0
$\begingroup$

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?

  • 3
    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 $120$. – 2017-02-12
  • 0
    @ Test 123, how to draw a figure? – 2017-02-12
  • 0
    Draw two vectors at the origin with some angle and check what happens when you take $-\vec{b}$. – 2017-02-12
  • 0
    Note that simply doubling $a$ changes nothing about what angles it makes with other vectors. – 2017-02-12
  • 0
    In fact, $\vec a \times \vec b = |\vec a||\vec b|\sin \theta$. – 2017-02-12
  • 0
    @ Rohan, sin or cos? – 2017-02-12
  • 0
    @S.Ramanujan It's $\sin$ but you consider the cross product not the inner product of the vectors – 2017-02-12
  • 0
    @ Test 123, here I am considering dot products, aren't I? – 2017-02-12
  • 0
    @S.Ramanujan Yes, the formula with cosine involves dot products – 2017-02-12

2 Answers 2

1

$\angle (2\vec{a},-2\vec{b})=\angle (\vec{a},-\vec{b})=180-\angle(\vec{a},\vec{b})$

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 $\theta=120$.

  • 0
    @ Test123, Why is $$(\angle 2a, -\angle 2b)=(\angle a, -\angle b)=\textrm 180^\circ-(\angle a, \angle b)$$? – 2017-02-12
  • 0
    @S.Ramanujan I explain each equality in the post. – 2017-02-12
  • 0
    @, Test123, I mean why is this equality true? – 2017-02-12
  • 0
    @S.Ramanujan Have you seen my answer? I say why the first equality holds and why the second equality holds. – 2017-02-12
3

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

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

  • 0
    Why not $5\pi/4$ ? – 2017-02-12
  • 0
    @A---B For the sake of simplicity, in analytic geometry, an angle is supposed be in $[0,180^\circ)$. – 2017-02-12
  • 0
    Why not $[0,\pi]$ ? – 2017-02-12
  • 0
    @A---B collinear vectors are supposed to have angle 0. – 2017-02-12
  • 0
    So it will not matter if they have opposite direction or not. – 2017-02-12
  • 0
    yeap, this is the convention – 2017-02-12