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To find the cross product \( \vec{u} \times \vec{v} \) of vectors \( \vec{u} = -8\vec{i} - 3\vec{j} + \vec{k} \) and \( \vec{v} = \langle -9, -7, 6 \rangle \), we can use the following determinant formula:

\[
\vec{u} \times \vec{v} = 
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
-8 & -3 & 1 \\
-9 & -7 & 6 \\
\end{vmatrix}
\]

We calculate this determinant by expanding along the first row:

\[
\vec{u} \times \vec{v} = \vec{i} \begin{vmatrix} -3 & 1 \\ -7 & 6 \end{vmatrix} - \vec{j} \begin{vmatrix} -8 & 1 \\ -9 & 6 \end{vmatrix} + \vec{k} \begin{vmatrix} -8 & -3 \\ -9 & -7 \end{vmatrix}
\]

Now, we compute each of these smaller 2x2 determinants:

1. For \( \vec{i} \):
   \[
   \begin{vmatrix} -3 & 1 \\ -7 & 6 \end{vmatrix} = (-3)(6) - (1)(-7) = -18 + 7 = -11
   \]

2. For \( -\vec{j} \):
   \[
   \begin{vmatrix} -8 & 1 \\ -9 & 6 \end{vmatrix} = (-8)(6) - (1)(-9) = -48 + 9 = -39
   \]

3. For \( \vec{k} \):
   \[
   \begin{vmatrix} -8 & -3 \\ -9 & -7 \end{vmatrix} = (-8)(-7) - (-3)(-9) = 56 - 27 = 29
   \]

Substituting these values back in, we get:

\[
\vec{u} \times \vec{v} = -11 \vec{i} + 39 \vec{j} + 29 \vec{k}
\]

So, the cross product is:

\[
\vec{u} \times \vec{v} = \langle -11, 39, 29 \rangle
\]

Would you like further clarification on any step?

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Here are five related questions to expand on this concept:

1. What is the significance of the direction of the cross product vector?
2. How can we determine if two vectors are parallel using the cross product?
3. How do we interpret the magnitude of a cross product geometrically?
4. What would change in the calculation if one of the vector components was zero?
5. Can you extend this cross product method to 4-dimensional vectors?

**Tip:** The cross product of two vectors is perpendicular to both original vectors in three-dimensional space.

Find the cross product \( \vec{u} \times \vec{v} \) where \( \vec{u} = -8\vec{i} - 3\vec{j} + \vec{k} \) and \( \vec{v} = \langle -9, -7, 6 \rangle \).

Learn how to find the cross product of vectors \( \vec{u} = -8\vec{i} - 3\vec{j} + \vec{k} \) and \( \vec{v} = \langle -9, -7, 6 \rangle \) using the determinant method. This example demonstrates the use of vector algebra and determinants, suitable for college-level students.

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