How do I prove that A^B is orthogonal to A?

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SUMMARY

The discussion focuses on proving that the cross product A^B is orthogonal to vector A, where A = (2, -2, 1) and B = (2, 0, -1). The method involves calculating the cross product A^B, denoted as C, and then demonstrating that the dot product of C with A equals zero, confirming orthogonality. Additionally, it addresses how to express (A^B)^B as a linear combination of A and B to show that it lies in the same plane as A and B.

PREREQUISITES
  • Understanding of vector operations, specifically cross products.
  • Familiarity with dot products and their geometric interpretation.
  • Knowledge of vector notation and operations in three-dimensional space.
  • Basic trigonometry, particularly the relationship between angles and the cosine function.
NEXT STEPS
  • Learn how to calculate the cross product of two vectors in three-dimensional space.
  • Study the properties of dot products and their implications for vector orthogonality.
  • Explore the vector triple product identity: a x (b x c) = b(ac) - c(ab).
  • Investigate the geometric interpretation of vectors and their relationships in a plane.
USEFUL FOR

This discussion is beneficial for students studying vector calculus, physics enthusiasts, and anyone needing to understand vector orthogonality and cross product applications in three-dimensional geometry.

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Homework Statement



If A = (2,-2,1) and B = (2, 0, -1)
show by explicit calculation that

i) A^B is orthogonal to A

ii) (A^B)^B lies in the same plane as A and B by expressing it as a linear combination of A and B

Homework Equations



A^B = |A||B|sin θ

The Attempt at a Solution



I know that when you do the cross product of two vectors the result will be a vector that is perpendicular to both and I can draw a diagram to demonstrate. However I can't show it by explicit calculation. :(

I thought maybe if I could prove that the angle between them was ∏/2...
 
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If you know how to calculate cross product ^ , you will do A^B and if you find correctrly the value ~(itwill be a vector the result ,lets call it C) then you take the dot product of C with A and if it is zero they are orthogonal.

since cos90 =0 and (A) dot (B)= A B cosθ

ii)i think you can use the identity : a x ( b x c ) = b(ac) -c(ab) , b=a here
 

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