Math Problem Statement

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

Solution

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

Cross Product Algebraic Properties:

  • The cross product u×vu \times v of two vectors uu and vv results in a vector that is perpendicular to both uu and vv.
  • If u×v=u×wu \times v = u \times w, we are given that the cross products of uu with vv and uu with ww are the same. This suggests that both vv and ww lie in a particular plane, but this doesn't necessarily mean that v=wv = w.

Step-by-step reasoning:

Given the equation u×v=u×wu \times v = u \times w, we can subtract both sides to get: u×vu×w=0u \times v - u \times w = 0 This simplifies to: u×(vw)=0u \times (v - w) = 0 This equation means that the vector uu is parallel to the vector vwv - w, or in other words, vwv - w is parallel to uu. This does not imply that v=wv = w, but rather that: vw=λuv - w = \lambda u for some scalar λ\lambda.

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

Geometrical interpretation:

The vectors vv and ww can be different, but they must lie in the same plane that contains uu. So, vv could be ww shifted by some multiple of uu, while still producing the same cross product with uu.

Counterexample:

Here’s a specific example to show that vwv \neq w is possible, even if u×v=u×wu \times v = u \times w.

Let: u=(100),v=(010),w=(011)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×v=(100)×(010)=(001)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×w=(100)×(011)=(001)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×v=u×wu \times v = u \times w, but clearly v=(010)v = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} is not equal to w=(011)w = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}.

Conclusion:

No, vv does not necessarily equal ww. The equation u×v=u×wu \times v = u \times w only implies that vv and ww differ by some multiple of uu, i.e., v=w+λuv = 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×v=u×wu \times v = u \times w but vwv \neq w?
  3. What does u×v=0u \times v = 0 imply about the relationship between vectors uu and vv?
  4. How would this analysis change if we considered the dot product instead of the cross product?
  5. In which special cases would v=wv = w hold if u×v=u×wu \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.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Cross Product
Linear Dependence
Vector Spaces

Formulas

u × v = u × w
u × (v - w) = 0
v = w + λu

Theorems

Cross Product Perpendicularity Theorem
Linear Dependence of Vectors

Suitable Grade Level

University/Advanced High School