Let u =u1ī + u2j +u3k , v = v1ī + v2j +v3k and w = w1ī + w2j + w3k 
Use the properties of determinent to prove (u×v).w = (v×w).u

To prove that \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = (\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u}\), we will use the properties of determinants and the fact that the cross product can be expressed as a determinant of a 3x3 matrix. Let’s break it down step-by-step.

### Step 1: Express the cross product using determinants
For two vectors \(\mathbf{u}\) and \(\mathbf{v}\), the cross product \(\mathbf{u} \times \mathbf{v}\) can be written as:

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

This determinant expands as follows:

\[
\mathbf{u} \times \mathbf{v} = \mathbf{i}(u_2 v_3 - u_3 v_2) - \mathbf{j}(u_1 v_3 - u_3 v_1) + \mathbf{k}(u_1 v_2 - u_2 v_1)
\]

Similarly, the cross product \(\mathbf{v} \times \mathbf{w}\) is:

\[
\mathbf{v} \times \mathbf{w} = \begin{vmatrix} 
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3 
\end{vmatrix}
\]

### Step 2: Dot product of cross product with third vector
The scalar triple product \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}\) is the dot product of \(\mathbf{u} \times \mathbf{v}\) with the vector \(\mathbf{w}\). This can also be written as a determinant:

\[
(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = \begin{vmatrix} 
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3 
\end{vmatrix}
\]

### Step 3: Determinant for \((\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u}\)
Similarly, the scalar triple product \((\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u}\) is:

\[
(\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u} = \begin{vmatrix} 
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3 \\
u_1 & u_2 & u_3 
\end{vmatrix}
\]

### Step 4: Symmetry of determinants
Notice that the two determinants:

\[
\begin{vmatrix} 
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3 
\end{vmatrix}
\quad \text{and} \quad
\begin{vmatrix} 
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3 \\
u_1 & u_2 & u_3 
\end{vmatrix}
\]

are simply the same determinant with the rows permuted cyclically. This means that the determinants are equal:

\[
\begin{vmatrix} 
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3 
\end{vmatrix}
= 
\begin{vmatrix} 
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3 \\
u_1 & u_2 & u_3 
\end{vmatrix}
\]

Thus, we conclude that:

\[
(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = (\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u}
\]

This completes the proof.

Would you like more details on any part of this proof, or do you have any other questions?

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Here are 5 related questions:
1. How can you interpret the scalar triple product geometrically?
2. What are the properties of determinants that make this proof work?
3. Can you explain the relationship between cross products and the area of parallelograms?
4. How does the scalar triple product relate to the volume of a parallelepiped?
5. What is the physical significance of the cross product in physics?

**Tip:** The scalar triple product can be used to calculate the volume of a parallelepiped formed by three vectors.

Let u = u1ī + u2j + u3k, v = v1ī + v2j + v3k and w = w1ī + w2j + w3k. Use the properties of determinant to prove (u×v)·w = (v×w)·u.

Explore the proof that (u × v)·w = (v × w)·u using determinants, cross products, and dot products in vector algebra. Learn the step-by-step solution, including key concepts like the scalar triple product and determinant symmetry. Suitable for undergraduate students.

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