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JEE Advanced 2013Paper 2
Exam
JEE Advanced
Year
2013
Paper
Paper 2
Subject
Mathematics
Chapter
Straight Lines

JEE Advanced 2013 · Paper 2 · Mathematics · Straight Lines

Paragraph: Let S=S_ 1 S_ 2 S_ 3 , where S_ 1 =\ z ∈ ℂ:|z| 0\ and S_...

Paragraph: Let S=S1∩S2∩S3S=S_{1} \cap S_{2} \cap S_{3}, where S1={z∈C:∣z∣<4},S2={z∈C:Im⁡[z−1+3i1−3i]>0}S_{1}=\{z \in \mathbb{C}:|z|<4\}, \quad S_{2}=\left\{z \in \mathbb{C}: \operatorname{Im}\left[\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right]>0\right\} and S3={z∈C:Re⁡z>0}S_{3}=\{z \in \mathbb{C}: \operatorname{Re} z>0\}.
Question: min⁡z∈S∣1−3i−z∣=\min _{z \in S}|1-3 i-\mathrm{z}|=

Options

  1. A2−32\dfrac{2 - \sqrt{3}}{2}
  2. B2+32\dfrac{2 + \sqrt{3}}{2}
  3. C3−32\dfrac{3 - \sqrt{3}}{2}
  4. D3+32\dfrac{3 + \sqrt{3}}{2}

Correct answer

C. 3−32\dfrac{3 - \sqrt{3}}{2}

Step-by-step solution

\mathop{\text{min}}_{\text{z} \in \text{s}} \left|1 - 3 \text{i} - \text{z}\right|perpendicularlengthofpoint(1,−3)fromlineperpendicular length of point (1, -3) from line\sqrt{3} \text{x} + \text{y} = 0<br/>∣3−33+1∣=∣3−32∣=3−32<br />\left|\dfrac{\sqrt{3} - 3}{\sqrt{3 + 1}}\right| = \left|\dfrac{\sqrt{3} - 3}{2}\right| = \dfrac{3 - \sqrt{3}}{2}
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