The foot of the perpendicular drawn from the origin to a plane is `(4,-2,-5)`. Find the equation of the plane in (i) vector form, ii) Cartesian form.

Explanation

Let O(0,0,0) be the origin and P(4,-2,-5) be the foot o the perpendicular drawn from O to the given plane.
i) Clearly, the required plane passes through the point P and it is perpendicular to OP.
`therefore veca=(4hati-2hatj)-5hatk)` and `vecn=vec(OP)(4hati-2hatj-5hatk)`.
So, the required equation of the plane is
`vecr.vecn=vecavecn`

`rArr vecr=(4hati-2hatj-5hatk)`
`=(4hati-2hatj-5hatk).(4hati-2hatj-5hatk)`
`=(16+4+25)=45`.
Hence, the vector equation of the plane is
`vecr.(4hati-2hatj-5hatk)=45`...................(i)
`vecr.(4hati-2hatj+zhatk)/(4hati-2hatj-5hatk)=45`.
`rArr 4x-2y-5z=45`, which is the required equation of the plane is Cartesian form.