What is the diameter of the sphere?

Direction: Consider the following for the two (02) items that follow :<br>Let 2x^{2}+2y^{2}+2z^{2}+3x+3y+3z-6=0 be a sphere.

  1. A. \frac{5\sqrt{3}}{4}
  2. B. \frac{5\sqrt{3}}{2}
  3. C. \frac{3\sqrt{5}}{4}
  4. D. \frac{3\sqrt{5}}{2}

Correct Answer: B. \frac{5\sqrt{3}}{2}

Explanation

Divide the equation by 2 to get standard form: x^2+y^2+z^2 + \frac{3}{2}x + \frac{3}{2}y + \frac{3}{2}z - 3 = 0. The center is (-u, -v, -w) = (-\frac{3}{4}, -\frac{3}{4}, -\frac{3}{4}). The radius is r = \sqrt{(-\frac{3}{4})^2 + (-\frac{3}{4})^2 + (-\frac{3}{4})^2 - (-3)} = \sqrt{\frac{27}{16} + 3} = \sqrt{\frac{75}{16}} = \frac{5\sqrt{3}}{4}. The diameter is 2r = 2 \times \frac{5\sqrt{3}}{4} = \frac{5\sqrt{3}}{2}.

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