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Angle between 2 vectors

  • Thread starter quietrain
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  • #1
654
2

Homework Statement


if |A+B| = |A| = |B|
then what is the angle between A and B?
A and B are vectors.

edit: assume they are non-zero vectors

The Attempt at a Solution



the only way this is possible is if the angle is 60degrees right? equilateral triangle.
but the ans given is 120 degrees. why?

if it is 120 degrees, then it would be isosceles triangle and the magnitude of A+B won't be equal to A or B anymore.

so is the ans given wrong?
thanks!
 
Last edited:

Answers and Replies

  • #2
chiro
Science Advisor
4,790
131

Homework Statement


if |A+B| = |A| = |B|
then what is the angle between A and B?
A and B are vectors.

The Attempt at a Solution



the only way this is possible is if the angle is 60degrees right? equilateral triangle.
but the ans given is 120 degrees. why?

if it is 120 degrees, then it would be isosceles triangle and the magnitude of A+B won't be equal to A or B anymore.

so is the ans given wrong?
thanks!
We can use norms to do this as well as inner products.

|A+B|^2 = <A+B,A+B> = <A,A+B> + <B,A+B> = <A,A> + 2<A,B> + <B,B> = <B,B> = <A,A>

From the above we have the following relationships:

<A,A> + 2<A,B> + <B,B> = <A,A> = <B,B>

2<A,B> + <B,B> = 0
2<A,B> + <A,A> = 0

<A,A> = <B,B> = 0

This implies that A and B are the zero vectors and that they have no angle since they have no length.
 
  • #3
654
2
We can use norms to do this as well as inner products.

|A+B|^2 = <A+B,A+B> = <A,A+B> + <B,A+B> = <A,A> + 2<A,B> + <B,B> = <B,B> = <A,A>

From the above we have the following relationships:

<A,A> + 2<A,B> + <B,B> = <A,A> = <B,B>

2<A,B> + <B,B> = 0
2<A,B> + <A,A> = 0

<A,A> = <B,B> = 0

This implies that A and B are the zero vectors and that they have no angle since they have no length.
oh erm, assuming A and B are non zero?
 
  • #4
chiro
Science Advisor
4,790
131
oh erm, assuming A and B are non zero?
Ohh **** I made a big mistake.

Ok so going from

2<A,B> + <A,A> = 0
2<A,B> + <B,B> = 0

So

2<A,B> = -<A,A> = -<B,B>

<A,B> = -<A,A>/2 = -<B,B>/2

<A,B> = |A||B|cos(theta)
= -|A|^2/2
= -|B|^2/2

But since |A| = |B| we can use

<A,B> = -|A||B|/2
= |A||B|cos(theta)
cos(theta) = -1/2

this implies theta = 120 degrees or 2(pi)/3

Hopefully thats right
 
  • #5
danago
Gold Member
1,122
4
Another way to arrive at the same answer is to use a geometric argument like you did with the equilateral triangle. The issue with your reasoning is that the angle between two vectors is usually defined as the angle formed when the vectors are connected "tail-to-tail", rather than "head-to-tail" like when they are arranged to form a trange:

[PLAIN]http://img822.imageshack.us/img822/1200/vectorsh.jpg [Broken]
 
Last edited by a moderator:
  • #6
654
2
ah i see thanks a lot everyone
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,793
922

Homework Statement


if |A+B| = |A| = |B|
then what is the angle between A and B?
A and B are vectors.

edit: assume they are non-zero vectors

The Attempt at a Solution



the only way this is possible is if the angle is 60degrees right? equilateral triangle.
but the ans given is 120 degrees. why?

if it is 120 degrees, then it would be isosceles triangle and the magnitude of A+B won't be equal to A or B anymore.

so is the ans given wrong?
thanks!
You are looking at the "parallellogram rule" wrong. If A and B have their endpoints at the point, P, then the vector from the "tip" of A to the "tip" of B would be B- A, not A+ B.

To get "A+ B" you have to move the "tail" of A to the "tip" of B (or vice-versa) and if you actually draw this, you will see that the angle between the two vectors is now 120 degrees, not 60.
 
  • #8
654
2
You are looking at the "parallellogram rule" wrong. If A and B have their endpoints at the point, P, then the vector from the "tip" of A to the "tip" of B would be B- A, not A+ B.

To get "A+ B" you have to move the "tail" of A to the "tip" of B (or vice-versa) and if you actually draw this, you will see that the angle between the two vectors is now 120 degrees, not 60.
yup thanks i understand now
 

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