Find the vector equation of the plane which is at a distance of `6/(sqrt(29))`from the origin and its normal vector from the origin is `2 hat i-3 hat j+4 hat k`. Also find its cartesian form.

Explanation

It is given that the distance from the origin of the plane is,

`d=6/sqrt29`

Also, it is given that its normal vector from the origin is,

`vecN=2hati-3hatj+4hatk`

It is known that the unit normal vector is defined as,

`hatn=(vecN)/abs(vecN)`

Therefore,

`hatn=(2hati-3hatj+4hatk)/sqrt(2^2+(-3)^2+4^2)`

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