Vectors Thinking Question: Proving Perpendicularity using the Cosine Law

  • Thread starter Thread starter DespicableMe
  • Start date Start date
  • Tags Tags
    Thinking Vectors
Click For Summary

Homework Help Overview

The discussion revolves around proving the perpendicularity of two vectors using the cosine law. Participants are tasked with expressing the cosine law in vector form and applying it to the condition |vector a - vector b| = |vector a + vector b|.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of the cosine law in terms of vectors and attempt to apply it to the given condition. There is an exploration of how to incorporate the equality of magnitudes into the proof of perpendicularity. Some participants suggest using coordinates for the vectors to facilitate the proof.

Discussion Status

There is an ongoing exploration of the problem, with participants questioning the correctness of initial attempts and suggesting alternative approaches. Guidance has been offered regarding the use of coordinates and the importance of squaring both sides of the equation to simplify the proof.

Contextual Notes

Participants note the challenge of integrating the condition |vector a - vector b| = |vector a + vector b| into their reasoning and the need for clarity in applying the cosine law to this specific scenario.

DespicableMe
Messages
40
Reaction score
0

Homework Statement



a) Write the cosine law using a vector operation.
b) If |vector a - vector b| = |vector a + vector b|, prove that a is perpendicular to b using the formula you found in a).

The Attempt at a Solution



The red period represents the dot product

a) magnitude c = SQUAREROOT (a.a + b.b - 2(a.b) )

b) magnitude c = SQUAREROOT (a.a + b.b - 2(a.b) )

= SQUAREROOT ( a2 + b2 - 2( |a||b|cos 90 )
= SQUAREROOT ( a2 + b2 )

And this is the Pythagorean theorem, which only applies for right angle triangles, so if mag. C is SQUARERT (a2+b2), then a and b are perpendicular.

DId I do this correctly?

I didn't know how to include |vector a - vector b| = |vector a + vector b| into the equation and it only asked to prove it using the formula that I found...
 
Physics news on Phys.org
DespicableMe said:

Homework Statement



a) Write the cosine law using a vector operation.
b) If |vector a - vector b| = |vector a + vector b|, prove that a is perpendicular to b using the formula you found in a).

The Attempt at a Solution



The red period represents the dot product

a) magnitude c = SQUAREROOT (a.a + b.b - 2(a.b) )

b) magnitude c = SQUAREROOT (a.a + b.b - 2(a.b) )

= SQUAREROOT ( a2 + b2 - 2( |a||b|cos 90 )
= SQUAREROOT ( a2 + b2 )

And this is the Pythagorean theorem, which only applies for right angle triangles, so if mag. C is SQUARERT (a2+b2), then a and b are perpendicular.

DId I do this correctly?

I didn't know how to include |vector a - vector b| = |vector a + vector b| into the equation and it only asked to prove it using the formula that I found...

The b part isn't correct. You need to start by assuming that |a - b| = |a + b|, and showing that it follows that a and b are perpendicular.

You could start by assigning coordinates to a and b, such as a = <a1, a2> and b = <b1, b2>.
 
Mark44 said:
The b part isn't correct. You need to start by assuming that |a - b| = |a + b|, and showing that it follows that a and b are perpendicular.

You could start by assigning coordinates to a and b, such as a = <a1, a2> and b = <b1, b2>.

I'm still kind of...
What can I do with the coordinates?

At one point, I did draw vector diagrams to show|a - b| = |a + b| but other than that, I didn't know how to use that in the equation.
 
Start by writing out the magnitudes using the definition of magnitude. Of course, it would be easier if you squared both sides first.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
15K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K