Find the vector equation of a plane passing through a point having position vector `(2hati-hatj+hatk)` and perpendicular to the vector `(4hati+2hatj-3hatk)`. Also, reduce it to Cartesian form.

Explanation

Here, `veca=(2hati-hatj+hatk)` and `vecn=(4hati+2hatj-3hatk)`.
Clearly, the required vector equation is `(vecr-veca).vecn=0`.
`rArr vecr.vecn=veca.vecn`
`rArr vecr.(4hati+2hatj-3hatk)=(2hati-hatj+hatk).(4hati+2hatj-3hatk)` =`(8-2-3)=3`.
Hence, the vector equation of the given plane is
`vecr.(4hati+2hatj-3hatk)=3`..............(1)
Reduction to Cartesian form:
Putting `vecr=(xhati+yhatj+zhatk)`, we get
`(xhati+yhatj+zhatk).(4hati+2hatj-3hatk)=3 rArr 4x+2y-3z=3`.
Hence, the equation ofthe given plane in Cartesian form is `4x+2y-3z=3`.