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