The sum of two vectors, A→ + B→

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Homework Statement



The sum of two vectors, A→ + B→, is perpendicular to their difference, A→ - B→. How do the vectors magnitude compare?


The Attempt at a Solution



SQRT[(A+B)^2 + (A-B)^2]
 
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Have you studied the scalar (dot) product of vectors?
 
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voko said:
Have you studied the scalar (dot) product of vectors?

I have but only very briefly-1 lecture class on that and that was 1 semester back. But, if you could give me a rough overview, I'll build on those knowledge.
 
What does perpendicularity mean for the dot product ? write it out as a vector expression, then use the distributive property of the dot product.
 
BvU said:
What does perpendicularity mean for the dot product ? write it out as a vector expression, then use the distributive property of the dot product.

It means A→.B→ = 0
 
Right. Now A+B is perpendicular to A-B, so: (A+B).(A-B)=0
 
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voko said:
Have you studied the scalar (dot) product of vectors?

BvU said:
What does perpendicularity mean for the dot product ? write it out as a vector expression, then use the distributive property of the dot product.

Dot product implies that the product of two vector A→.B→ = 0

Let A→+B→ = R1 [itex]\wedge[/itex] A→-B→= R2

R1.R2 = 0

(A→+B→).(A→-B→) = A^2→ - B^2→ = 0

A=B
 
Bingo.
 
See, it was not that hard :)
 
  • #10
voko said:
See, it was not that hard :)

It wasn't but my interpretation was different. I went in with the assumption
1) A right angle triangle exists.
2) the length parallel to the y-axis = r1
3) length perpendicular to r1 = r2
4) find the resultant
 
  • #11
I do not understand how your assumption is related to the problem.
 
  • #12
voko said:
I do not understand how your assumption is related to the problem.

I interpreted the question wrongly and build assumptions on the wrong premise.
 

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