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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2024 · Mathematics · Three Dimensional Geometry

Paragraph: Let 2 x^2+2 y^2+2 z^2+3 x+3 y+3 z-6=0 be a sphere....

Paragraph: Let 2x2+2y2+2z2+3x+3y+3z−6=02 x^2+2 y^2+2 z^2+3 x+3 y+3 z-6=0 be a sphere.
Question: What is the diameter of the sphere?

Options

  1. A534\frac{5 \sqrt{3}}{4}
  2. B532\frac{5 \sqrt{3}}{2}
  3. C354\frac{3 \sqrt{5}}{4}
  4. D352\frac{3 \sqrt{5}}{2}

Correct answer

B. 532\frac{5 \sqrt{3}}{2}

Step-by-step solution

(b) Given sphere : 2x2+2y2+2z2+3x+3y+3z−6=02 x^2+2 y^2+2 z^2+3 x+3 y+3 z-6=0....(i) Divide equation (1) by (2) ⇒x2+y2+z2+32x+32y+32z−3=0\Rightarrow x^2+y^2+z^2+\frac{3}{2} x+\frac{3}{2} y+\frac{3}{2} z-3=0 radius of  sphere =(34)2+(34)2+(34)2+3=2716+3=754radius =534Diameter of sphere =532\begin{aligned} & \text{ radius\text{ of } sphere }=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^2+3}=\sqrt{\frac{27}{16}+3}=\frac{\sqrt{75}}{4} \\ & \text{radius }=\frac{5 \sqrt{3}}{4} \\ & \text{Diameter of sphere }=\frac{5 \sqrt{3}}{2}\end{aligned}$$
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