Question

The ratio of the distances from $\left( { - 1,1,3} \right)$ and $\left( {3,2,1} \right)$ to the plane $2x + 5y - 7z + 9 = 0$ is
A. $2:1$
B. $1:3$
C. $1:1$
D. $1:2$

Answer 1

First, we will use the perpendicular distance from a point $\left( {m,n,q} \right)$ to the plane $Ax + By + Cz + D = 0$ is $P = \dfrac{{\left| {Am + Bn + Cq + D} \right|}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}$. Then we will substitute the two points in the above equation to find their perpendicular distances ${P_1}$ and ${P_2}$ respectively. Then we will find the ratio of the above perpendicular distances to the plane.

Complete step by step solution:
Given the equation of the plane is $2x + 5y - 7z + 9 = 0$.

We know that the distance from a point $\left( {m,n,q} \right)$ to the plane $Ax + By + Cz + D = 0$ is $P = \dfrac{{\left| {Am + Bn + Cq + D} \right|}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}$.

First, we will find the distance from a point $\left( { - 1,1,3} \right)$ to the plane $2x + 5y - 7z + 9 = 0$.

$${P_1} = \dfrac{{\left| {2\left( { - 1} \right) + 5\left( 1 \right) - 7\left( 3 \right) + 9} \right|}}{{\sqrt {{2^2} + {5^2} + {{\left( { - 7} \right)}^2}} }} \\\ = \dfrac{{\left| { - 2 + 5 - 21 + 9} \right|}}{{\sqrt {4 + 25 + 49} }} \\\ = \dfrac{{\left| { - 9} \right|}}{{\sqrt {78} }} \\\ = \dfrac{9}{{\sqrt {78} }} \\\$$

Now, we will find the distance from point $\left( {3,2,1} \right)$ to the given plane.

$${P_2} = \dfrac{{\left| {2\left( 3 \right) + 5\left( 2 \right) - 7\left( 1 \right) + 9} \right|}}{{\sqrt {{2^2} + {5^2} + {{\left( { - 7} \right)}^2}} }} \\\ = \dfrac{{\left| {6 + 10 - 7 + 9} \right|}}{{\sqrt {4 + 25 + 49} }} \\\ = \dfrac{{\left| { - 9} \right|}}{{\sqrt {78} }} \\\ = \dfrac{{18}}{{\sqrt {78} }} \\\$$

Dividing ${P_1}$ by ${P_2}$, we get

$$\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{\dfrac{9}{{\sqrt {78} }}}}{{\dfrac{{18}}{{\sqrt {78} }}}} \\\ = \dfrac{9}{{\sqrt {78} }} \times \dfrac{{\sqrt {78} }}{{18}} \\\ = \dfrac{9}{{18}} \\\ = \dfrac{1}{2} \\\$$

Thus, the ratio of the distances from $\left( { - 1,1,3} \right)$ and $\left( {3,2,1} \right)$ to the plane $2x + 5y - 7z + 9 = 0$ is $1:2$.

Hence, the option D is correct.

Note:
In this question, some students use the distance formula instead of the perpendicular formula, which is wrong. Also, we calculate the ratio of the distance of the points from the plane by dividing ${P_1}$ with ${P_2}$, where ${P_1}$ is the distance of the closet point to the plane.