OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24

OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24. Step by step Solutions of OP Malhotra SK Gupta, Anubhuti Gangal S.Chand ISC Class-12 Mathematics with Exe-24(a), Exe-24(b), Exe-24(c), Exe-24(d), Exe-24(e), and Exe-24(f), Self Revision and Chapter Test . Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24

Class:	12th
Subject:	Mathematics
Chapter  :	Ch-24 The Plane  of Section -B

Board	ISC
Writer	OP Malhotra, SK Gupta, Anubhuti Gangal
Publications	S.Chand Publications 2020-21

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-: Included Topics :-

Exe-24(a),

 Exe-24(b), 

Exe-24(c), 

Exe-24(d), 

Exe-24(e), 

Exe-24(f) 

Self Revision

Chapter Test

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OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24

Plane :-

A plane is determined uniquely if any one of the following is known:

(i) The normal to the plane and its distance from the origin is given, i.e., equation ofa plane in normal form.

(ii) It passes through a point and is perpendicular to a given direction.

(iii) It passes through three given non collinear points.

Equations of a Plane in Normal form  :-

Vector form: The equation of plane in normal form is given by 𝑟⃗ ⋅𝑛⃗ =𝑑, where 𝑛⃗  is a vector which is normal to the plane.
Cartesian form: The equation of the plane is given by ax + by + cz = d, where a, b and c are the direction ratios of plane and d is the distance of the plane from origin.
Another equation of the plane is lx + my + nz = p, where l, m, and n are direction cosines of the perpendicular from origin and p is a distance of a plane from origin.
Note: If d is the distance from the origin and l, m and n are the direction cosines of the normal to the plane through the origin, then the foot of the perpendicular is (ld, md, nd).

 

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Exe-24(a)

OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24

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Exe-24(b)

OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24

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Exe-24(c)

OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24

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Exe-24(d)

 The Plane ISC Class-12 Maths Solutions Ch-24

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Exe-24(e)

OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24

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Equation of a Plane Perpendicular to a given Vector and Passing Through a given Point
Vector form:

Let a plane passes through a point A with position vector 𝑎⃗  and perpendicular to the vector 𝑛⃗ , then (𝑟⃗ − 𝑎⃗ )⋅𝑛⃗ =0
This is the vector equation of the plane.
Cartesian form:

Equation of plane passing through point (x₁, y₁, z₁) is given by
a (x – x₁) + b (y – y₁) + c (z – z₁) = 0 where, a, b and c are the direction ratios of normal to the plane.

Equation of Plane in Intercept Form: If a, b and c are x-intercept, y-intercept and z-intercept, respectively made by the plane on the coordinate axes, then equation of plane is  = 𝑥/𝑎+𝑦/𝑏+𝑧/𝑐=1

Cartesian form: If the equation of planes are a₁x + b₁y + c₁z = d₁ and a₂x + b₂y + c₂z = d₂, then equation of any plane passing through the intersection of planes is a₁x + b₁y + c₁z – d₁ + λ (a₂x + b₂y + c₂z – d₂) = 0
where, λ is a constant and calculated from given condition.

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Exe-24(f)

OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24

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Self Revision

 The Plane ISC Class-12 Maths Solutions Ch-24

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Chapter Test

OP Malhotra The Plane ISC Class-12 Maths Solutions Ch-24

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-: End of   The Plane S. Chand ISC Class-12 Maths Solution :-

Return to :-  OP Malhotra S. Chand ISC Class-12 Maths Solutions

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