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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2022 · Mathematics · Three Dimensional Geometry

The plane 6 x+k y+3 z-12=0 where meets the coordinate axes at A, B...

The plane 6x+ky+3z−12=06 x+k y+3 z-12=0 where meets the coordinate axes at A,BA, B and CC respectively. The equation of the sphere passing through the origin and A,B,CA, B, C is x2+y2+z2−2x−3y−4z=0x^2+y^2+z^2-2 x-3 y-4 z=0. If pp is the perpendicular distance from the centre of the sphere to the plane, then which one of the following is correct?

Options

  1. A0<p<0.50 < p < 0.5
  2. B0.5<p<10.5 < p < 1
  3. C1<p<1.51 < p < 1.5
  4. Dp>1.5p > 1.5

Correct answer

B. 0.5<p<10.5 < p < 1

Step-by-step solution

Perpendicular distance from centre (1,32,2)\left(1, \frac{3}{2}, 2\right) to theplane 6x+4y+3z−12=0P=6+4(32)+3(2)−1236+16+9=6+6+6−1261=661=0.74\begin{aligned} & 6 x+4 y+3 z-12=0 \\ & P=\frac{6+4\left(\frac{3}{2}\right)+3(2)-12}{\sqrt{36+16+9}} \\ & =\frac{6+6+6-12}{\sqrt{61}}=\frac{6}{\sqrt{61}}=0.74\end{aligned}
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