Find the vector and Cartesian equations of the plane passing throgh the point (2,-1,1) and perpendicular to the line having direction ratios 4,2,-3.

Answer

`vecr.(4hati+2hatj-3hatk)=3, 4x+2y-3z=3`.

Explanation

Let the equation of the plane be `a(x-2)+b(y+1)+c(z-1)=0`, where a=4,b=2 and c=3.
So, the required equation is
`4(x-2)+2(y+1)-3(z-1)=0 rArr 4x+2y-3z=3`.
In vector form, we have `vecr.(4hati+2hatj-3hatk)=3`.