Find the angle between 2 vectors

In summary: This should work, now that I see your drawing, but for some reason, this equation gives a negative value for ##\cos(\theta)## while the way I did it gives a number that is equal in absolute value. I don't see why that's happening, though. In other words, if r and s are positive numbers, then |a-2b|^2 should give a positive number for ##\cos(\theta)##.
  • #1
Helly123
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Homework Statement


|a| = 2
|b| = ## \sqrt3##
|a - 2b| = 2

Angle between a and b

Homework Equations

The Attempt at a Solution


##\theta## is angle between a and b
So angle between a and -2b is 180-##\theta## [/B]
##|a-2b|^2## = |a|^2 + |2b|^2 -2|a||2b|cos(180-##\theta##)
##2^2## = 2^2 + (2##\sqrt3##)^2 -2(2)(2##\sqrt3##)( - 1)cos(##\theta##)
Then find the angle

What is wrong? Can you help me?
 
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  • #2
Helly123 said:

Homework Statement


|a| = 2
|b| = ## \sqrt3##
|a - 2b| = 2

Angle between a and b

Homework Equations

The Attempt at a Solution


##\theta## is angle between a and b
So angle between a and -2b is 180-##\theta## [/B]
##|a-2b|^2## = |a|^2 + |2b|^2 -2|a||2b|cos(180-##\theta##)
##2^2## = 2^2 + (2##\sqrt3##)^2 -2(2)(2##\sqrt3##)( - 1)cos(##\theta##)
Then find the angle

What is wrong? Can you help me?

Why bother finding the angle between ##\vec{a}## and ##-2 \vec{b}##? However, since you have found that angle, you must then use it correctly, not the way you have done it.
 
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  • #3
Helly123 said:

Homework Statement


|a| = 2
|b| = ## \sqrt3##
|a - 2b| = 2

Angle between a and b

Homework Equations

The Attempt at a Solution


##\theta## is angle between a and b
So angle between a and -2b is 180-##\theta## [/B]
##|a-2b|^2## = |a|^2 + |2b|^2 -2|a||2b|cos(180-##\theta##)
##2^2## = 2^2 + (2##\sqrt3##)^2 -2(2)(2##\sqrt3##)( - 1)cos(##\theta##)
Then find the angle

What is wrong? Can you help me?
What's wrong is that you have used the Law of Cosines on the wrong triangle. Instead of doing this, use the idea that ##|a - 2b|^2 = (a - 2b) \cdot (a - 2b)##, and expand this out. You will also need one of the definitions of the dot product -- the one that involves the angle and the magnitudes of the vectors.
 
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  • #4
Mark44 said:
What's wrong is that you have used the Law of Cosines on the wrong triangle. Instead of doing this, use the idea that ##|a - 2b|^2 = (a - 2b) \cdot (a - 2b)##, and expand this out. You will also need one of the definitions of the dot product -- the one that involves the angle and the magnitudes of the vectors.
But it still makes more sense to me. Because the |a| |b| and |a-2b| can be seen as triangle right? |a| is a length of vector a with magnitude 2 right?
Although, i find that expanding (a-2b)^2 also can be the way. Maybe if being elaborated, you mean that :
(a-2b)•(a-2b) = |a-2b|.|a-2b|cos0
Thus, |a-2b|^2 = (a-2b)•(a-2b)
 
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  • #5
Ray Vickson said:
Why bother finding the angle between ##\vec{a}## and ##-2 \vec{b}##? However, since you have found that angle, you must then use it correctly, not the way you have done it.
Because i thought finding angle between ##\vec a## and ##\vec { -2b}## i can find angle between a and b. Since 180 - angle(a-2b) = angle(a+b) under the condition, if it's true that ##\vec b## has same degree with regards to x-axis as ##\vec {-2b}## just different direction going left or right and different lenght
 
  • #6
Helly123 said:
Because i thought finding angle between ##\vec a## and ##\vec { -2b}## i can find angle between a and b. Since 180 - angle(a-2b) = angle(a+b) under the condition, if it's true that ##\vec b## has same degree with regards to x-axis as ##\vec {-2b}## just different direction going left or right and different lenght
I don't see how this works at all. Knowing the angle between ##\vec a - \vec {2b}## and either ##\vec a## or ##\vec b## doesn't help you find the angle between ##\vec a## and ##\vec b##.

You're given |a| and |b|, so if you can get ##a \cdot b##, you can find ##\theta##. If you do what I said, of expanding ##(a - 2b) \cdot (a - 2b)##, you will be able to solve for ##a \cdot b## in terms of everything else, and from that you'll be able to get ##\cos(\theta)## and hence ##\theta##.
 
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  • #7
Mark44 said:
I don't see how this works at all. Knowing the angle between ##\vec a - \vec {2b}## and either ##\vec a## or ##\vec b## doesn't help you find the angle between ##\vec a## and ##\vec b##.

You're given |a| and |b|, so if you can get ##a \cdot b##, you can find ##\theta##. If you do what I said, of expanding ##(a - 2b) \cdot (a - 2b)##, you will be able to solve for ##a \cdot b## in terms of everything else, and from that you'll be able to get ##\cos(\theta)## and hence ##\theta##.
I meant something like this
2m7gpxd.jpg

Can i say that |a+b|^2 = |a|^2 + |b|^2 - 2|a||b|cos##\theta##

|a-2b|^2 = |a|^2 + |2b|^2 - 2|a||2b|(-1)cos##\theta##
 

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  • 2m7gpxd.jpg
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  • #8
Helly123 said:
I meant something like this View attachment 224578
Can i say that |a+b|^2 = |a|^2 + |b|^2 - 2|a||b|cos##\theta##
Sure, because here you're using the Law of Cosines.
Helly123 said:
|a-2b|^2 = |a|^2 + |2b|^2 - 2|a||2b|(-1)cos##\theta##
This should work, now that I see your drawing, but for some reason, this equation gives a negative value for ##\cos(\theta)## while the way I did it gives a number that is equal in absolute value. I don't see why that's happening, though. In other words, if r and s are positive numbers, your way gives the cosine as -r/s and my way gives the cosine as r/s.
 
  • #9
Mark44 said:
Sure, because here you're using the Law of Cosines.

This should work, now that I see your drawing, but for some reason, this equation gives a negative value for ##\cos(\theta)## while the way I did it gives a number that is equal in absolute value. I don't see why that's happening, though. In other words, if r and s are positive numbers, your way gives the cosine as -r/s and my way gives the cosine as r/s.
Yes. But not only negative. My way give wrong value for ##\theta##. It gives cos##\theta## = ##\frac{-12}{8\sqrt 3}##...
 
  • #10
Helly123 said:
Yes. But not only negative. My way give wrong value for ##\theta##. It gives cos##\theta## = ##\frac{-12}{8\sqrt 3}##...
This simplifies to ##\cos(\theta) = \frac{-3}{2\sqrt 3} = \frac{-\sqrt 3} 2##. I get ##\cos(\theta) = \frac {\sqrt 3} 2##.
 
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  • #11
Helly123 said:
I meant something like this View attachment 224578
Can i say that |a+b|^2 = |a|^2 + |b|^2 - 2|a||b|cos##\theta##

|a-2b|^2 = |a|^2 + |2b|^2 - 2|a||2b|(-1)cos##\theta##

No, you cannot say that. The correct evaluation is ##|a+b|^2 = |a|^2+|b|^2 + 2 |a| |b| \cos \theta,## where ##\theta## is the angle between ##a## and ##b##. Note the + sign in front of the ##2 |a| |b| \cos \theta.## Similarly, ##|a-b|^2 = |a|^2+|b|^2 - 2 |a| |b| \cos \theta.##

Therefore, ##|a-2b|^2 = |a|^2 + |2b|^2 - 2 |a| |2b| \cos \theta##.
 
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  • #12
Ray Vickson said:
No, you cannot say that. The correct evaluation is ##|a+b|^2 = |a|^2+|b|^2 + 2 |a| |b| \cos \theta,## where ##\theta## is the angle between ##a## and ##b##. Note the + sign in front of the ##2 |a| |b| \cos \theta.## Similarly, ##|a-b|^2 = |a|^2+|b|^2 - 2 |a| |b| \cos \theta.##

Therefore, ##|a-2b|^2 = |a|^2 + |2b|^2 - 2 |a| |2b| \cos \theta##.
Why the law of cosinus is ##c^2 = a^2 + b^2 - 2 ab \cos \theta##
How can we get conclusion like that?
 
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  • #13
Helly123 said:
Why the law of cosinus is ##c^2 = a^2 + b^2 - 2 ab \cos \theta##
How can we get conclusion like that?
In the vector addition for the law of cosines, ## \vec{b}=\vec{a}+\vec{c} ## or ## \vec{a}=\vec{b}+\vec{c} ## is what the triangles look like if you make the triangle into a vector sum. The vectors get added by connecting tail to head. Either ## \vec{b} ## or ## \vec{a} ## winds up being the resultant. Thereby ## |\vec{c}|=|\vec{a}-\vec{b}| ##. ## \\ ## ( Note: If you want to find ## \vec{a}+\vec{b} ## , you need to pull ## \vec{a} ## and ## \vec{b} ## apart, so that their tails are not connected.).
 
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  • #14
Helly123 said:
Why the law of cosinus is ##c^2 = a^2 + b^2 - 2 ab \cos \theta##
How can we get conclusion like that?
$$
|\mathbf{a} - 2 \mathbf{b}|^2 = (\mathbf{a} - 2 \mathbf{b}) \cdot (\mathbf{a} - 2 \mathbf{b})\\
= \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot ( 2\mathbf{b} ) + (2 \mathbf{b}) \cdot (2 \mathbf{b})

$$
By definition, for two vectors ##\mathbf{u}## and ##\mathbf{v}## their inner product is ##\mathbf{u} \cdot \mathbf{v} = |u| |v| \cos \theta##, where ##\theta## is the angle between the two vectors.
 
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  • #15
Charles Link said:
In the vector addition for the law of cosines, ## \vec{b}=\vec{a}+\vec{c} ## or ## \vec{a}=\vec{b}+\vec{c} ## is what the triangles look like if you make the triangle into a vector sum. The vectors get added by connecting tail to head. Either ## \vec{b} ## or ## \vec{a} ## winds up being the resultant. Thereby ## |\vec{c}|=|\vec{a}-\vec{b}| ##. ## \\ ## ( Note: If you want to find ## \vec{a}+\vec{b} ## , you need to pull ## \vec{a} ## and ## \vec{b} ## apart, so that their tails are not connected.).
In vector addition for the law of cosines, we define c with a and b. So, it supposed to be ##\vec c = \vec a + \vec b## right?
a+ b is head of vec a meets tail of vec b right? Or, when
you said that ##\vec c = \vec a - \vec b##. Do you mean that is when head of vec a meets the tail of vec b?
 
  • #16
Ray Vickson said:
$$
|\mathbf{a} - 2 \mathbf{b}|^2 = (\mathbf{a} - 2 \mathbf{b}) \cdot (\mathbf{a} - 2 \mathbf{b})\\
= \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot ( 2\mathbf{b} ) + (2 \mathbf{b}) \cdot (2 \mathbf{b})

$$
By definition, for two vectors ##\mathbf{u}## and ##\mathbf{v}## their inner product is ##\mathbf{u} \cdot \mathbf{v} = |u| |v| \cos \theta##, where ##\theta## is the angle between the two vectors.
When i examined it carefully. When i realized you substituted the ##\vec a ##. ##\vec {2b}## of #15 into |##\vec a##| |##\vec 2b##|cos##\theta## and resulted the formula in #11. It makes senses...
 
  • #17
Ray Vickson said:
No, you cannot say that. The correct evaluation is ##|a+b|^2 = |a|^2+|b|^2 + 2 |a| |b| \cos \theta,## where ##\theta## is the angle between ##a## and ##b##. Note the + sign in front of the ##2 |a| |b| \cos \theta.## Similarly, ##|a-b|^2 = |a|^2+|b|^2 - 2 |a| |b| \cos \theta.##

Therefore, ##|a-2b|^2 = |a|^2 + |2b|^2 - 2 |a| |2b| \cos \theta##.
I think yes. Thanks
 

1. What is the formula for finding the angle between 2 vectors?

The formula for finding the angle between 2 vectors is given by the dot product formula: θ = cos^-1 ((u · v) / (|u| * |v|)), where u and v are the two vectors and |u| and |v| represent their magnitudes.

2. Can the angle between 2 vectors be negative?

No, the angle between 2 vectors is always positive. However, the direction of the angle (clockwise or counterclockwise) can be indicated by the sign of the dot product in the formula.

3. How do I find the angle between 2 vectors in 3-dimensional space?

The formula for finding the angle between 2 vectors in 3-dimensional space is slightly different. It is given by: θ = cos^-1 ((u · v) / (√(u1^2 + u2^2 + u3^2) * √(v1^2 + v2^2 + v3^2))), where u and v are the two vectors with their respective components (u1, u2, u3) and (v1, v2, v3).

4. What is the difference between the dot product and cross product when finding the angle between 2 vectors?

The dot product is used to find the angle between 2 vectors, while the cross product is used to find the perpendicular vector to those 2 vectors. The dot product formula involves multiplying the corresponding components of the 2 vectors and adding them together, while the cross product formula involves finding the determinant of a 3x3 matrix.

5. How can I use the angle between 2 vectors in real-life applications?

The angle between 2 vectors can be used in various real-life applications, such as calculating the forces acting on an object in physics, determining the direction and speed of an object's movement in navigation, or finding the similarity between two texts in natural language processing. It is a fundamental concept in mathematics and has many practical uses in different fields.

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