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प्रश्न
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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उत्तर
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`
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