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UPSC NDA II 2015 Mathematics Paper

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Question : 93 of 120

Marks: +1, -0

The radius of the sphere

3x^{2} + 3y^{2} + 3z^{2} - 8x + 4y + 8z - 15 = 0

2
3
4
5

Solution:

Concept:

The radius of a sphere is found from its equation in standard form.

Explanation:

Divide the given equation

3x^{2} + 3y^{2} + 3z^{2} - 8x + 4y + 8z - 15 = 0

by

3

.

We get

x^{2} + y^{2} + z^{2} - \frac{8}{3}x + \frac{4}{3}y + \frac{8}{3}z - 5 = 0

.

Compare with the standard form

x^{2} + y^{2} + z^{2} + 2ux + 2vy + 2wz + d = 0

.

Thus

2u = -\frac{8}{3} \Rightarrow u = -\frac{4}{3}

.

2v = \frac{4}{3} \Rightarrow v = \frac{2}{3}

.

2w = \frac{8}{3} \Rightarrow w = \frac{4}{3}

.

d = -5

.

Radius

r = \sqrt{u^{2} + v^{2} + w^{2} - d}

.

Substitute:

r = \sqrt{\left(-\frac{4}{3}\right)^{2} + \left(\frac{2}{3}\right)^{2} + \left(\frac{4}{3}\right)^{2} - (-5)}

.

r = \sqrt{\frac{16}{9} + \frac{4}{9} + \frac{16}{9} + 5} = \sqrt{\frac{36}{9} + 5} = \sqrt{4 + 5} = \sqrt{9} = 3

.

Answer:

3

(Option B).

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