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.

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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.