Find the vector and Cartesian equations of the plane which passes through the point (5,2-4) and perpendicular to the line with direction ratios 2,3,-1.

Explanation

The plane passes through the point A(5,2,-4) whose position vector is `veca=(5hati+2hatj-4hatk)` and the normal vector `vecn` perpendicular to the plane is `vecn=(2hati+3hatj-hatk)`.
So, the vector equation of the plane is
`vecr.vecn=veca.vecn`
`rArr vecr.(2hati+3hatj-hatk)=(5hati+2hat-4hatk).(2hati+3hatj-hatk)`.
Hence, the required vector equation of the plane is
`vecr=(2hati+3hatj-hatk)=20`.............(i)
Cartesian Form:
Taking `vecr=(xhati+yhatj+zhatk)`, equation (i) may be written as
`(xhati+yhatj+zhatk).(2hati+3hatj-hatk)=20 rArr 2x+3y-z=20`.
Hence, the requried Cartesian equation is `2x+3y-z=20`.