Question

Find the distance between the parallel planes $\overline{r}.\left(2\overline{i}-3\overline{j}+6\overline{k}\right)=5$ and $\overline{r}.\left(6\overline{i}-9\overline{j}+18\overline{k}\right)+20=0$.

Answer 1

Plane 1 : $\overrightarrow{r}.(2\hat{i}-3\hat{j}+6\hat{k})=5$

Cartesian form equation,

$2x-3y+6z-5=0$

Plane 2 : $\overrightarrow{r}.(6\hat{i}-9\hat{j}+18\hat{k})+20=0$

Cartesian form equation,

$6x-9y+18z+20=0$

Taking plane 1,

Putting $y$ and $z$ coordinates = $0$

$2x=5$

$x=\frac{5}{2}$

coordinates = $(\frac{5}{2},0,0)$

Distance of point $(\frac{5}{2},0,0)$ from the plane $6x-9y+18z+20=0$

$=\left|\frac{6\left(\frac{5}{2}\right)-9\left(0\right)+18\left)\right(0)+20}{\sqrt{(6{)}^2+(9{)}^2+(18{)}^2}}\right|$

$=\frac{35}{\sqrt{441}}$

$=\frac{35}{21}$

$=\frac{5}{3}$ units