Question

Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector `2hati + hatj - 2hatk`.

Answer 1

If `hatn` is a unit vector along the normal and p is the length of the perpendicular from origin to the plane, then the vector equation of the plane is `barr.hatn` = p.

Here, `barn = 2hati + hatj - 2hatk` and p = 42

∴ `|barn| = sqrt(2^2 + 1^2 + (-2)^2`

= `sqrt(9)`

= 3

`hatn = barn/|barn|`

= `(1)/(3)(2hati + hatj - 2hatk)`

∴ The vector equation of the required plane is

`barr.[1/3(2hati + hatj - 2hatk)]` = 42

i.e. `barr.(2hati + hatj - 2hatk)` = 126.