Can Shifting Vectors Affect Their Exponential Distance Sum Equality?

[thatboi]

Hi all,
Suppose I had some some n-dimensional vectors $\vec{a}_{1}, \vec{a}_{2}, \vec{b}_{1},\vec{b}_{2}$ that satisfied $e^{||\vec{a}_{1}||^2}+e^{||\vec{a}_{2}||^2}=e^{||\vec{b}_{1}||^2}+e^{||\vec{b}_{2}||^2}$. Now suppose there was another non-zero n-dimensional vector $\vec{A}$. Is there anything I can say about the equation $e^{||\vec{a}_{1}-\vec{A}||^2}+e^{||\vec{a}_{2}-\vec{A}||^2}=e^{||\vec{b}_{1}-\vec{A}||^2}+e^{||\vec{b}_{2}-\vec{A}||^2}$? For example, is the equation satisfied for $\vec{a}_{i} \neq \vec{b}_{j}$ for $i,j = {1,2}$. Also I mean $||\cdot||$ as in the L2 norm.

 

[scottdave]

Can you find an example that fails your equation? Start with a simple (1-dimensional) case.

 

[thatboi]

Can you find an example that fails your equation? Start with a simple (1-dimensional) case.

I suppose even in the 1-dimensional case, my second equation is satisfied as long as $a_{1}=b_{1}$, $a_{2}=b_{2}$ or vice versa right. I was just wondering if there was some other solutions $a_{1},b_{1}$ that satisfied the set of equations above.

 

[scottdave]

I suppose even in the 1-dimensional case, my second equation is satisfied as long as $a_{1}=b_{1}$, $a_{2}=b_{2}$ or vice versa right. I was just wondering if there was some other solutions $a_{1},b_{1}$ that satisfied the set of equations above.

What happens when $a_{1}= -b_{1}$, $a_{2}= -b_{2}$ for example?