Vector Multiplication: Finding the Correct Solution

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SUMMARY

The discussion centers on vector multiplication, specifically the cross product of the unit vectors ex, ey, and ez. The user initially attempted to solve the expression (ex + ez) x (3ey - 4ez) but made a sign error in their calculations. The correct solution is -3ex + 4ey + 3ez, which highlights the importance of understanding the properties of cross products among orthogonal vectors. The user was reminded that the order of multiplication affects the sign of the result.

PREREQUISITES
  • Understanding of vector algebra and cross products
  • Familiarity with unit vectors in three-dimensional space
  • Knowledge of the properties of orthogonal vectors
  • Basic skills in simplifying algebraic expressions
NEXT STEPS
  • Study the properties of cross products in vector algebra
  • Learn about the right-hand rule for determining the direction of cross products
  • Explore common mistakes in vector multiplication and how to avoid them
  • Practice solving vector multiplication problems with varying combinations of unit vectors
USEFUL FOR

This discussion is beneficial for students studying physics or mathematics, particularly those focusing on vector calculus and linear algebra. It is also useful for educators teaching these concepts and anyone looking to improve their understanding of vector operations.

Roodles01
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Homework Statement


Still having a little trouble so here's the problem.

(ex + ez) x (3ey - 4ez)


The Attempt at a Solution



(ex * ez) + (ex * (-4ez)) + (ez * 3ey) + ( ez * (-4ez)

now, these are all orthogonal to each other, so, for example, if I have ex * ey then I should end up with ez, shouldn't I?
So here is my solution.

= 3ez - 4ey + 3ex - 4ez
= 3ex - 4 ey + ez

The solution shown to be correct is -3ex + 4ey + 3ez

so what have I done wrong, please?
 
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Roodles01 said:
now, these are all orthogonal to each other, so, for example, if I have ex * ey then I should end up with ez, shouldn't I?
I assume by * you mean X (vector product). So yes, ##e_x \times e_y = e_z##. What about ##e_y \times e_x##, ##e_x \times e_z##, and the other combinations? (You are making a sign error.)
 
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Thank you.
A simple thing becomes clear again.
 

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