NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3
NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 contains all the questions with detailed solutions. Students are advised to solve these questions for better understanding of the concepts in exercise 11.3.
Krati Saraswat16 Feb, 2024
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NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 (Three-Dimensional Geometry)
NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 of Three-Dimensional Geometry is prepared by the academic team of PW. We have prepared NCERT Solutions for all exercise of chapter 11. Given below is step by step solutions to all questions given in the NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 of Three Dimensional Geometry.NCERT Solutions for Class 12 Maths Chapter 11 Miscellaneous Exercise
NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 Overview
NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 covers these important topics. Students are encouraged to review each topic thoroughly in order to fully understand the concepts taught in the chapter and make optimal use of the provided solutions. These NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 are the outcome of the dedicated effort that the Physics Wallah teachers have been doing to aid students in understanding the ideas covered in this chapter. After going over NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 and rehearsing these questions, students will be able to score good marks in their examinations.NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3
Solve The Following Questions of NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3: Question 1. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a) z = 2 (b) x + y + z = 1 (c) 2x + 3y - z = 5 (d) 5y + 8 = 0 Solution : (The equation of the plane is z = 2 or 0 x + 0 y + z = 2 … (1) The direction ratios of normal are 0, 0, and 1. ∴ √0 2 + 0 2 + 1 2 = 1 Dividing both sides of equation (1) by 1, we obtain 0.x + 0.y + 1.z = 2 This 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 the perpendicular drawn from the origin. Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units. (b) x + y + z = 1 … (1) The direction ratios of normal are 1, 1, and 1. ∴ √(1)² + (1)² + (1)² = √3 Dividing both sides of equation (1) by √3, we obtain 1/√3x + 1/√3y + 1/√3z = 1/√3 ....(2) 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. Therefore, the direction cosines of the normal are 1/√3,1/√3 and 1/√3the distance of normal from the origin is 1/√3 units. (c) 2 x + 3 y − z = 5 … (1) The direction ratios of normal are 2, 3, and −1. ∴ √(2)² + (3)² + (-1)² = √14 Dividing both sides of equation (1) by √14 , we obtain 2/√14x + 3/√14y - 1/√14z = 5/√14 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. Therefore, the direction cosines of the normal to the plane are 2/√14x + 3/√14y - 1/√14z and the distance of normal from the origin is 5/√14 units. (d) 5 y + 8 = 0 ⇒ 0 x − 5 y + 0 z = 8 … (1) The direction ratios of normal are 0, −5, and 0. ∴√0 +(-5)² + 0 = 5 Dividing both sides of equation (1) by 5, we obtain -y = 8/5 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. Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of normal from the origin is 8/5 units.NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.1
Question 2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector
Solution :
The normal vector is,
It is known that the equation of the plane with position vector is given by, .
= d
.png)
This is the vector equation of the required plane.
NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2
Question 3. Find the Cartesian equation of the following planes: (a)
(c)
Solution :
(a) It is given that equation of the plane is
For any arbitrary point P (
x
,
y
,
z
) on the plane, position vector is given by,
Substituting the value of in equation (1), we obtain
This is the Cartesian equation of the plane.
(b)
For any arbitrary point P (
x
,
y
,
z
) on the plane, position vector is given by,
Substituting the value of in equation (1), we obtain
This is the Cartesian equation of the plane.
(c)
For any arbitrary point P (
x
,
y
,
z
) on the plane, position vector is given by,
Substituting the value of in equation (1), we obtain
This is the Cartesian equation of the given plane.
Question4. 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 + 4z - 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Solution :
(a)
Let the coordinates of the foot of perpendicular P from the origin to the plane be (
x
1
,
y
1
,
z
1
).
2
x
+ 3
y
+ 4
z
− 12 = 0
⇒ 2
x
+ 3
y
+ 4
z
= 12 … (1)
The direction ratios of normal are 2, 3, and 4.
∴ √ 2² + 3² + 4² = √29
Dividing both sides of equation (1) by √29, 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
(b)
Let the coordinates of the foot of perpendicular P from the origin to the plane be (
x
1
,
y
1
,
z
1
).
3y + 4z - 6 = 0
⇒ 0x + 3y + 4y = 6 … (1)
The direction ratios of the normal are 0, 3, and 4.
∴ √0 + 3² + 4² = 5
Dividing both sides of equation (1) by 5, we obtain
0x + 3/5y + 4/5z = 6/5
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
(c)
Let the coordinates of the foot of perpendicular P from the origin to the plane be (
x
1
,
y
1
,
z
1
).
x + y + z = 1… (1)
The direction ratios of the normal are 1, 1, and 1.
√1² + 1² + 1² = √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 (
x
1
,
y
1
,
z
1
).
5y + 8 = 0
⇒ 0
x
− 5
y
+ 0
z
= 8 … (1)
The direction ratios of the normal are 0, −5, and 0.
√0 + (-5)
2
+ 0 = 5
Dividing both sides of equation (1) by 5, we obtain
-y = 8/5
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
.png)
Question5. Find the vector and Cartesian equations of the planes
(a) that passes through the point (1, 0, −2) and the normal to the plane is
.
(b) that passes through the point (1, 4, 6) and the normal vector to the plane is
.
Solution :
(a)
The position vector of point (1, 0, −2) is
The normal vector N perpendicular to the plane is
The vector equation of the plane is given by,
is the position vector of any point P (
x
,
y
,
z
) in the plane.
Therefore, equation (1) becomes
This is the Cartesian equation of the required plane
.
(b)
The position vector of the point (1, 4, 6) is
The normal vector perpendicular to the plane is
The vector equation of the plane is given by,
is the position vector of any point P (
x
,
y
,
z
) in the plane.
Therefore, equation (1) becomes
This is the Cartesian equation of the required plane.
Question 6. Find the equations of the planes that passes through three points:
(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)
(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)
Solution :
We know that through three collinear points A, B, C i.e., through a straight line, we can pass an infinite number of planes.
(a) The given points are A (1, 1, −1), B (6, 4, −5), and C (−4, −2, 3).
Since A, B, C are collinear points, there will be infinite number of planes passing through the given points.
(b) The given points are A (1, 1, 0), B (1, 2, 1), and C (−2, 2, −1).
Therefore, a plane will pass through the points A, B, and C.
It is known that the equation of the plane through the points,
is
This is the Cartesian equation of the required plane.
Question7. Find the intercepts cut off by the plane 2x + y - z = 5
Solution :
2x + y - z = 5 .........(1)
Dividing both sides of equation (1) by 5, we obtain
It is known that the equation of a plane in intercept form is x/a + y/b + z/c = 1, where
a
,
b
,
c
are the intercepts cut off by the plane at
x
,
y
, and
z
axes respectively.
Therefore, for the given equation,
a = 5/2, b = 5 and c = -5
Thus, the intercepts cut off by the plane are 5/2, 5 and -5.
Question8. Find the equation of the plane with intercept 3 on the y- axis and parallel to ZOX plane.
Solution :
The equation of the plane ZOX is
y
= 0
Any plane parallel to it is of the form,
y
=
a
Since the
y
-intercept of the plane is 3,
∴
a
= 3
Thus, the equation of the required plane is
y
= 3
Question9. Find the equation of the plane through the intersection of the planes 3x - y + 2z - 4 = 0 and x + y + z - 2 = 0 and the point (2, 2, 1).
Solution :
The equation of any plane through the intersection of the planes,
3
x
−
y
+ 2
z
− 4 = 0 and
x
+
y
+
z
− 2 = 0, is
The plane passes through the point (2, 2, 1). Therefore, this point will satisfy equation (1).
Question10. Find the vector equation of the plane passing through the intersection of the planes and through the point (2, 1, 3).
Solution :
The equations of the planes are
The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,
The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,
Substituting in equation (3), we obtain
This is the vector equation of the required plane.
Question11. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0
Solution :
The equation of the plane through the intersection of the planes, x + y + z = 1 and 2x + 3y + 4z = 5, is
(x + y + z - 1) + λ(2x + 3y + 4z - 5)
⇒(2λ + 1)x + (3λ + 1)y + (4λ + 1)z - (5λ + 1) = 0 .....(1)
The direction ratios,
a
1
,
b
1
,
c
1
, of this plane are (2λ + 1), (3λ + 1), and (4λ + 1).
The plane in equation (1) is perpendicular to x - y + z = 0
Its direction ratios,
a
2
,
b
2
,
c
2
, are 1, −1, and 1.
Since the planes are perpendicular,
Substituting λ = -1/3 in equation (1), we obtain
This is the required equation of the plane.
Question12. Find the angle between the planes whose vector equations are
Solution :
The equations of the given planes are
It is known that if n1 and n2 are normal to the planes, then the angle between them, Q, is given by,
Question13. In the following cases, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angle between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x - y - 10z + 4 = 0
(b) 2x + y + 3z - 2 = 0 and x - 2y + 5 = 0
(c) 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1
(d) 2x - y + 3z - 1 = 0 and 2x - y + 3z + 3 = 0
(e) 4x + 8y + z - 8 = 0 and y + z - 4 = 0
Solution :
The direction ratios of normal to the plane,
, are
a
1
,
b
1
,
c
1
and
.
(a)
The equations of the planes are 7
x
+ 5
y
+ 6
z
+ 30 = 0 and
3
x
−
y
− 10
z
+ 4 = 0
Here,
a
1
= 7,
b
1
=5,
c
1
= 6
Therefore, the given planes are not parallel.
The angle between them is given by,
(b)
The equations of the planes are 2x + y + 3z - 2 = 0 and x - 2y + 5 = 0
Thus, the given planes are perpendicular to each other.
(c)
The equations of the given planes are 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1
Here,
Thus, the given planes are not perpendicular to each other.
Thus, the given planes are parallel to each other.
(d)
The equations of the planes are and 2x - y + 3z - 1 = 0 and 2x - y + 3z + 3 = 0
Thus, the given lines are parallel to each other.
(e)
The equations of the given planes are
4x + 8y + z - 8 = 0 and y + z - 4 = 0
Therefore, the given lines are not perpendicular to each other.
Therefore, the given lines are not parallel to each other.
The angle between the planes is given by,
Question14. In the following cases find the distances of each of the given points from the corresponding given plane:
(a) Point (0, 0, 0) Plane 3x - 4y + 12z = 3
(b) Point (3, −2, 1)
Plane 2x - y + 2z + 3 =
0
(c) Point (2, 3, −5)
Plane x + 2y - 2z = 9
(d) Point (−6, 0, 0)
Plane 2x - 3y + 6z - 2 = 0
Solution :
It is known that the distance between a point,
p
(
x
1
,
y
1
,
z
1
), and a plane,
Ax
+
By
+
Cz
=
D
, is given by,
(a)
The given point is (0, 0, 0) and the plane is 3x - 4y + 12z = 3
(b)
The given point is (3, − 2, 1) and the plane is 2x - y + 2z + 3 = 0
(c)
The given point is (2, 3, −5) and the plane is x + 2y - 2z = 9
(d)
The given point is (−6, 0, 0) and the plane is 2x - 3y + 6z - 2 = 0
NCERT Solutions For Class 12 Maths Chapter 11 Exercise 11.3 FAQs
Who invented 3D geometry?
Euclid of Alexandria, who is said to be a student at the Academy by Plato was the one who wrote a treatise in 13 books (chapters).
What is the use of three-dimensional geometry?
The three-dimensional geometry is used for the representation of a point, line, or a plane.
What are the applications of three-dimensional geometry in real life?
Applications of geometry in the real world include the computer-aided design (CAD) for construction blueprints, the design of assembly systems in manufacturing such as automobiles, nanotechnology, computer graphics, visual graphs, video game programming, and virtual reality creation.
Who is the real father of geometry?
Euclid was a Greek mathematician and is also known as the 'father of Geometry'. He compiled elements which have several geometric theories. These are still used by mathematicians all around the world.
Which shape is three-dimensional?
A cube, rectangular prism, sphere, cone, and cylinder are the basic three-dimensional figures we see around us.
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