How to prove this without computation

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SUMMARY

In R3, if vectors a and b are orthogonal, then for any nonzero vector u, the projection of the projection of u onto b, denoted as proja(projbu), equals zero. This conclusion is derived from the property that the dot product of orthogonal vectors a and b is zero, indicating that the component of u projected onto b is a scalar multiple of b. Consequently, since a is orthogonal to b, the projection of u onto b does not influence the projection onto a, confirming that proja(projbu) = 0.

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  • Understanding of vector projections in R3
  • Knowledge of orthogonal vectors and their properties
  • Familiarity with dot product calculations
  • Basic linear algebra concepts
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  • Explore the implications of the dot product in vector spaces
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Students and educators in mathematics, particularly those studying linear algebra, as well as anyone interested in understanding vector projections and orthogonality in three-dimensional space.

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Homework Statement


Prove: In R3if a and b are orthogonal vectors, then for every nonzero vector u, we have proja(projbu)=0

Homework Equations


I know the dot product of a and b is zero

The Attempt at a Solution



Based on the definition of projection I eventually got to
(ab) in the numerator which is zero and was done. So how can I write this out with words only? [/B]
 
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Loststudent22 said:

Homework Statement


Prove: In R3if a and b are orthogonal vectors, then for every nonzero vector u, we have proja(projbu)=0

Homework Equations


I know the dot product of a and b is zero

The Attempt at a Solution



Based on the definition of projection I eventually got to
(ab) in the numerator which is zero and was done. So how can I write this out with words only? [/B]

You haven't defined projbu but I suppose it is the projection on b of u. I'm not sure why you want to write it "in words" instead of equations, but you could observe that the projection on b of u is a multiple of b. Then since a is orthogonal to b, then...
 
then that component is zero? Yeah I will add the computation also but I wasn't really sure what else to say.

yeah its the projection on b of u
 

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