Proveing Perpendicular of two vectors by applying Pythagoras rule

  • #1
Given that a=(a1,a2,a3) and b=(b1,b2,b3) by applying the Pythagoras rule, Prove that a1b1+a2b2+a3b3=0 if a and b perpendicular



The Scalar product a.b = |a||b|CosQ -------------(1)



if two vectors are perpendicular; Q=90degrees
then CosQ=0;
from (1)

a.b=0
(a1,a2,a3).(b1,b2,b3)=0
a1b1+a2b2+a3b3=0

is this a correct solution for the above Question?
please can someone give me an opinion! :confused:
 
Last edited:

Answers and Replies

  • #2
I like Serena
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The proof is correct, but they are asking for a proof by Pythagorean rule, which this isn't.

What would the Pythagorean rule be for a triangle with vectors a and b as its sides?
 
  • #3
according to the Pythagorean rule;
for a Right triangle
|a|2+|b|2=|c|2

can you give me a hint or something to solve my question properly, :smile:
 
  • #4
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according to the Pythagorean rule;
for a Right triangle
|a|2+|b|2=|c|2

can you give me a hint or something to solve my question properly, :smile:

Can you relate c to a and b?

(I think that is a hint. :wink:)
 
  • #5
vectors are realy hard lesson to me, so please show me how to prove that equation, I dont know where to start:confused: please help me!!
 
  • #6
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All right.

c is the vector that starts from the end of vector a, and ends at the end of vector b.
This means that c = b - a.
Do you know this type of vector algebra?

So you have |a|2 + |b|2 = |b - a|2

Fill in the definition of the norm and work out the algebra?
 
  • #7
Thank you so much I like Serena, finally I was able to proved that:smile:, can you explain how did you get the equation |a|2+|b|2=|b-a|2 more?:smile:
 
  • #9
I know about addition and subtraction of vectors, what i wanted to understand is how did you get c=b-a instead of c=a+b,
can you explain little bit more, what did you mean by,
"c is the vector that starts from the end of vector a, and ends at the end of vector b.
This means that c = b - a" ?
 
  • #10
I like Serena
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I know about addition and subtraction of vectors, what i wanted to understand is how did you get c=b-a instead of c=a+b,

You can use c=a+b as well.
On the wiki link in my previous post you can see that you'll get a different triangle, but that does not matter for Pythagoras, since the angle is 90 degrees.


can you explain little bit more, what did you mean by,
"c is the vector that starts from the end of vector a, and ends at the end of vector b.
This means that c = b - a" ?

That is exactly what it means.

As opposed to c = a + b, where c is the vector that begins at the beginning of a, and ends at the end of b (the head-to-tail construction of vector addition).
In this case vector b would be shifted so its tail is at the head of a.
 
  • #11
Thank you very much I like Serena. I got the idea, you are a really good helper:smile: hope to catch you again,:smile: bye...
 
  • #12
ehild
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I know about addition and subtraction of vectors, what i wanted to understand is how did you get c=b-a instead of c=a+b,
can you explain little bit more, what did you mean by,
"c is the vector that starts from the end of vector a, and ends at the end of vector b.
This means that c = b - a" ?

a,b vectors determine a parallelogram, and one diagonal is c1=a+b, the other is c2=b-a, see attachment. It is all the same which one you select. You get Pythagoras equation only in case of a1b1+a2b2+a3b3=0

ehild
 

Attachments

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  • #13
wow... thank you ehild:) your attachment really helpful for me, now I got a more clear idea about |c| = |b - a|.
 

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