📖 generic · 12th TN - English Medium · MATHEMATICS-VOLUME 1 · Page 263question

the plane from the origin are given

Chapter 8: Chapter 6 · MATHEMATICS-VOLUME 1

the plane from the origin are given (a) Vector equation of a plane in normal form Theorem . The equation of the plane at a distance p from the origin and perpendicular to the unit normal vector ˆ d is r d Vector - - Proof Consider a plane whose perpendicular distance from the origin is p . Let A be the foot of the perpendicular from O to the plane. Let ˆ d be the unit normal vector in the direction of OA Then OA pd If r  is the position vector of an arbitrary point P on the plane, then AP is perpendicular to OA Therefore, AP OA  pd pd ⇒ pd ⇒ which gives r d = p .

... ( ) The above equation is called the vector equation of the plane in normal form . (b) Cartesian equation of a plane in normal form Let , l m n be the direction cosines of ˆ d . Then we have ˆ li mj nk Thus, equation ( ) becomes li mj nk = p If P is ( x,y,z ), then xi yj zk Therefore, ) ( xi yj zk li mj nk = p or lx my nz ...

( ) Equation ( ) is called the Cartesian equation of the plane in normal form . Remark (i) If the plane passes through the origin, then p = . So, the equation of the plane is . lx my nz (ii) If d is normal vector to the plane, then ˆ  is a unit normal to the plane.

So, the vector equation of the plane is or r d q , where q p d . The equation r d q is the vector equation of a plane in standard form. Note In the standard form  r d q d need not be a unit normal and q need not be the perpendicular distance.

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