TITLE: Find Value of $\vec a \cdot \vec b+\vec b \cdot \vec c+\vec a \cdot \vec c$?
QUESTION [0 upvotes]: Let $\vec a, \vec b,\vec c$ be three unit vector such that $\vec a \cdot \vec b+\vec b \cdot \vec c-\vec a \cdot \vec c=\frac{3}{2}$ then value of $\vec a \cdot \vec b+\vec b \cdot \vec c+\vec a \cdot \vec c$


My Approach:
let $\vec a \cdot \vec b=\vec b \cdot \vec c=\frac{1}{2}$ and $\vec a \cdot \vec c =\frac{-1}{2}$
So,
$\vec a \cdot \vec b+\vec b \cdot \vec c+\vec a \cdot \vec c=\frac{1}{2}$
I got the answer but how do i prove that $\vec a \cdot \vec b=\vec b \cdot \vec c=\frac{1}{2}$ and $\vec a \cdot \vec c =\frac{-1}{2}$
or
Is this question wrong?

REPLY [2 votes]: A solution without complex numbers. From the comment of achille hui it follows that $b=a+c$ since $|a-b+c|^2=0$.
Now this implies from the condition and the fact the all vectors are unit vectors:
$|a+c|^2-a \cdot c=3/2$
Therefore expanding we get $a \cdot c=-1/2$
The wanted expression is instead:
$|a+c|^2+a\cdot c=1+1+3a \cdot c=1/2$
That is equal to the result of Jack D'Aurizio.
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