. . Intercept form of the equation of a plane Let the plane r n q meets the coordinate axes at , , A B C respectively such that the intercepts on the axes are OA a OB b OC . Now position vector of the point A is ˆ ai .
Since A lies on the given plane, we have ai n q which gives q i n . Similarly, since the vectors ˆ bj and ˆ ck lie on the given plane, we have ˆ q j n and ˆ q k n . Substituting xi yj zk in r n q , we get xi n yj n zk n q So q q q q Dividing by q , we get, = . This is called the intercept form of equation of the plane having intercepts , , a b c on the , , x y z axes respectively.
Theorem . The general equation ax by cz of first degree in , , x y z represents a plane. Proof The equation ax by cz can be written in the vector form as follows ) ( xi yj zk ai bj ck or r n Since this is the vector form of the equation of a plane in standard form, the given equation ax by cz represents a plane. Here ai bj ck is a vector normal to the plane.
Note In the general equation ax by cz of a plane, , , a b c are direction ratios of the normal to the plane. Fig. . ( a , , ) ( , b , ) ( , , c ) O B C Vector - - Example .
Find the vector and Cartesian form of the equations of a plane which is at a distance of units from the origin and perpendicular to Let p = If ˆ d is the unit normal vector in the direction of the vector then ( ) If r is the position vector of an arbitrary