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RE: Math Contest #25 [2 SBI]

in #puzzle5 years ago

I read this only now...
Anyway in order to be orthogonal the scalar product must be zero.
The scalar product is 23a^2 + 2a , so "a" has to be -2/23 and x is free. But x has to be different from 0, otherwise the first vector is not defined. Another solution is a = 0 and x different from 0 but we get a null vector for the first one and it is not so much interesting.
Bye