Find the vector equation of a plane which is at a distance of 7 units from the
origin and normal to the vector `3hati+5hatj-6hatk`

Answer

The distance of the required plane from the origin is p = 7. A vector normal to the plane = `3hati+5hatj-6hatk`
`therefore` The unit vector normal to the plane `=(1)/abs(3hati+5hatj-6hatk) (3hati+5hatj-6hatk) = (1)/sqrt(9+25+36) (3hati+5hatj-6hatk) = (3hati+5hatj-6hatk)/sqrt70`
Hence, the vector equation of the plane is `vecr.((3hati+5hatj-6hatk)/sqrt70) = 7`