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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2025 · Mathematics · Three Dimensional Geometry

For the following two (02) items: Suppose S is the sphere with the...

For the following two (02) items:
Suppose SS is the sphere with the smallest radius that passes through the points A(1,0,0)A(1, 0, 0), B(0,1,0)B(0, 1, 0) and C(0,0,1)C(0, 0, 1).

What is the radius of SS?

Options

  1. A13\sqrt{\dfrac{1}{3}}
  2. B23\sqrt{\dfrac{2}{3}}
  3. C13\dfrac{1}{3}
  4. D11

Correct answer

B. 23\sqrt{\dfrac{2}{3}}

Step-by-step solution

The points A(1,0,0)A(1, 0, 0), B(0,1,0)B(0, 1, 0), and C(0,0,1)C(0, 0, 1) form an equilateral triangle in 3D space.

The side length of this triangle is AB=(1−0)2+(0−1)2+(0−0)2=2AB = \sqrt{(1-0)^2 + (0-1)^2 + (0-0)^2} = \sqrt{2}.

The sphere with the smallest radius passing through three given points has its center at the circumcenter of the triangle formed by these points, and its radius is equal to the circumradius of the triangle.

For an equilateral triangle of side length aa, the circumradius is given by R=a3R = \dfrac{a}{\sqrt{3}}.

Substituting a=2a = \sqrt{2}, we get R=23=23R = \dfrac{\sqrt{2}}{\sqrt{3}} = \sqrt{\dfrac{2}{3}}.

Alternatively, the circumcenter of the equilateral triangle is its centroid G(13,13,13)G\left(\dfrac{1}{3}, \dfrac{1}{3}, \dfrac{1}{3}\right).

The radius is the distance AG=(1−13)2+(0−13)2+(0−13)2=49+19+19=69=23AG = \sqrt{\left(1 - \dfrac{1}{3}\right)^2 + \left(0 - \dfrac{1}{3}\right)^2 + \left(0 - \dfrac{1}{3}\right)^2} = \sqrt{\dfrac{4}{9} + \dfrac{1}{9} + \dfrac{1}{9}} = \sqrt{\dfrac{6}{9}} = \sqrt{\dfrac{2}{3}}.

Answer: 23\sqrt{\dfrac{2}{3}}
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