Find the length of the perpendicular drawn from the origin to the plane 2x `\ 3y+6z+21=0.`

Explanation

We know that the distance of a point `left(x_{1}, y_{1}, z_{1})` from a plane `A x+B y+C z-D=0` is given by `|frac{A x_{1}+B y_{1}+C z_{1}-D}{sqrt{A^{2}+B^{2}+C^{2}}}right|`.

Therefore, the distance of origin from the plane `2 x-3 y+6 z+21` is

`|frac{2(0)-3(0)+6(0)+21}{sqrt{2^{2}+(-3)^{2}+6^{2}}} |`
`=frac{21}{sqrt{4+9+36}}`

`=frac{21}{sqrt{49}}`
`=frac{21}{7}=3`

We know that the length of perpendicular drawn from the point to the plane is same as the shortest distance of point from the plane.


Hence, the length of the perpendicular drawn from the orlgin to the plane `2 x` `-3 y+6 z+21=0` is `3` .