# 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-x

_{1})+B(y-y

_{1})+C(z-z

_{1})=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 (x

_{1}, y

_{1}, z

_{1}), (x

_{2}, y

_{2}, z

_{2}) and (x

_{3}, y

_{3}, z

_{3}) 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. n_{1}=d_{1} and r. n_{2}=d_{2} are the vector equation of two planes then equation of the plane passing through the intersection of these two planes is given by

r.( n_{1}+λ n_{2})=d_{1}+λd_{2}**Cartesian Form:** If A

_{1}x+B

_{1}y+C

_{1}z=d

_{1}and A

_{2}x+B

_{2}y+C

_{2}z=d

_{2}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

(A

_{1}x+B

_{1}y+C

_{1}z-d

_{1})+λ(A

_{2}x+B

_{2}y+C

_{2}z-d

_{2})=0

In general, if P

_{1}and P

_{2}are the equations of two planes then the equation of the plane passing through the intersection of P

_{1}and P

_{2}is given by

P

_{1}+λP

_{2}=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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