RD Sharma Class 12 Solutions Chapter 29 - The Plane (Ex 29.9) Exercise 29.9

What is a plane and how can it be determined in mathematical language?

A plane can be determined very easily, If any of the following is known,

• The plane's normal and its distance from the origin are supplied, i.e., the plane's equation in normal form.

• It is perpendicular to a specified direction and goes through a point.

• It goes through three non-collinear sites that are specified.

Let’s now try to understand the equation of a plane in very simple language:

•  The vector form of a plane equation in normal form is r. n^=d

• Where r is the point's position vector on the specified plane, n is the normal unit vector from the origin to the plane, and d is the normal unit vector's length from the origin to the plane.

•  lx+my+nz=d is the Cartesian form of the plane equation in normal form.

• P(x,y,z) is any point on the provided plane, n is the normal unit vector from the origin to the plane, and l,m,n are the normal unit vector's direction cosines.

• If the vector equation of the plane is r.(ai+bj+ck)=d, then the Cartesian equation of the plane is (ax+by+cz)=d. The direction ratios of the normal to the plane are a,b,c.

Equation of a plane passing through three collinear points

The vector form of the equation of the plane is:

Important: (r −a ).[(b −a)×(c−a)]=0 

Here a, b, c are the position vectors of the three non-collinear points on the plane, whereas, r  is the position vector of any point on the given plane.