# Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is 2i-3j+6k - Mathematics

Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is 2hati-3hatj+6hatk

#### Solution

Given:

Normal vector, hatn=2hati-3hatj+6hatk

Perpendicular distance, d = 5 units

The vector equation of a plane that is at a distance of 5 units from the origin and has its normal vector hatn=2hati-3hatj+6hatk is as follows:

vecr.hatn = d

vecr.(2hati-3hatj+6hatk)=5

Concept: Vector and Cartesian Equation of a Plane
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