What Is the Angle Between Two Vectors if Their Magnitudes Are Equal?

[quietrain]

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!

 

[chiro]

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.

 

[quietrain]

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?

 

[chiro]

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 that's right

 

[danago]

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

 

[quietrain]

ah i see thanks a lot everyone

 

[HallsofIvy]

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.

 

[quietrain]

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