Question

Let $\vec{v}$ be a vector in the plane such that $\left|v-i\right|=\left|v-2i\right|=\left|v-j\right|$ Then, $\left|v\right|$ lies in the interval

• A. (0, 1]
• B. (1, 2)
• C. (2, 3]
• D. (3, 4]

Answer 1

Answer: (C)

We have,
$\left|v-i\right|=\left|v-2i\right|=\left|v-j\right|$
Clearly, v is circumcentre of $\Delta ABC$

where $A(1,0),B(2,0),C(0,1)$ and $O(x,y)$
$\therefore O{A}^2=O{B}^2=O{C}^2$
$(x-1{)}^2+(y-0{)}^2=(x-2{)}^2+y^2$
$=x^2+(y-1{)}^2$
$\Rightarrow x^2-2x+1+y^2=x^2-4x+4+4+y^2$
$=x^2+y^2-2y+1$
$\Rightarrow 2x=3$
$\Rightarrow x=\frac{3}{2}$
$\Rightarrow x^2-2x+1+y^2$
$=x^2+y^2-2y+1$
$\Rightarrow x=y=3/2$
$\therefore (x,y)=\left(\frac{3}{2},\frac{3}{2}\right)$
$\Rightarrow \vec{v}=\frac{3}{2}\hat{i}+\frac{3}{2}\hat{j}$
$\left|\vec{v}\right|=\sqrt{\frac{9}{4}+\frac{9}{4}}$
$=\frac{3\sqrt{2}}{2}$ lie in $(2,3]$.