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The problem:

If the angle between $\vec a$ and $\vec b$ is $60^\circ $, find the angle between $2\vec a$ and $-2\vec b$.

My Attempt

The angle between $\vec a$ and $\vec b$ is given by: $$\cos \theta =\dfrac {\vec a\times \vec b}{ |\vec a|\times |\vec b|}.$$

But, how to solve this?

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    $\begingroup$ Draw a picture. The angle between $\vec{a}$ and $-\vec{b}$ is $180-60=120$. The angle between $2\vec{a}$ and $-2\vec{b}$ will be the same, namely $120$. $\endgroup$ – Test123 Feb 12 '17 at 14:07
  • $\begingroup$ @ Test 123, how to draw a figure? $\endgroup$ – pi-π Feb 12 '17 at 14:09
  • $\begingroup$ Draw two vectors at the origin with some angle and check what happens when you take $-\vec{b}$. $\endgroup$ – Test123 Feb 12 '17 at 14:10
  • $\begingroup$ Note that simply doubling $a$ changes nothing about what angles it makes with other vectors. $\endgroup$ – Kaynex Feb 12 '17 at 14:13
  • $\begingroup$ In fact, $\vec a \times \vec b = |\vec a||\vec b|\sin \theta$. $\endgroup$ – Rohan Feb 12 '17 at 14:20
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$\angle (2\vec{a},-2\vec{b})=\angle (\vec{a},-\vec{b})=180-\angle(\vec{a},\vec{b})$

The explanation why each equality holds:

For the first equality: $$ \cos{\angle (2\vec{a},-2\vec{b})}=\frac{4\vec{a}\vec{b}}{2||\vec{a}||2||\vec{b}||}= \cos{\angle (\vec{a},-\vec{b})} $$ For the second equality: $$ \cos{\angle (\vec{a},-\vec{b})}=-\frac{\vec{a} \vec{b}}{||\vec{a}||||\vec{b}||}=-\cos{\angle (\vec{a},\vec{b})}\Rightarrow \angle (\vec{a},-\vec{b})=180-\angle (\vec{a},\vec{b})=120 $$ Note that when we refer to the angle of two vectors we consider the angle that is between $0$ and $180$ so when we found $\cos{\theta}=-\cos{60}=-\frac 1 2$ we have that $\theta=120$.

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  • $\begingroup$ @ Test123, Why is $$(\angle 2a, -\angle 2b)=(\angle a, -\angle b)=\textrm 180^\circ-(\angle a, \angle b)$$? $\endgroup$ – pi-π Feb 12 '17 at 14:37
  • $\begingroup$ @S.Ramanujan I explain each equality in the post. $\endgroup$ – Test123 Feb 12 '17 at 14:49
  • $\begingroup$ @, Test123, I mean why is this equality true? $\endgroup$ – pi-π Feb 12 '17 at 14:52
  • $\begingroup$ @S.Ramanujan Have you seen my answer? I say why the first equality holds and why the second equality holds. $\endgroup$ – Test123 Feb 12 '17 at 14:53
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Let $\bar{\theta}$ be the angle between $2\vec a$ and $-2\vec b$. So $$\cos \bar{\theta} =\dfrac {\vec 2a\times \vec -2b}{ |2\vec a|\times |-2\vec b|} = \dfrac {-4(\vec a\times \vec b)}{ 4(|\vec a|\times |\vec b|)} = -\dfrac {\vec a\times \vec b}{ |\vec a|\times |\vec b|} = -\cos(\theta) = -\frac{1}{2},$$

what implies

$$\bar{\theta}=\arccos(-1/2)=120^\circ.$$

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  • $\begingroup$ Why not $5\pi/4$ ? $\endgroup$ – A---B Feb 12 '17 at 14:22
  • $\begingroup$ @A---B For the sake of simplicity, in analytic geometry, an angle is supposed be in $[0,180^\circ)$. $\endgroup$ – Filburt Feb 12 '17 at 14:31
  • $\begingroup$ Why not $[0,\pi]$ ? $\endgroup$ – A---B Feb 12 '17 at 14:38
  • $\begingroup$ @A---B collinear vectors are supposed to have angle 0. $\endgroup$ – Filburt Feb 12 '17 at 14:40
  • $\begingroup$ So it will not matter if they have opposite direction or not. $\endgroup$ – A---B Feb 12 '17 at 15:01

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