Question

Find the distance between the planes $r(2\hat{i}-\hat{j}+3\hat{k})-4=0$ and $r(6\hat{i}-3\hat{j}+9\hat{k})+13=0$.

• A. $\frac{5}{3\sqrt{14}}$
• B. $\frac{10}{3\sqrt{14}}$
• C. $\frac{25}{3\sqrt{14}}$
• D. None of these

Answer 1

The correct option is C

$\frac{25}{3\sqrt{14}}$

Explanation for the correct option:

Step 1. Simplify those two equations

Given equations are

$\begin{array}{rcl}r(2\hat{i}-\hat{j}+3\hat{k})-4& =& 0\end{array}$…..(i)

$\begin{array}{rcl}r(6\hat{i}-3\hat{j}+9\hat{k})+13& =& 0\end{array}$…..(ii)

Equation (ii) can be written as

$r(2\hat{i}-\hat{j}+3\hat{k})+\frac{13}{3}=0$ …..(iii)

Step:2 Calculate the distance between two planes.

Distance between two lines are given by$\frac{{\mathbf{a}}_{\mathbf{1}}\mathbf{-}{\mathbf{a}}_{\mathbf{2}}}{\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}}$,

We have,

$\begin{array}{rcl}{a}_1& =& \frac{13}{3},\\ a_2& =& -4,\\ x& =& 2\\ y& =& -1\\ z& =& 3\end{array}$

Put all the values in distance formula

$\therefore$the distance between these two parallel lines $=\frac{{\displaystyle \frac{13}{3}-(-4)}}{\sqrt{2^2+{(-1)}^2+3^2}}$

$=\frac{{\displaystyle \frac{13+12}{3}}}{\sqrt{14}}$

$=\frac{\mathbf{25}}{\mathbf{3}\sqrt{\mathbf{14}}}$

Hence, Option $\left(C\right)$ is the correct answer.