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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

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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`

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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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