Equation of a Plane in Normal Form, Equation of a Plane Perpendicular to a given Vector Video Lecture | Mathematics (Maths) Class 12 - JEE

equation of a plane in normal form
consider a plane whose perpendicular
distance from the origin is D where D is
not equal to 0 if vector o n is the
normal from the origin to the plane and
n cap is the unit vector normal along o
n vector then vector o n is equal to D
into n cap let P be any point on the
plane then vector NP is perpendicular to
vector o n therefore where dot product
is 0 that is vector n p dot vector o n
is equal to 0 let R be the position
vector of the point P then from the
triangle o NP we have vector o n plus
vector NP is equal to vector o P
therefore NP is equal to our vector
minus D into n cap now n P vector dot o
n vector is equal to 0 therefore our
vector minus D into n cap dot d into n
cap is equal to 0 on simplification we
get our vector dot n cap is equal to D
Cartesian form of the equation of a
plane in normal form vector form of the
equation of the plane in normal form is
vector r dot n cap is equal to d now our
vector is the position vector of the
point P therefore our vector is equal to
X I cap plus YJ cap plus ZK cap and let
l m n be the direction cosines of unit
vector n then
ten cup is equal to Li cup + MJ cap + n
K cup substituting the corresponding
values in the vector equation of the
plane we get the Cartesian equation of
the plane in the normal form as LX plus
my plus NZ is equal to d