B Can Shifting Vectors Affect Their Exponential Distance Sum Equality?

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