Question

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

• A. $A$ is empty
• B. $A$ contains exactly one element
• C. $A$ has $6$ elements
• D. $A$ has infinitely many elements

Answer 1

The correct option is B $A$ contains exactly one element

Let $\overrightarrow{p}=(\frac{a_1}{\sqrt{2}},\frac{a_2}{\sqrt{4}},\frac{a_3}{\sqrt{8}})\&\overrightarrow{q}=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{4}},\frac{1}{\sqrt{8}})$ and angle between them is $\theta$

Given that ${\left(\sum _{i=1}^3\frac{a_i}{2^i}\right)}^2=\sum _{i=1}^3\frac{a_i^2}{2^i}.$

$\Rightarrow \frac{7}{8}{\left(\sum _{i=1}^3\frac{a_i}{2^i}\right)}^2=\frac{7}{8}\sum _{i=1}^3\frac{a_i^2}{2^i}$

$\Rightarrow \frac{7}{8}{(\frac{a_1}{2}+\frac{a_2}{4}+\frac{a_3}{8})}^2=\frac{7}{8}(\frac{a_1^2}{2}+\frac{a_2^2}{4}+\frac{a_3^2}{8})\Rightarrow \frac{7}{8}{(\overrightarrow{p}\cdot \overrightarrow{q})}^2=|\overrightarrow{p}{|}^2|\overrightarrow{q}{|}^2\Rightarrow \frac{7}{8}|\overrightarrow{p}{|}^2|\overrightarrow{q}{|}^2{\cos }^2\theta =|\overrightarrow{p}{|}^2|\overrightarrow{q}{|}^2\Rightarrow |\overrightarrow{p}|=0\text{or}{\cos }^2\theta =\frac{8}{7}$

${\cos }^2\theta =\frac{8}{7}$ is not possible. $\Rightarrow |\overrightarrow{p}|=0\Rightarrow \overrightarrow{p}=(0,0,0)\Rightarrow \overrightarrow{a}=(0,0,0)$
Therefore, the equation has only one solution.