point ( , , ) x y z on the plane, then using r d , the vector equation of the plane in normal form is ( ) Substituting xi yj zk in the above equation, we get ( ) xi yj zk Applying dot product in the above equation and simplifying, we get , which is the Cartesian equation of the required plane. Example . If the Cartesian equation of a plane is = − , find the vector equation of the plane in the standard form. If xi yj zk is the position vector of an arbitrary point ( , , ) x y z on the plane, then the given equation can be written as ) ( ) xi yj zk = − or ) ( ) xi yj zk ⋅− .
That is, ( ) ⋅− which is the vector equation of the given plane in standard form. Example . Find the direction cosines of the normal to the plane and length of the perpendicular from the origin to the plane ( ) Let q = If ˆ d is the unit vector in the direction of the vector , then ( ) Now, dividing the given equation by , we get ˆ = which is the equation of the plane in the normal form r d From this equation, we infer that ( ) is a unit vector normal to the plane from the origin. Therefore, the direction cosines of ˆ d are and the length of the perpendicular from the origin to the plane is .
Vector - - Applications of Vector Algebra Example . Find the vector and Cartesian equations of the plane passing through the point with position vector and normal to vector i If the position vector of the given point is , then the equation of the plane passing through a point and normal