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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2018 · Mathematics · Three Dimensional Geometry

What is the distance of the point (2,3,4) from the plane 3 x-6 y+2...

What is the distance of the point (2,3,4)(2,3,4) from the plane 3x−6y+2z+11=?3 \mathrm{x}-6 \mathrm{y}+2 \mathrm{z}+11=?

Options

  1. A1 unit
  2. B2 unit
  3. C3 unit
  4. D4 unit

Correct answer

A. 1 unit

Step-by-step solution

 Distance =∣3×2−6×3+2×4+1132+(−6)2+(2)2∣=∣6−18+8+1149∣=1\begin{aligned} \text{ Distance } & =\left|\frac{3 \times 2-6 \times 3+2 \times 4+11}{\sqrt{3^2+(-6)^2+(2)^2}}\right| \\ & =\left|\frac{6-18+8+11}{\sqrt{49}}\right|=1\end{aligned}
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