�Mathematics –Class XII��Unit IV-Vector & 3D

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Chapter 11 -Three Dimensional Geometry

Sub Topic-Equation of Plane in normal form

LINES

• Introduction & Definition.
• Normal form of equation of a plane.
• Cartesian form of equation of plane.
• Assignment

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Outline:

INTRODUCTION

A plane is a rectangular figure with infinite length and infinite breadth.

or

A plane is defined as a surface such that the line joining any two points on it lies totally on the surface.

Example: Black Board, Paper etc.

PLANE

• A particular plane can be specified in several ways.
• One and only one plane can be drawn perpendicular to a given direction at a given distance.
• Now we will find the equation of a plane which is at a distance of ‘d’ from the origin and perpendicular to a given direction.

BACK

Normal Form of Equation of a Plane

• Let a plane be at a distance of ‘d’ from the origin and n be the unit vector along normal of the plane and N be it’s foot.
• P be any point in the plane with position vector r .

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CONTINUE ….

Continued ---

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• NP = OP - ON

Where OP = r , ON = d n

Since ON is perpendicular to the plane , ON is also perpendicular to NP.

Hence NP . ON = 0 ( 1 )

⇒ ( r – d n ) .d n = 0

⇒ r . n = d ----- ( 2 )

Which is the vector form of equation of a plane in scalar form.

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Continue

Cartesian form of equation of a plane.

• Let OP = xi + yj + zk and l, m, n be the direction

cosines of unit vector n, then n = li + mj + nk

Hence from equation ( 2 ), we have

lx + my + nz = d ----( 3 )

which is the Cartesian form of equation of a plane .

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Continue

Cartesian Form of Equation of a Plane

Let A(x₁, y₁, z₁) and P(x , y , z) and the direction ratios of N are n₁ , n₂ and n₃ . Then

a = x₁i + y₁j + z₁k ,

r = xi + yj + zk

Hence r – a = (x – x₁)i + ( y – y₁)j +(z – z₁)k

N = n₁i + n₂j + n₃k

( r – a ) . N = (x-x₁)n₁+ (y-y₁)n₂ + (z-z₃)k = 0

Which is the required equation of the plane.

N

A

P

N

a

r

O

BACK

Q1.Find the equation of a plane passing through the point (2, -1, 1) and perpendicular to the vector �4i + 2j – 3k

Q2.Find the normal vector and normal form of the given plane �x + 2y + 3z – 6 = 0

Q3.Find the equation of the plane passing through the points (-1,2,-5) and perpendicular to the line,

x – 4 ₌ y – 3 ₌ z – 2

1 4 5

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