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

Explanation

The equation of plane is 2 x-3 y+4 z-6=0
`rightarrow 2 x-3 y+4 z=6`

It is known that the distance of a plane A x+B y+C z=D from a point `(x_{1}, y_{1}, z_{1})` is given by


`d=frac{|A x_{1}+B y_{1}+C z_{1}-D|}{sqrt{A^{2}+B^{2}+C^{2}}}`


So, the distance of the plane 2 x-3 y+4 z-6=0 from the origin is
`d= frac{|2 times 0-3 times 0+4 times 0-6|}{sqrt{2^{2}+(-3^{2})+4^{2}}}`
`=frac{|-6|}{sqrt{4+9+16}}`
`=frac{6}{sqrt{29}}` units.

Therefore, the dlstance of the plane 2 x-3 y+4 z-6=0 from the orlgin is `frac{6}{sqrt{29}}` unlts.