Various forms of a Plane

Created: Thursday, 15 September 2011 07:21 | Published: Thursday, 15 September 2011 07:21 | Written by Super User | Print | Email

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+λ 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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