Q. 45.0( 1 Vote )

Find the vector and Cartesian equations of a plane which is at a distance of 6 units from the origin and which has a normal with direction ratios 2, -1, -2.

Answer :

Given :


d = 6


direction ratios of are (2, -1, -2)



To Find : Equation of 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 vector equation of the plane.


Now,



= (x × 2) + (y × (-1)) + (z × (-2))


= 2x - y – 2z


Therefore equation of the plane is



This is Cartesian equation of the plane.


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