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 $\frac{|6x_1-9y_1+18z_1+20|}{\sqrt{6^2+{\left(-9\right)}^2+{\left(18\right)}^2}}$
(Using distance P\left(x_{1}, y_{1}, z_{1}\right) to plane $ax+by+cz+d=0))$
$=\frac{|3\left(2x_1-3y_1+6z_1-5\right)+15+20|}{21}$
$=\frac{0+15+20}{21}=\frac{35}{21}=\frac{5}{3}$
Short Cut Method : Using fact : Distance between the parallel planes

$\vec{r}\cdot \vec{n}=d_1$ and $\vec{r}\cdot \vec{n}=d_2$
$\therefore d=\frac{|d_1-d_2|}{|\vec{n}|}$
$=\frac{20+15}{|6\hat{i}+\left(-9\right)\hat{j}+18\hat{k}|}=\frac{35}{21}=\frac{5}{3}$