Question

Find the distance of the plane $2x-3y+4z-6=0$ from the origin.

Answer 1

Given equation of plane is

$2x-3y+4z-6=0$

The direction ratios of the plane
are $2,-3,4$
So, the direction cosines are

$\frac{2}{\sqrt{2^2+(-3{)}^2+4^2}},-\frac{3}{\sqrt{2^2+(-3{)}^2+4^2}},\frac{4}{\sqrt{2^2+(-3{)}^2+4^2}}$

$=\frac{2}{\sqrt{29}},-\frac{3}{\sqrt{29}},\frac{4}{\sqrt{29}}$

Dividing the equation of plane by $\sqrt{29}$,we get

$\frac{2}{\sqrt{29}}x-\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z-\frac{6}{\sqrt{29}}=0$

$\Rightarrow \frac{2}{\sqrt{29}}x-\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z=\frac{6}{\sqrt{29}}$

$\Rightarrow (x\hat{i}+y\hat{j}+z\hat{z}).(\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k})$

$=\frac{6}{\sqrt{29}}$

$\Rightarrow \overrightarrow{r}.(\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k})=\frac{6}{\sqrt{29}}$

Vector equation of a plane is

$\overrightarrow{r}.\hat{n}=d$

So, $d=\frac{6}{\sqrt{29}}$

Hence, the distance of the plane $2x-3y+4z-6=0$ from the origin is $\frac{6}{\sqrt{29}}$