Question

The distance between the planes $\vec{r}\cdot\left(2 \hat{i}-3 \hat{j}+6 \hat{k}\right)=5 $ and $\vec{r}\cdot\left(6 \hat{i}-9 \hat{j}+18 \hat{k}\right)+20=0$ is

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{5}{3}$
• D. $\frac{7}{3}$

Answer 1

Answer: (C)

Given equation of planes are $\vec{r}\cdot\left(2 \hat{i}-3 \hat{j}+6 \hat{k}\right)-5 =0$
or $2x-3y+6z-5=0 \, \dots (i)$
and $\vec{r}\cdot\left(6 \hat{i}-9\hat{j}+18\hat{k}\right)+20=0$
or $6x -9y +18z +20=0$
Let $(x_{1}, y_{1}, z_{1})$ is a point lie in the plane (i)
$\therefore 2x_{1}-3y_{1}+6z_{1}-5=0$
Now distance from $(x_{1}, y_{1}, z_{1})$ to the plane
$6x - 9y + 18z + 20 = 0$ is $\frac{\left|6x_{1}-9y_{1}+18z_{1}+20\right|}{\sqrt{6^{2}+\left(-9\right)^{2}+\left(18\right)^{2}}}$
(Using distance P\left(x_{1}, y_{1}, z_{1}\right) to plane $ax+by+cz +d =0 ))$
$=\frac{\left|3\left(2x_{1}-3y_{1}+6z_{1}-5\right)+15+20\right|}{21}$
$=\frac{0+15+20}{21}=\frac{35}{21}=\frac{5}{3}$
Short Cut Method : Using fact : Distance between the parallel planes

$\vec{r}\cdot\vec{n}=d_{1}$ and $ \vec{r}\cdot\vec{n}=d_{2}$
$\therefore d=\frac{\left|d_{1}-d_{2}\right|}{\left|\vec{n}\right|}$
$=\frac{20+15}{\left|6 \hat{i}+\left(-9\right)\hat{j}+18 \hat{k}\right|}=\frac{35}{21}=\frac{5}{3}$