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On the Exercise they give that:

$||\vec{a}|| = ||\vec{c}|| = 5$

$||\vec{b}||= 1$

$ \alpha = \angle(\vec{a},\vec{b}) = \frac\pi8 $

$|| \vec{a} + \vec{b} + \vec{c}|| = || \vec{a} - \vec{b} + \vec{c}|| $

And the exercise wants to know the angle $\theta = \angle(\vec{b},\vec{c},)$

The answer is $\frac78\pi$, but I don't know how to get there, can someone please help me?

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Since $|| \vec{a} + \vec{b} + \vec{c}|| = || \vec{a} - \vec{b} + \vec{c}|| $ we can get from here that: $$(\vec{a}+\vec{c}).\vec{b}=0\to \vec{b}.\vec{c}=-\vec{a}.\vec{b}$$ also $$\vec{a}.\vec{b}=|\vec{a}||\vec{b}|\cos{\theta_{ab}}=5\cos\frac{\pi}{8}$$therefore $$\vec{b}.\vec{c}=-5\cos\frac{\pi}{8}=5\cos\frac{7\pi}{8}=|\vec{b}||\vec{c}|\cos{\theta_{bc}}$$. So we have $$\theta_{bc}=\frac{7\pi}{8}$$

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  • $\begingroup$ Why $$\vec{b}.\vec{c}=-5\cos\frac{\pi}{8}=5\cos\frac{7\pi}{8}=|\vec{b}||\vec{c}|\cos{\theta_{bc}}$$ ? $\endgroup$ – Fernando Gardeazabal Jan 24 '18 at 18:23
  • $\begingroup$ The 2nd equality comes from $-\cos \theta=\cos (\pi-\theta)$ and the 3rd one from definition of inner product... $\endgroup$ – Mostafa Ayaz Jan 24 '18 at 18:25
  • $\begingroup$ Sorry , I have a lot of trouble understanding this, where does the 7 come from? $\endgroup$ – Fernando Gardeazabal Jan 24 '18 at 18:53
  • $\begingroup$ $$\frac{7\pi}{8}=\pi-\frac{\pi}{8}$$ $\endgroup$ – Mostafa Ayaz Jan 24 '18 at 19:02
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Guide:

if $\| u + v\| = \| u - v\|$, this means that $\langle u, v \rangle = 0$.

Hence $\langle a+c, b \rangle = 0$.

$\langle a,b\rangle +\langle c,b\rangle =0 $

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After squaring of the both sides we obtain: $$\vec{a}\vec{b}+\vec{c}\vec{b}=0,$$ wshich gives $$5\cos\frac{\pi}{8}+5\cos\theta=0,$$ which gives $$\theta=\frac{7\pi}{8}.$$

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