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UPSC NDA I 2022 Mathematics Solved Paper

Question Numbers: 91-93

Consider the following for the next three (03) items that follow

The Plane

6x + ky + 3z - 12 = 0

where

k \neq 0

meets the coordinate axes at A, B and C respectively. The equation of the sphere passing through the origin and A, B, C is

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

Question : 92 of 120

Marks: +1, -0

If p is the perpendicular distance from the centre of the sphere to the plane, then which one of the following is correct ?

$0 < p < 0.5$
$0.5 < p < 1$
$1 < p < 1.5$
$p > 1.5$

Validate

Solution:

Concept:

The perpendicular distance from a point to a plane is calculated using the standard distance formula. The center of the sphere is derived from its general equation.

Explanation:

The sphere equation is

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

Its center is

(1,\frac{3}{2},2)

, obtained from

(-u,-v,-w)

The plane equation is

6x+4y+3z-12=0

Distance

p = \frac{|6(1)+4(\frac{3}{2})+3(2)-12|}{\sqrt{6^{2}+4^{2}+3^{2}}}

Numerator:

6 + 6 + 6 - 12 = 6

Denominator:

\sqrt{36+16+9}=\sqrt{61}\approx 7.81

p = \frac{6}{\sqrt{61}} \approx 0.768

This value is greater than

0.5

and less than

1

Hence, it lies in the range

0.5 < p < 1

Answer:

Option B:

0.5 < p < 1

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