NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 – Free PDF Download
Exercise 11.3 of NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry is based on the following topics:
- Plane
- Equation of a plane in normal form
- Equation of a plane perpendicular to a given vector, passing through the given point
- Equation of a plane passing through three non-collinear points
- Intercept form of the equation of a plane
- A plane passing through the intersection of two given planes
- Coplanarity of Two Lines
- Angle between Two Planes
- Distance of a Point from a Plane
- Angle between a Line and a Plane
Solving the problems of this exercise will help the students understand the problem-solving methods related to the topics mentioned above.
NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry Exercise 11.3
Access Other Exercises of Class 12 Maths Chapter 11
Exercise 11.1 Solutions: 5 Questions
Exercise 11.2 Solutions: 17 Questions
Miscellaneous Exercise on Chapter 11 Solutions: 23 Questions
Access Answers to NCERT Class 12 Maths Chapter 11 Exercise 11.3
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:
(a) z = 2
Given:
The equation of the plane, z = 2 or 0x + 0y + z = 2 …. (1)
Direction ratio of the normal (0, 0, 1)
By using the formula,
√[(0)2 + (0)2 + (1)2] = √1
= 1
Now,
Divide both sides of equation (1) by 1, and we get
0x/(1) + 0y/(1) + z/1 = 2
So this is of the form lx + my + nz = d
Where l, m and n are the direction cosines, and d is the distance.
∴ The direction cosines are 0, 0, 1.
Distance (d) from the origin is 2 units.
(b) x + y + z = 1
Given:
The equation of the plane, x + y + z = 1…. (1)
Direction ratio of the normal (1, 1, 1)
By using the formula,
√[(1)2 + (1)2 + (1)2] = √3
Now,
Divide both sides of equation (1) by √3, and we get
x/(√3) + y/(√3) + z/(√3) = 1/√3
So this is of the form lx + my + nz = d
Where l, m and n are the direction cosines, and d is the distance.
∴ The direction cosines are 1/√3, 1/√3, 1/√3.
Distance (d) from the origin is 1/√3 units.
(c) 2x + 3y – z = 5
Given:
The equation of the plane, 2x + 3y – z = 5…. (1)
Direction ratio of the normal (2, 3, -1)
By using the formula,
√[(2)2 + (3)2 + (-1)2] = √14
Now,
Divide both sides of equation (1) by √14, and we get
2x/(√14) + 3y/(√14) – z/(√14) = 5/√14
So this is of the form lx + my + nz = d
Where l, m and n are the direction cosines, and d is the distance.
∴ The direction cosines are 2/√14, 3/√14, -1/√14.
Distance (d) from the origin is 5/√14 units.
(d) 5y + 8 = 0
Given:
The equation of the plane, 5y + 8 = 0
-5y = 8 or
0x – 5y + 0z = 8…. (1)
Direction ratio of the normal (0, -5, 0)
By using the formula,
√[(0)2 + (-5)2 + (0)2] = √25
= 5
Now,
Divide both sides of equation (1) by 5, and we get
0x/(5) – 5y/(5) – 0z/(5) = 8/5
So this is of the form lx + my + nz = d
Where l, m and n are the direction cosines, and d is the distance.
∴ The direction cosines are 0, -1, 0.
Distance (d) from the origin is 8/5 units.
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:
3. Find the Cartesian equation of the following planes.
(a) 
Solution:
Given:
The equation of the plane.
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 + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Solution:
(a) 2x + 3y + 4z – 12 = 0
Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x, y, z).
2x + 3y + 4z = 12 …. (1)
Direction ratios are (2, 3, 4)
√[(2)2 + (3)2 + (4)2] = √(4 + 9 + 16)
= √29
Now,
Divide both sides of equation (1) by √29, and we get
2x/(√29) + 3y/(√29) + 4z/(√29) = 12/√29
So this is of the form lx + my + nz = d
Where l, m and n are the direction cosines, and d is the distance.
∴ The direction cosines are 2/√29, 3/√29, 4/√29
Coordinate of the foot (ld, md, nd) =
= [(2/√29) (12/√29), (3/√29) (12/√29), (4/√29) (12/√29)]
= 24/29, 36/29, 48/29
(b) 3y + 4z – 6 = 0
Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x, y, z).
0x + 3y + 4z = 6 …. (1)
Direction ratios are (0, 3, 4).
√[(0)2 + (3)2 + (4)2] = √(0 + 9 + 16)
= √25
= 5
Now,
Divide both sides of equation (1) by 5, and we get
0x/(5) + 3y/(5) + 4z/(5) = 6/5
So this is of the form lx + my + nz = d
Where l, m and n are the direction cosines, and d is the distance.
∴ The direction cosines are 0/5, 3/5, 4/5
Coordinate of the foot (ld, md, nd) =
= [(0/5) (6/5), (3/5) (6/5), (4/5) (6/5)]
= 0, 18/25, 24/25
(c) x + y + z = 1
Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x, y, z).
x + y + z = 1 …. (1)
Direction ratios are (1, 1, 1).
√[(1)2 + (1)2 + (1)2] = √(1 + 1 + 1)
= √3
Now,
Divide both sides of equation (1) by √3, and we get
1x/(√3) + 1y/(√3) + 1z/(√3) = 1/√3
So this is of the form lx + my + nz = d
Where l, m and n are the direction cosines, and d is the distance.
∴ The direction cosines are 1/√3, 1/√3, 1/√3.
Coordinate of the foot (ld, md, nd) =
= [(1/√3) (1/√3), (1/√3) (1/√3), (1/√3) (1/√3)]
= 1/3, 1/3, 1/3
(d) 5y + 8 = 0
Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x, y, z).
0x – 5y + 0z = 8 …. (1)
Direction ratios are (0, -5, 0).
√[(0)2 + (-5)2 + (0)2] = √(0 + 25 + 0)
= √25
= 5
Now,
Divide both sides of equation (1) by 5, and we get
0x/(5) – 5y/(5) + 0z/(5) = 8/5
So this is of the form lx + my + nz = d
Where l, m and n are the direction cosines, and d is the distance.
∴ The direction cosines are 0, -1, 0.
Coordinate of the foot (ld, md, nd) =
= [(0/5) (8/5), (-5/5) (8/5), (0/5) (8/5)]
= 0, -8/5, 0
5. 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:
x – 1 – 2y + 8 + z – 6 = 0
x – 2y + z + 1 = 0
x – 2y + z = -1
∴ The required Cartesian equation of the plane is x – 2y + z = -1
x – 1 – 2y + 8 + z – 6 = 0
x – 2y + z + 1 = 0
x – 2y + z = -1
∴ The required Cartesian equation of the plane is x – 2y + z = -1
6. Find the equations of the planes that pass through three points.
(a) (1, 1, –1), (6, 4, –5), (–4, –2, 3)
(b) (1, 1, 0), (1, 2, 1), (–2, 2, –1)
Solution:
Given:
The points are (1, 1, -1), (6, 4, -5), (-4, -2, 3).
Let,
= 1(12 – 10) – 1(18 – 20) -1 (-12 + 16)
= 2 + 2 – 4
= 0
The value of the determinant is 0.
∴ The points are collinear, as there will be infinite planes passing through the given 3 points.
(b) (1, 1, 0), (1, 2, 1), (–2, 2, –1)
7. Find the intercepts cut off by the plane 2x + y – z = 5.
Solution:
Given:
The plane 2x + y – z = 5
Let us express the equation of the plane in intercept form
x/a + y/b + z/c = 1
Where a, b and c are the intercepts cut off by the plane at the x, y and z axes, respectively.
2x + y – z = 5 …. (1)
Now divide both sides of equation (1) by 5, and we get
2x/5 + y/5 – z/5 = 5/5
2x/5 + y/5 – z/5 = 1
x/(5/2) + y/5 + z/(-5) = 1
Here, a = 5/2, b = 5 and c = -5
∴ The intercepts cut off by the plane are 5/2, 5 and -5.
8. Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane.
Solution:
We know that the equation of the plane ZOX is y = 0
So, the equation of plane parallel to ZOX is of the form y = a
Since the y-intercept of the plane is 3, a = 3
∴ The required equation of the plane is y = 3
9. 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:
Given:
The equation of the plane passes through the intersection of the plane is given by
(3x – y + 2z – 4) + λ (x + y + z – 2) = 0, and the plane passes through the points (2, 2, 1).
So, (3 × 2 – 2 + 2 × 1 – 4) + λ (2 + 2 + 1 – 2) = 0
2 + 3λ = 0
3λ = -2
λ = -2/3 …. (1)
Upon simplification, the required equation of the plane is given as
(3x – y + 2z – 4) – 2/3 (x + y + z – 2) = 0
(9x – 3y + 6z – 12 – 2x – 2y – 2z + 4)/3 = 0
7x – 5y + 4z – 8 = 0
∴ The required equation of the plane is 7x – 5y + 4z – 8 = 0
10. Find the vector equation of the plane passing through the intersection of the planes 
and through the point (2, 1, 3).
Solution:
The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,
11. 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:
Let the equation of the plane that passes through the two-given planes
x + y + z = 1 and 2x + 3y + 4z = 5 is
(x + y + z – 1) + λ (2x + 3y + 4z – 5) = 0
(2λ + 1) x + (3λ + 1) y + (4λ + 1) z -1 – 5λ = 0…… (1)
So, the direction ratio of the plane is (2λ + 1, 3λ + 1, 4λ + 1).
And direction ratio of another plane is (1, -1, 1).
Both planes are ⊥.
So by substituting in a1a2 + b1b2 + c1c2 = 0
(2λ + 1 × 1) + (3λ + 1 × (-1)) + (4λ + 1 × 1) = 0
2λ + 1 – 3λ – 1 + 4λ + 1 = 0
3λ + 1 = 0
λ = -1/3
Substitute the value of λ in equation (1), and we get
x – z + 2 = 0
∴ The required equation of the plane is x – z + 2 = 0
12. Find the angle between the planes whose vector equations are
Solution:
13. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles 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 = 0
(d) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0
Solution:
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
Given:
The equation of the given planes are
7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
Two planes are ⊥ if the direction ratio of the normal to the plane is
a1a2Â + b1b2Â + c1c2Â = 0
21 – 5 – 60
-44 ≠0
Both the planes are not ⊥ to each other.
Now, two planes are || to each other if the direction ratio of the normal to the plane is
∴ The angle is cos-1 (2/5)
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
Given:
The equation of the given planes are
2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
Two planes are ⊥ if the direction ratio of the normal to the plane is
a1a2Â + b1b2Â + c1c2Â = 0
2 × 1 + 1 × (-2) + 3 × 0
= 0
∴ The given planes are ⊥ to each other.
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
Given:
The equation of the given planes are
2x – 2y + 4z + 5 =0 and x – 2y + 5 = 0
We know that two planes are ⊥ if the direction ratio of the normal to the plane is
a1a2Â + b1b2Â + c1c2Â = 0
6 + 6 + 24
36 ≠0
∴ Both the planes are not ⊥ to each other.
Now let us check, if both planes are || to each other if the direction ratio of the normal to the plane is
∴ The given planes are || to each other.
(d) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
Given:
The equation of the given planes are
2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
We know that two planes are ⊥ if the direction ratio of the normal to the plane is
a1a2Â + b1b2Â + c1c2Â = 0
2 × 2 + (-1) × (-1) + 3 × 3
14 ≠0
∴ Both the planes are not ⊥ to each other.
Now, let us check if two planes are || to each other if the direction ratio of the normal to the plane is
∴ The given planes are || to each other.
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0
Given:
The equation of the given planes are
4x + 8y + z – 8 = 0 and y + z – 4 = 0
We know that two planes are ⊥ if the direction ratio of the normal to the plane is
a1a2Â + b1b2Â + c1c2Â = 0
0 + 8 + 1
9 ≠0
∴ Both the planes are not ⊥ to each other.
Now, let us check if two planes are || to each other if the direction ratio of the normal to the plane is
∴ Both planes are not || to each other.
Now, let us find the angle between them, which is given as
∴ The angle is 45o.
14. In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0, 0, 0) 3x – 4y + 12 z = 3
(b) (3, -2, 1) 2x – y + 2z + 3 = 0
(c) (2, 3, -5) x + 2y – 2z = 9
(d) (-6, 0, 0) 2x – 3y + 6z – 2 = 0
Solution:
(a) Point Plane
(0, 0, 0) 3x – 4y + 12 z = 3
We know that, distance of point P(x1, y1, z1) from the plane Ax + By + Cz – D = 0 is given as
Given point is (0, 0, 0), and the plane is 3x – 4y + 12z = 3
= |3/√169|
= 3/13
∴ The distance is 3/13.
(b) Point Plane
(3, -2, 1) 2x – y + 2z + 3 = 0
We know that, distance of point P(x1, y1, z1) from the plane Ax + By + Cz – D = 0 is given as
Given point is (3, -2, 1), and the plane is 2x – y + 2z + 3 = 0
= |13/√9|
= 13/3
∴ The distance is 13/3.
(c) Point Plane
(2, 3, -5) x + 2y – 2z = 9
We know that, distance of point P(x1, y1, z1) from the plane Ax + By + Cz – D = 0 is given as
Given point is (2, 3, -5), and the plane is x + 2y – 2z = 9
= |9/√9|
= 9/3
= 3
∴ The distance is 3.
(d) Point Plane
(-6, 0, 0) 2x – 3y + 6z – 2 = 0
We know that, distance of point P(x1, y1, z1) from the plane Ax + By + Cz – D = 0 is given as
Given point is (-6, 0, 0), and the plane is 2x – 3y + 6z – 2 = 0
= |14/√49|
= 14/7
= 2
∴ The distance is 2.
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