 Given u = <3, −1, 5> and v = <6, 10, 4>, find the cross product u × v

To find the cross product \( \mathbf{u} \times \mathbf{v} \) where \( \mathbf{u} = \langle 3, -1, 5 \rangle \) and \( \mathbf{v} = \langle 6, 10, 4 \rangle \), we can use the determinant method. The formula for the cross product in three dimensions is given by:

\[
\mathbf{u} \times \mathbf{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3
\end{vmatrix}
\]

Substituting the components of \( \mathbf{u} \) and \( \mathbf{v} \):

\[
\mathbf{u} \times \mathbf{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & -1 & 5 \\
6 & 10 & 4
\end{vmatrix}
\]

Calculating the determinant, we expand it as follows:

\[
\mathbf{u} \times \mathbf{v} = \mathbf{i} \begin{vmatrix} -1 & 5 \\ 10 & 4 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 5 \\ 6 & 4 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & -1 \\ 6 & 10 \end{vmatrix}
\]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -1 & 5 \\ 10 & 4 \end{vmatrix} = (-1)(4) - (5)(10) = -4 - 50 = -54 \)
2. \( \begin{vmatrix} 3 & 5 \\ 6 & 4 \end{vmatrix} = (3)(4) - (5)(6) = 12 - 30 = -18 \)
3. \( \begin{vmatrix} 3 & -1 \\ 6 & 10 \end{vmatrix} = (3)(10) - (-1)(6) = 30 + 6 = 36 \)

Now substituting back, we have:

\[
\mathbf{u} \times \mathbf{v} = -54 \mathbf{i} - (-18) \mathbf{j} + 36 \mathbf{k}
\]
\[
= -54 \mathbf{i} + 18 \mathbf{j} + 36 \mathbf{k}
\]

Thus, the cross product \( \mathbf{u} \times \mathbf{v} \) is:

\[
\mathbf{u} \times \mathbf{v} = \langle -54, 18, 36 \rangle
\]

### Summary
The cross product \( \mathbf{u} \times \mathbf{v} = \langle -54, 18, 36 \rangle \).

Do you want further details or explanations? Here are some related questions:

1. What is the geometric interpretation of the cross product?
2. How does the cross product relate to the area of a parallelogram?
3. Can you calculate the magnitude of the cross product?
4. What are the properties of the cross product?
5. How can the cross product be applied in physics?

**Tip:** Remember that the cross product is anti-commutative, meaning \( \mathbf{u} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{u}) \).

Given u = <3, −1, 5> and v = <6, 10, 4>, find the cross product u × v.

Explore the cross product calculation for vectors u = <3, -1, 5> and v = <6, 10, 4>. This guide provides a detailed solution using the determinant method, key formulas, and explanations suitable for students in Grades 10-12.

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