To calculate the cross product \(\vec{C} \times \vec{A}\), we'll use the determinant method involving the unit vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\).

Given:
\[
\vec{A} = 3.00 \hat{i} - 4.00 \hat{j} - 2.00 \hat{k}
\]
\[
\vec{C} = -6.00 \hat{i} - 9.00 \hat{j} - 1.00 \hat{k}
\]

The cross product \(\vec{C} \times \vec{A}\) is given by:

\[
\vec{C} \times \vec{A} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-6.00 & -9.00 & -1.00 \\
3.00 & -4.00 & -2.00
\end{vmatrix}
\]

Expanding the determinant:

\[
\vec{C} \times \vec{A} = \hat{i} \left((-9.00)(-2.00) - (-1.00)(-4.00)\right) - \hat{j} \left((-6.00)(-2.00) - (-1.00)(3.00)\right) + \hat{k} \left((-6.00)(-4.00) - (-9.00)(3.00)\right)
\]

Simplifying each component:

\[
\hat{i}: (-9.00)(-2.00) - (-1.00)(-4.00) = 18.00 - 4.00 = 14.00
\]

\[
\hat{j}: (-6.00)(-2.00) - (-1.00)(3.00) = 12.00 - (-3.00) = 15.00
\]

\[
\hat{k}: (-6.00)(-4.00) - (-9.00)(3.00) = 24.00 - (-27.00) = 51.00
\]

Thus, the cross product \(\vec{C} \times \vec{A}\) is:

\[
\vec{C} \times \vec{A} = 14.00 \hat{i} - 15.00 \hat{j} + 51.00 \hat{k}
\]

Would you like further explanation or details?

Here are 5 related questions:
1. What is the significance of the cross product in vector operations?
2. How does the direction of \(\vec{C} \times \vec{A}\) relate to \(\vec{C}\) and \(\vec{A}\)?
3. How do you find the magnitude of the cross product \(\vec{C} \times \vec{A}\)?
4. What does it mean if the cross product of two vectors is zero?
5. Can you use the right-hand rule to verify the direction of the cross product?

**Tip:** The cross product of two vectors results in a vector that is perpendicular to the plane formed by the original vectors.

Learn how to calculate the cross product CxA using determinants with a step-by-step solution. This problem covers vector algebra concepts, including the cross product formula using determinants. Suitable for advanced high school students and anyone studying vector algebra.

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