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

​

Chapter 11 -Three Dimensional Geometry

Sub Topic-Various Form of Equation of Plane

Outline:�

• Equation of a plane passing through a point and perpendicular to a given vector.
• Equation of a plane passing through a point, containing a line and parallel to a line.
• Equation of a plane passing through three points.
• Assignment

​

Equation of a Plane Passing through a given point and ⊥ to a given vector�

• Let a plane pass through given point ‘A’ whose position vector is a and ⊥ N .

Let r be the position vector of any point in the plane.

Then AP ⊥ N .

Hence AP . N = 0

But AP = OP - OA

= r - a

Hence ( r – a ) . N = 0

N

A

P

O

N

a

r

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

Equation of a plane passing through a point, containing a line and parallel to a given line

Let equations of two lines are

r = a + λ b ------- (1)

and r = a^| + μ b’ ------- (2)

Let a plane contains a line (1), then the plane is also passing through the point A whose position vector is a.

Let P be any point on the plane whose position vector is r, then AP totally lies in the plane.

Hence b and b’ are parallel.

We know that b x b’ is perpendicular to both b and b^|

So normal of the plane is parallel to b x b’ . Hence AP, b, b’are coplanar.

Therefore equation of the required plane is

[ AP , b , b’ ] = 0 ⇒ ( r - a ) . b x b’ = 0 which is in vector form.

​

​

​

BACK

Cartesian form of equation of a Plane

​

( x – x₁) ( y – y₁) ( z – z₁)

b₁ b₂ b_{3 = 0}

​

b₁^’ b₂^’ b₃^’

​

​

​

BACK

Equation of a plane passing through three points

• Let a plane is passing through the given points A, B and C such that OA = a, OB = b, OC = c.
• Let P be any point in the plane such that OP = r .
• We know that AB x AC is perpendicular to both AB, AC.
• Hence AB x AC is parallel to the normal of the plane.
• Equation of the plane in vector form be

( r – a ) .( AB x AC ) = 0

(r – a ). [(b – a) x ( c – a )] = 0

​

BACK

ASSIGNMENT

Q1.Find the equation of the plane passing through the points A(1, 1, 1), B( 1, -1, 1), C( -7, -3, -5)

Q2.Find the equation of the plane passing through the points P(2, 2, -1), Q( 3, 4, 2), R( 7, 0, 6).

Q3.Find the equation of a plane passing through the point (2, -1, 1) and perpendicular to the vector �4i + 2j – 3k.

Q4.Find the equation of the plane containing the line

x – 4 ₌ y – 3 ₌ z – 2

1 4 5

and passing through the line

x – 3 ₌ y – 2 ₌ z

1 -4 5

​

​

​

​

​