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NDA 2012
Exam
NDA
Year
2012
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2012 · Mathematics · Three Dimensional Geometry

What is the diameter of the sphere x^2+y^2+z^2-4 x+6 y-8 z-7=0

What is the diameter of the sphere x2+y2+z2−4x+6y−8z−7=0x^2+y^2+z^2-4 x+6 y-8 z-7=0

Options

  1. A4 units
  2. B5 units
  3. C6 units
  4. D12 units

Correct answer

C. 6 units

Step-by-step solution

Compare with x2+y2+z2+2ux+2vy+2wz+d=0u=−2,v=3,w=−4,d=−7 radius =u2+v2+w2−d=4+9+16+7=36=6.\begin{aligned} & x^2+y^2+z^2+2 u x+2 v y+2 w z+d=0 \\ & u=-2, v=3, w=-4, d=-7 \\ & \text{ radius }=\sqrt{u^2+v^2+w^2-d} \\ & =\sqrt{4+9+16+7} \\ & =\sqrt{36}=6. \end{aligned}
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