If u × v = u × w and u ̸= 0, then does v = w? If so, justify your answer algebraically. If
not, give a specific example of three nonzero vectors u, v, and w such that u × v = u × w
but v ̸= w

To determine whether \( u \times v = u \times w \) implies \( v = w \), let's break this problem down algebraically and geometrically.

### Cross Product Algebraic Properties:

- The **cross product** \( u \times v \) of two vectors \( u \) and \( v \) results in a vector that is **perpendicular** to both \( u \) and \( v \).
- If \( u \times v = u \times w \), we are given that the cross products of \( u \) with \( v \) and \( u \) with \( w \) are the same. This suggests that both \( v \) and \( w \) lie in a particular plane, but this doesn't necessarily mean that \( v = w \).

### Step-by-step reasoning:

Given the equation \( u \times v = u \times w \), we can subtract both sides to get:
\[
u \times v - u \times w = 0
\]
This simplifies to:
\[
u \times (v - w) = 0
\]
This equation means that the vector \( u \) is **parallel** to the vector \( v - w \), or in other words, \( v - w \) is **parallel** to \( u \). This does not imply that \( v = w \), but rather that:
\[
v - w = \lambda u
\]
for some scalar \( \lambda \).

Thus, the most general conclusion we can make is:
\[
v = w + \lambda u
\]
This shows that \( v \) and \( w \) are not necessarily equal, but they differ by a scalar multiple of the vector \( u \).

### Geometrical interpretation:
The vectors \( v \) and \( w \) can be different, but they must lie in the same plane that contains \( u \). So, \( v \) could be \( w \) shifted by some multiple of \( u \), while still producing the same cross product with \( u \).

### Counterexample:
Here’s a specific example to show that \( v \neq w \) is possible, even if \( u \times v = u \times w \).

Let:
\[
u = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad v = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \quad w = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}
\]
Now calculate the cross products:

\[
u \times v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
\]
\[
u \times w = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
\]
So, \( u \times v = u \times w \), but clearly \( v = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \) is not equal to \( w = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \).

### Conclusion:
No, \( v \) does not necessarily equal \( w \). The equation \( u \times v = u \times w \) only implies that \( v \) and \( w \) differ by some multiple of \( u \), i.e., \( v = w + \lambda u \), but they are not necessarily the same vector.

---

Do you want further details, or have any questions? Here are five related questions to explore:

1. How does the geometric interpretation of the cross product help understand vector equations?
2. Can you provide another example where \( u \times v = u \times w \) but \( v \neq w \)?
3. What does \( u \times v = 0 \) imply about the relationship between vectors \( u \) and \( v \)?
4. How would this analysis change if we considered the dot product instead of the cross product?
5. In which special cases would \( v = w \) hold if \( u \times v = u \times w \)?

**Tip:** The cross product between two vectors results in a vector perpendicular to both. This makes it a valuable tool for determining orthogonality or relationships between planes and vectors.

If u × v = u × w and u ≠ 0, then does v = w? If so, justify your answer algebraically. If not, give a specific example of three nonzero vectors u, v, and w such that u × v = u × w but v ≠ w.

Explore whether u × v = u × w implies that v = w with a detailed algebraic solution. Learn key vector algebra concepts like the cross product, linear dependence, and vector spaces. Suitable for advanced high school or university-level students.

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