Find the vector and Cartesian equations of a plane which is at a distance of from the origin and whose normal vector from the origin is

Given :




To Find : Equation of a plane


Formulae :


1) Unit Vector :


Let be any vector


Then the unit vector of is



Where,


2) Dot Product :


If are two vectors




then,



3) Equation of plane :


Equation of plane which is at a distance of 5 units from the origin and having as a unit vector normal to it is



Where,


For given normal vector



Unit vector normal to the plane is






Equation of the plane is





This is a vector equation of the plane.


Now,



= (x × 2) + (y × (-3)) + (z × 4)


= 2x - 3y + 4z


Therefore equation of the plane is


2x - 3y + 4z = 6


This is the Cartesian equation of the plane.


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