NCERT solution Class 12 three dimensional geometry

NCERT Solutions

Class 12 Maths

Three Dimensional Geometry

NCERT Questions

Ex.11.1 Q.1
Ex.11.1 Q.2
Ex.11.1 Q.3
Ex.11.1 Q.4
Ex.11.1 Q.5
Ex.11.2 Q.1
Ex.11.2 Q.2
Ex.11.2 Q.3
Ex.11.2 Q.4
Ex.11.2 Q.5
Ex.11.2 Q.6
Ex.11.2 Q.7
Ex.11.2 Q.8
Ex.11.2 Q.9
Ex.11.2 Q.11
Ex.11.2 Q.11
Ex.11.2 Q.12
Ex.11.2 Q.13
Ex.11.2 Q.14
Ex.11.2 Q.15
Ex.11.2 Q.16
Ex.11.2 Q.17
Ex.11.3 Q.1
Ex.11.3 Q.2
Ex.11.3 Q.3
Ex.11.3 Q.4
Ex.11.3 Q.5
Ex.11.3 Q.6
Ex.11.3 Q.7
Ex.11.3 Q.8
Ex.11.3 Q.9
Ex.11.3 Q.10
Ex.11.3 Q.11
Ex.11.3 Q.12
Ex.11.3 Q.13
Ex.11.3 Q.14
Ex.Misc. Q.1
Ex.Misc.Q.2
Ex.Misc.Q.3
Ex.Misc.Q.4
Ex.Misc.Q.5
Ex.Misc.Q.6
Ex.Misc.Q.7
Ex.Misc.Q.8
Ex.Misc.Q.9
Ex.Misc.Q.10
Ex.Misc.Q.11
Ex.Misc.Q.12
Ex.Misc.Q13
Ex.Misc.Q.14
Ex.Misc.Q.15
Ex.Misc.Q.16
Ex.Misc.Q.17
Ex.Misc.Q.18
Ex.Misc.Q.19
Ex.Misc.Q.20
Ex.Misc.Q.21
Ex.Misc.Q.22
Ex.Misc.Q.23

Ex.11.3 Q.4

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

(a) 2x + 3y + 4z – 12 = 0

(b) 3y + 4x – 6 = 0

(d) 5y + 8 = 0

(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁, y₁, z₁).

Given, 2x + 3y + 4z − 12 = 0

=> 2x + 3y + 4z = 12 ..........1

The direction ratios of normal are 2, 3, and 4.

So, √(2²+ 3² + 4²} = √(4+ 9 + 16) = √29

Dividing both sides of equation 1 by √29, we get

This equation is of the form lx + my + nz = d, where l, m, n are the

direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁, y₁, z₁).

Given, 3y + 4z – 6 = 0

=> 0x + 3y + 4z = 6 ……..1

The direction ratios of the normal are 0, 3, and 4.

So, √(0²+ 3² + 4²) = √(0+ 9 + 16) = √25 = 5

Dividing both sides of equation 1 by 5, we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of

normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

Given, x + y + z = 1 ……….1

The direction ratios of the normal are 1, 1, and 1.

So, √(1²+ 1² + 1²) = √(1+ 1 + 1) = √3

Dividing both sides of equation 1 by √3, we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the

direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁, y₁, z₁).

Given, 5y + 8 = 0

=> 0x − 5y + 0z = 8 ........1

The direction ratios of the normal are 0, −5, and 0.

So, √{1²+ (-5)² + 1²} = √(0+ 25 + 0) = √25 = 5

Dividing both sides of equation 1 by 5, we get

-y =

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines

of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

Three Dimensional Geometry Class 12 Maths NCERT Chapter 11 NCERT Solutions

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