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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. What is the equation of the line through the origin and the centre of the sphere?

Options

  1. Ax=y=zx=y=z
  2. B2x=3y=4z2 x=3 y=4 z
  3. C6x=3y=4z6 x=3 y=4 z
  4. D6x=4y=3z6 x=4 y=3 z

Correct answer

D. 6x=4y=3z6 x=4 y=3 z

Step-by-step solution

Equation of line passing through (0,0,0)and(1,32,2)and \left(1, \frac{3}{2}, 2\right) is x−01=y−032=z−026x=4y=3z\frac{x-0}{1}=\frac{y-0}{\frac{3}{2}}=\frac{z-0}{2}6 x=4 y=3 z For question 94 and 95 Since point (2,3,−6)(2,3,-6) lies plane 2xk+2y3+z3=2∴4k+63−63=2⇒4k=2⇒k=2\begin{aligned} & \quad \frac{2 \mathrm{x}}{\mathrm{k}}+\frac{2 \mathrm{y}}{3}+\frac{\mathrm{z}}{3}=2 \\ & \therefore \quad \frac{4}{\mathrm{k}}+\frac{6}{3}-\frac{6}{3}=2 \\ & \Rightarrow \frac{4}{\mathrm{k}}=2 \Rightarrow \mathrm{k}=2\end{aligned} So, equation of plane is x+2y3+z3=2⇒3x+2y+z=6\begin{aligned} & x+\frac{2 y}{3}+\frac{z}{3}=2 \\ & \Rightarrow 3 x+2 y+z=6\end{aligned}
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