Let A be the set of vectors a = (a_1, a_2, a_ | vedclass

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(B) Given the equation: $\left(\frac{a_1}{2} + \frac{a_2}{4} + \frac{a_3}{8}\right)^2 = \frac{a_1^2}{2} + \frac{a_2^2}{4} + \frac{a_3^2}{8}$.Expanding

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$A$ contains exactly one element

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(B) Given the equation: $\left(\frac{a_1}{2} + \frac{a_2}{4} + \frac{a_3}{8}\right)^2 = \frac{a_1^2}{2} + \frac{a_2^2}{4} + \frac{a_3^2}{8}$.Expanding the left side: $\frac{a_1^2}{4} + \frac{a_2^2}{16} + \frac{a_3^2}{64} + \frac{a_1 a_2}{4} + \frac{a_2 a_3}{16} + \frac{a_3 a_1}{8} = \frac{a_1^2}{2} + \frac{a_2^2}{4} + \frac{a_3^2}{8}$.Multiplying by $64$ to clear denominators: $16a_1^2 + 4a_2^2 + a_3^2 + 16a_1 a_2 + 4a_2 a_3 + 8a_3 a_1 = 32a_1^2 + 16a_2^2 + 8a_3^2$.Rearranging terms: $16a_1^2 + 12a_2^2 + 7a_3^2 - 16a_1 a_2 - 4a_2 a_3 - 8a_3 a_1 = 0$.This can be rewritten as: $8(a_1 - a_2)^2 + (2a_2 - a_3)^2 + 2(a_3 - 2a_1)^2 + 4a_3^2 = 0$.Since the sum of squares is zero,each term must be zero: $a_1 - a_2 = 0$,$2a_2 - a_3 = 0$,$a_3 - 2a_1 = 0$,and $a_3 = 0$.This implies $a_1 = a_2 = a_3 = 0$.Thus,the set $A$ contains exactly one element,which is the zero vector $(0, 0, 0)$.

Let $A$ be the set of vectors $a = (a_1, a_2, a_3)$ satisfying $\left(\sum_{i=1}^3 \frac{a_i}{2^i}\right)^2 = \sum_{i=1}^3 \frac{a_i^2}{2^i}$. Then,

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