Question

The distance of the plane $6\mathrm{x}-3\mathrm{y}+2\mathrm{z}+14=0$ from the origin is

• A. $2$
• B. $1$
• C. $14$
• D. $8$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

Finding the distance of the plane from the origin:

Given, $6\mathrm{x}-3\mathrm{y}+2\mathrm{z}+14=0\dots \dots \left(1\right)$

This equation is of the format, $\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\dots \dots \left(2\right)$

Thus, the distance of a plane $\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0$from the origin is given by $\frac{\left|\mathrm{d}\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}$.

Comparing the equation $\left(1\right)\text{and}\left(2\right)$ we get the values as

$\mathrm{a}=6,\mathrm{b}=-3,\mathrm{c}=2,\mathrm{d}=14$

Substituting the values we get,

$\text{Distance}=\frac{\left|14\right|}{\sqrt{36+9+4}}$

$=\frac{14}{\sqrt{49}}$

$=\frac{14}{7}$

$=\mathrm{2}$

Hence, option (A) is the correct answer.