Question

The distance of the point $\left(3,8,2\right)$ from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3x+2y-2z=0$ is

• A. $2$
• B. $3$
• C. $6$
• D. $7$

Answer 1

The correct option is D

$7$

Explanation for the correct answer:

Step 1: Find the equation of the parallel plane to $3x+2y-2z=0$ and passing through $\left(3,8,2\right)$.

Let Equation of the plane parallel to given plane be $3x+2y-2z+c=0$

This plane passes through $\left(3,8,2\right)$

Therefore

$$\begin{gathered} 3\left(3\right)+2\left(8\right)-2\left(2\right)+c=0 \\ \Rightarrow 9+16-4+c=0 \\ \Rightarrow c=-21 \end{gathered}$$

So equation of the plane is $3x+2y-2z-21=0$

Step 2: Finding the coordinates of the point of intersection of plane and line.

We now find the intersection of this plane and the given line

let $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}=k$

therefore,

$$\begin{gathered} x=2k+1 \\ y=4k+3 \\ z=3k+2 \end{gathered}$$

These points should satisfy equation of the plane, hence

$$\begin{gathered} 3\left(2k+1\right)+2\left(4k+3\right)-2\left(3k+2\right)-21=0 \\ \Rightarrow 6k+3+8k+6-6k-4-21=0 \\ \Rightarrow 8k-16=0 \\ \Rightarrow k=2 \end{gathered}$$

So the coordinates of the point are $x=5,y=11,z=8$

Step 3: Finding the distance.

So the distance between $\left(5,11,8\right)$and$\left(3,8,2\right)$ is given by:

$\sqrt{{(a_1-a_2)}^2+{(b_1-b_2)}^2+{(c_1-c_2)}^2}$

$$\begin{gathered} =\sqrt{{\left(5-3\right)}^2+{\left(11-8\right)}^2+{\left(8-2\right)}^2} \\ =\sqrt{4+9+36} \\ =\sqrt{49} \\ =7 \end{gathered}$$

Hence, option (D) is the correct answer.