Maximum value of $|\overrightarrow{a}-2\overrightarrow{b}|^2+|\overrightarrow{b}-2\overrightarrow{c}|^2+|\overrightarrow{c}-2\overrightarrow{a}|^2$

Question

I have a question the goes, if $|\overrightarrow{a}|=|\overrightarrow{b}| = 1$ and $|\overrightarrow{c}|=2$, then what is the maximum value of :

$|\overrightarrow{a}-2\overrightarrow{b}|^2+|\overrightarrow{b}-2\overrightarrow{c}|^2+|\overrightarrow{c}-2\overrightarrow{a}|^2$ ?

Here's what I tried:

$|\overrightarrow{a}-2\overrightarrow{b}|^2+|\overrightarrow{b}-2\overrightarrow{c}|^2+|\overrightarrow{c}-2\overrightarrow{a}|^2 = 30-4\cos\theta_1-8\cos\theta_2-8\cos\theta_3$

where $\theta_1=$ angle between $\overrightarrow{a}$ and $\overrightarrow{b}$, $\theta_2=$ angle between $\overrightarrow{b}$ and $\overrightarrow{c}$, $\theta_3=$ angle between $\overrightarrow{c}$ and $\overrightarrow{a}$

and for maximum value, I put $\cos\theta_1=\cos\theta_2=\cos\theta_3=-1$, and got the maximum value of $50$ but the answer says $42$. Where did I go wrong and what would be the correct approach? Was I wrong in putting $\cos\theta_1=\cos\theta_2=\cos\theta_3=-1$? Why?

Answer 1

By putting all $\cos \theta_i=-1$ you imply that all the vectors make an angle $\pi$ with each other !

start off with $${|\vec{a}+\vec{b}+\vec{c}|}^2\ge 0$$ to get $$-\sum \vec{a}\cdot \vec{b}\le 3$$ therefore $$\sum {|\vec{a}-\vec{2b}|}^2=30-4\sum \vec{a}\cdot \vec{b}\le 42$$