Question

The distance of the point $\left(1,-2,3\right)$ from the plane $\mathrm{x}-\mathrm{y}+\mathrm{z}=5$ measured parallel to the line $\left(\frac{x}{2}\right)=\left(\frac{y}{3}\right)=\left(\frac{z}{-6}\right)$ is:

• A. $\frac{1}{7}$
• B. $7$
• C. $\frac{7}{5}$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for correct option(s):

Finding the distance of the point from the plane.

The line parallel to $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{-6}$ and passing through $(1,-2,3)$ is $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}-3}{-6}=\mathrm{t}\left(\mathrm{i}\right)$

Cross multiplying and simplifying $\left(i\right)$,

$\Rightarrow \mathrm{x}=2\mathrm{t}+1,\mathrm{y}=3\mathrm{t}-2,\mathrm{z}=3-6\mathrm{t}$

Applying values of $x,y\mathrm{and}z$ to the plane,

$\mathrm{x}-\mathrm{y}+\mathrm{z}=5$

Simplifying,

$\begin{array}{rcl}& \Rightarrow & 2\mathrm{t}+1+2-3\mathrm{t}+3-6\mathrm{t}=5\\ & \Rightarrow & 6-7\mathrm{t}=5\\ & \Rightarrow & \mathrm{t}=\frac{-1}{7}\end{array}$

Hence, putting the value of $t$, we get the value of $x,y\mathrm{and}z$ as: $\left(\frac{5}{7},-\frac{17}{7},\frac{27}{7}\right)$

Therefore, the distance is given by:

$$\begin{gathered} \text{Distance}=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}} \\ =\sqrt{1} \\ =1 \end{gathered}$$

The distance of the point from the plane measured parallel to the line is $\mathbf{1}$ unit.

Hence, the correct answer is option (D).