Articles

Various forms of a Plane

Plane - Introduction

A plane can be determined uniquely if anyone of the following is known:
(i)     The normal to the plane and its distance from origin is given.
(ii)     It passes through a point and is perpendicular to a given direction.
(iii)    It passes through three given non collinear points.



Equation of plane in normal form

Vector Form:  If r is the position vector of a point P in the plane, d is the perpendicular distance from origin and ň is the unit normal to the plane then its vector equation is given by
                                     r. ň = d

Cartesian Form:
  If P(x, y, z) is a point in the plane, d is the perpendicular distance from origin and <l, m, n> are the direction cosines of ň, then the Cartesian form of the plane is given by

                                  lx +my +nz =d

Note:   If <a, b, c> are the direction ratios of the normal to the plane then the equation is ax+ by+ cz = d



Equation of a plane perpendicular to a given vector and passing through a given point

Vector Form:  If a is the position vector of a given point and N is the perpendicular vector then its equation is given by
                              ( r- a). N=0

Cartesian Form:
  If A(x1, y1, z1) is the given point and P(x, y, z) is a general point in the plane and A, B and C are the direction ratios of N then the Cartesian equation is given by

                         A(x-x1)+B(y-y1)+C(z-z1)=0



Equation of a plane passing through three non collinear points

Vector Form:  If a, b and c are the position vectors of three points and r be any point in the plane, then the equation of the plane passing through three given points is
                   ( r- a).[( b- a) X ( c- a)]=0

Cartesian Form:
  If (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) are the three given points then equation of the plane is

                          



Intercept Form of a plane

If the plane makes intercepts a, b and c on x, y and z axes respectively then its equation in intercept form is given by
                    


Here coordinates of A, B and C are A(a,0,0), B(0,b,0) and C(0,0,c) respectively.



Intersection of two planes

Vector Form:  If r. n1=d1 and r. n2=d2 are the vector equation of two planes then equation of the plane passing through the intersection of these two planes is given by
                             r.( n1 n2)=d1+λd2

Cartesian Form:
  If A1x+B1y+C1z=d1 and A2x+B2y+C2z=d2 are the equations of two planes in the Cartesian form then the equation of the plane passing through the intersection of the given planes is

                  (A1x+B1y+C1z-d1)+λ(A2x+B2y+C2z-d2)=0

In general, if P1 and P2 are the equations of two planes then the equation of the plane passing through the intersection of P1 and P2 is given by

                                P1+λP2=0

Example: Find the equation of the plane through the intersection of the planes x+y+z-6=0 and 2x+3y+4z+5=0 ant the point (1, 1, 1)


Solution: Equation of the plane passing through x+y+z-6=0 and 2x+3y+4z+5=0 is given by

                  (x+y+z-6) + λ (2x+3y+4z+5)=0        ……………(1)

Passes through (1, 1, 1)

(1+1+1-6) + λ (2+3+4+5) = 0
λ=3/14

 Substitute the value of λ in (1), so that the equation is

(x+y+z-6)+3/14(2x+3y+4z+5) = 0
20x+23y+26z-69=0, which is the required equation.
                      
                                                                               
      

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