How do I calculate the angle between this two vectors?

Question

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?

Answer 1

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 $$\theta_{bc}=\frac{7\pi}{8}$$

Answer 2

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 $

Answer 3

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 $$\theta=\frac{7\pi}{8}.$$