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Let $\overrightarrow{\mathrm{v}}$ be a vector in the plane such that $|\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{i}}|=|\overrightarrow{\mathrm{v}}-\overrightarrow{2 \mathrm{i}}|=|\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{j}}| .$ Then $|\overrightarrow{\mathrm{v}}|$ lies in the interval -
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The correct answer is:
$(2,3]$
$\forall \mathrm{A} \equiv(1,0), \quad \mathrm{B} \equiv(0,1), \quad \mathrm{C}(2,0)$
Let $V(x, y)$
$\mathrm{VA}=\mathrm{VB}=\mathrm{VC}$
$(x-1)^{2}+y^{2}=x^{2}+(y-1)^{2}=(x-2)^{2}+y^{2}$
$(x, y)=\left(\frac{3}{2}, \frac{3}{2}\right)$
$V=\frac{3 i+3 j}{2}$
$|\mathrm{v}|=\frac{3}{\sqrt{2}} \in(2,3)$
Let $V(x, y)$
$\mathrm{VA}=\mathrm{VB}=\mathrm{VC}$
$(x-1)^{2}+y^{2}=x^{2}+(y-1)^{2}=(x-2)^{2}+y^{2}$
$(x, y)=\left(\frac{3}{2}, \frac{3}{2}\right)$
$V=\frac{3 i+3 j}{2}$
$|\mathrm{v}|=\frac{3}{\sqrt{2}} \in(2,3)$
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