Proveing Perpendicular of two vectors by applying Pythagoras rule

[harshakantha]

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!

 

[I like Serena]

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?

 

[harshakantha]

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,

 

[I like Serena]

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,

Can you relate c to a and b?

(I think that is a hint. )

 

[harshakantha]

vectors are really hard lesson to me, so please show me how to prove that equation, I don't know where to start please help me!

 

[I like Serena]

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?

 

[harshakantha]

Thank you so much I like Serena, finally I was able to proved that, can you explain how did you get the equation |a|²+|b|²=|b-a|² more?

 

[I like Serena]

Please say a little more about what you do not understand.
You leave me guessing.

I substituted c = b -a, but I presume you already got that?

So is your question about vector algebra?
What do you know about vector addition and subtraction?

Here's a link: http://mathforum.org/~klotz/Vectors/subtraction.html

Or perhaps this section of the wiki page will help you?
http://en.wikipedia.org/wiki/Euclidean_vector#Addition_and_subtraction

 

[harshakantha]

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" ?

 

[I like Serena]

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.

 

[harshakantha]

Thank you very much I like Serena. I got the idea, you are a really good helper hope to catch you again, bye...

 

[ehild]

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

 

[harshakantha]

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