Find the angle between 2 vectors

[Helly123]

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?

 

[Ray Vickson]

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.

 

[Mark44]

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.

 

[Helly123]

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)

 

[Helly123]

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

 

[Mark44]

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 $\theta$.

 

[Helly123]

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

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$\theta$

 

[Mark44]

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.

|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.

 

[Helly123]

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$\theta$ = $\frac{-12}{8\sqrt 3}$...

 

[Mark44]

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 $\cos(\theta) = \frac {\sqrt 3} 2$.

 

[Ray Vickson]

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, $|a-2b|^2 = |a|^2 + |2b|^2 - 2 |a| |2b| \cos \theta$.

 

[Helly123]

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?

 

[Charles Link]

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.).

 

[Ray Vickson]

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.

 

[Helly123]

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?

 

[Helly123]

$$
|\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...

 

[Helly123]

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