\sqrt{3}}{2} \, \mathbf{\hat{y}}.$

Final Answer:

$\mathbf{a} \times \mathbf{b} = \frac{9\sqrt{3}}{2} \, \mathbf{\hat{y}}.$

Would you like to delve deeper into cross products or clarify anything?

Related Questions:

1. How is the cross product direction determined using the right-hand rule?
2. What would $\mathbf{a} \cdot \mathbf{b}$ (dot product) equal in this case?
3. How does the result change if $\theta$ is increased to $\pi/2$?
4. Why is the result of $\mathbf{a} \times \mathbf{b}$ perpendicular to both vectors?
5. What is the geometric interpretation of the magnitude of $\mathbf{a} \times \mathbf{b}$?

Tip:

To quickly verify cross products, remember that the resulting vector is always orthogonal to the plane containing the original vectors.