NCERT Solutions for Class 12 Maths Chapter 11 Miscellaneous Exercise – Free PDF Download

The Miscellaneous Exercise of NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry is based on the following topics:

1. Direction Cosines and Direction Ratios of a Line
2. Equation of a Line in Space
3. Angle between Two Lines
4. Shortest Distance between Two Lines
5. Plane
6. Coplanarity of Two Lines
7. Angle between Two Planes
8. Distance of a Point from a Plane
9. Angle between a Line and a Plane

Solve all the problems of this exercise to get thorough with the concepts and topics covered in the entire chapter.

NCERT Solution Class 12 Chapter 11 – Three Dimensional Geometry Miscellaneous Exercise

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Access Other Exercises of Class 12 Maths Chapter 11

Exercise 11.1 Solutions 5 Questions

Exercise 11.2 Solutions 17 Questions

Exercise 11.3 Solutions 14 Questions

Access Answers to NCERT Class 12 Maths Chapter 11 Miscellaneous Exercise

1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, –1), (4, 3, –1).

Solution:

Let us consider OA as the line joining the origin (0, 0, 0) and point A (2, 1, 1).

And let BC be the line joining the points B (3, 5, −1) and C (4, 3, −1)

So the direction ratios of OA = (a₁, b₁, c₁) ≡ [(2 – 0), (1 – 0), (1 – 0)] ≡ (2, 1, 1)

And the direction ratios of BC = (a₂, b₂, c₂) ≡ [(4 – 3), (3 – 5), (-1 + 1)] ≡ (1, -2, 0)

Given:

OA is ⊥ to BC

Now we have to prove that:

a₁a₂ + b₁b₂ + c₁c₂ = 0

Let us consider LHS: a₁a₂ + b₁b₂ + c₁c₂

a₁a₂ + b₁b₂ + c₁c₂ = 2 × 1 + 1 × (−2) + 1 × 0

= 2 – 2

= 0

We know that R.H.S is 0

So LHS = RHS

∴ OA is ⊥ to BC

Hence proved.

2. If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are (m₁n₂ – m₂n₁), (n₁l₂ – n₂l₁), (l₁m₂ – l₂m₁)

Solution:

Let us consider l, m, n to be the direction cosines of the line perpendicular to each of the given lines.

Then, ll₁ + mm₁ + nn₁ = 0 … (1)

And ll₂ + mm₂ + nn₂ = 0 … (2)

Upon solving (1) and (2) by using cross-multiplication, we get

Thus, the direction cosines of the given line are proportional to

(m₁n₂ – m₂n₁), (n₁l₂ – n₂l₁), (l₁m₂ – l₂m₁)

So, its direction cosines are

We know that

(l₁² + m₁² + n₁²) (l₂² + m₂² + n₂²) – (l₁l₂ + m₁m₂ + n₁n₂)²

= (m₁n₂ – m₂n₁)² + (n₁l₂ – n₂l₁)² + (l₁m₂ – l₂m₁)² … (3)

It is given that the given lines are perpendicular to each other.

So, l₁l₂ + m₁m₂ + n₁n₂ = 0

Also, we have

l₁² + m₁² + n₁² = 1

And, l₂² + m₂² + n₂² = 1

Substituting these values in equation (3), we get

(m₁n₂ – m₂n₁)² + (n₁l₂ – n₂l₁)² + (l₁m₂ – l₂m₁)² = 1

λ = 1

Hence, the direction cosines of the given line are (m₁n₂ – m₂n₁), (n₁l₂ – n₂l₁), (l₁m₂ – l₂m₁)

3. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Solution:

Angle between the lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ is given by

Given:

a₁ = a, b₁ = b, c₁ = c

a₂ = b – c, b₂ = c – a, c₂ = a – b

Substituting the values in the above equation, we get,

= 0

Cos θ = 0

So, θ = 90° [Since, cos 90 = 0]

Hence, the angle between the given pair of lines is 90°.

4. Find the equation of a line parallel to x – axis and passing through the origin.

Solution:

We know that equation of a line passing through (x₁, y₁, z₁) and parallel to a line with direction ratios a, b, c is

Given: the line passes through the origin, i.e. (0, 0, 0)

x₁ = 0, y₁ = 0, z₁ = 0

Since the line is parallel to the x-axis,

a = 1, b = 0, c = 0

∴ Equation of the line is given by

5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (–4, 3, –6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Solution:

We know that the angle between the lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ is given by

So now, a line passing through A (x₁, y₁, z₁) and B (x₂, y₂, z₂) has direction ratios (x₁ – x₂), (y₁ – y₂), (z₁ – z₂)

The direction ratios of a line joining the points A (1, 2, 3) and B (4, 5, 7)

= (4 – 1), (5 – 2), (7 – 3)

= (3, 3, 4)

∴ a₁ = 3, b₁ = 3, c₁ = 4

The direction ratios of a line joining the points C (-4, 3, -6) and B (2, 9, 2)

= (2 – (-4)), (9 – 3), (2-(-6))

= (6, 6, 8)

∴ a₂ = 6, b₂ = 6, c₂ = 8

Now let us substitute the values in the above equation. We get,

6. If the lines 

 and 
 are perpendicular, find the value of k.

Solution:

We get

x₂ = 1, y₂ = 2, z₂ = 3

And a₂ = 3k, b₂ = 1, c₂ = -5

Since the two lines are perpendicular,

a₁a₂ + b₁b₂ + c₁c₂ = 0

(-3) × 3k + 2k × 1 + 2 × (-5) = 0

-9k + 2k – 10 = 0

-7k = 10

k = -10/7

∴ The value of k is -10/7.

7. Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane

Solution:

8. Find the equation of the plane passing through (a, b, c) and parallel to the plane

Solution:

The equation of a plane passing through (x₁, y₁, z₁) and perpendicular to a line with direction ratios A, B, and C is given as

A (x – x₁) + B (y – y₁) + C (z – z₁) = 0

It is given that the plane passes through (a, b, c)

So, x₁ = a, y₁ = b, z₁ = c

Since both planes are parallel to each other, their normal will be parallel

Direction ratios of normal = (1, 1, 1)

So, A = 1, B =1, C = 1

The equation of the plane in Cartesian form is given as

A (x – x₁) + B (y – y₁) + C (z – z₁) = 0

1(x – a) + 1(y – b) + 1(z – c) = 0

x + y + z – (a + b + c) = 0

x + y + z = a + b + c

∴ The required equation of the plane is x + y + z = a + b + c

9. Find the shortest distance between lines
 and 

Solution:

10. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ – plane.

Solution:

We know that two vectors are equal if their corresponding components are equal

So,

0 = 5 – 2λ

5 = 2λ

λ = 5/2

y = 1 + 3λ … (5)

And,

z = 6 – 5λ … (6)

Substituting the value of λ in equations (5) and (6), we get

y = 1 + 3λ

= 1 + 3 × (5/2)

= 1 + (15/2)

= 17/2

And

z = 6 – 5λ

= 6 – 5 × (5/2)

= 6 – (25/2)

= – 13/2

∴ The coordinates of the required point are (0, 17/2, -13/2).

11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX – plane.

Solution:

We know that two vectors are equal if their corresponding components are equal.

So,

x = 5 – 2λ … (5)

0 = 1 + 3λ

-1 = 3λ

λ = -1/3

And,

z = 6 – 5λ … (6)

Substituting the value of λ in equations (5) and (6), we get

x = 5 – 2λ

= 5 – 2 × (-1/3)

= 5 + (2/3)

= 17/3

And

z = 6 – 5λ

= 6 – 5 × (-1/3)

= 6 + (5/3)

= 23/3

∴ The coordinates of the required point are (17/3, 0, 23/3).

12. Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane 2x + y + z = 7.

Solution:

We know that the equation of a line passing through two points A (x₁, y₁, z₁) and B (x₂, y₂, z₂) is given as

It is given that the line passes through points A (3, –4, –5) and B (2, –3, 1)

So, x₁ = 3, y₁ = -4, z₁ = -5

And, x₂ = 2, y₂ = -3, z₂ = 1

Then the equation of the line is

So, x = -k + 3 |, y = k – 4 |, z = 6k – 5 … (1)

Now let (x, y, z) be the coordinates of the point where the line crosses the given plane 2x + y + z + 7 = 0

By substituting the values of x, y, and z in equation (1) in the equation of a plane, we get

2x + y + z + 7 = 0

2(-k + 3) + (k – 4) + (6k – 5) = 7

5k – 3 = 7

5k = 10

k = 2

Now substituting the value of k in x, y, and z, we get,

x = – k + 3 = – 2 + 3 = 1

y = k – 4 = 2 – 4 = – 2

z = 6k – 5 = 12 – 5 = 7

∴ The coordinates of the required point are (1, -2, 7).

13. Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Solution:

We know that the equation of a plane passing through (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.

It is given that the plane passes through (-1, 3, 2)

So, the equation of the plane is given by

A (x + 1) + B (y – 3) + C (z – 2) = 0 ……… (1)

Since this plane is perpendicular to the given two planes. So, their normal to the plane would be perpendicular to the normal of both planes.

We know that

So, the required normal is the cross-product of the normal of planes

x + 2y + 3z = 5 and 3x + 3y + z = 0

Hence, the direction ratios are = -7, 8, -3

∴ A = -7, B = 8, C = -3

Substituting the obtained values in equation (1), we get

A (x + 1) + B (y – 3) + C (z – 2) = 0

-7(x + 1) + 8(y – 3) + (-3) (z – 2) = 0

-7x – 7 + 8y – 24 – 3z + 6 = 0

-7x + 8y – 3z – 25 = 0

7x – 8y + 3z + 25 = 0

∴ The equation of the required plane is 7x – 8y + 3z + 25 = 0.

14. If the points (1, 1, p) and (–3, 0, 1) be equidistant from the plane

 , then find the value of p.

Solution:

20 – 12p = ± 8

20 – 12p = 8 or, 20 – 12p = -8

12p = 12 or, 12p = 28

p = 1 or, p = 7/3

∴ The possible values of p are 1 and 7/3.

15. Find the equation of the plane passing through the line of intersection of the planes  and  and parallel to x-axis.

Solution:

Since this plane is parallel to the x-axis.

So, the normal vector of plane (1) will be perpendicular to the x-axis.

The direction ratios of the normal (a_1, b_1, c₁) ≡ [(1 – 2λ), (1 – 3λ), (1 +)]

The direction ratios of x-axis (a_2, b_2, c₂) ≡ (1, 0, 0)

Since the two lines are perpendicular,

a₁a₂ + b₁b₂ + c₁c₂ = 0

(1 – 2λ) × 1 + (1 – 3λ) × 0 + (1 + λ) × 0 = 0

(1 – 2λ) = 0

λ = 1/2

Substituting the value of λ in equation (1), we get

16. If O is the origin and the coordinates of P be (1, 2, –3), then find the equation of the plane passing through P and perpendicular to OP.

Solution:

We know that the equation of a plane passing through (x₁, y₁, z₁) and perpendicular to a line with direction ratios A, B, and C is given as

A(x – x₁) + B(y – y₁) + C (z – z₁) = 0

It is given that the plane passes through P (1, 2, 3)

So, x₁ = 1, y₁ = 2, z₁ = – 3

The normal vector to the plane is =

Where O (0, 0, 0), P (1, 2, -3)

So, the direction ratios of
is = (1 – 0), (2 – 0), (-3 – 0)

= (1, 2, – 3)

Where, A = 1, B = 2, C = -3

The equation of a plane in Cartesian form is given as

1(x – 1) + 2(y – 2) – 3(z – (-3)) = 0

x – 1 + 2y – 4 – 3z – 9 = 0

x + 2y – 3z – 14 = 0

∴ The equation of the required plane is x + 2y – 3z – 14 = 0

Solution:

Since this plane is perpendicular to the plane

So, the normal vector of plane (1) will be perpendicular to the normal vector of the plane (2).

Direction ratios of Normal of plane (1) = (a_1, b_1, c₁) ≡ [(1 – 2λ), (2 – λ), (3 + λ)]

Direction ratios of Normal of plane (2) = (a_2, b_2, c₂) ≡ (-5, -3, 6)

Since the two lines are perpendicular,

a₁a₂ + b₁b₂ + c₁c₂ = 0

(1 – 2λ) × (-5) + (2 – λ) × (-3) + (3 + λ) × 6 = 0

-5 + 10λ – 6 + 3λ + 18 + 6λ = 0

19λ + 7 = 0

λ = -7/19

By substituting the value of λ in equation (1), we get

18. Find the distance of the point (–1, –5, –10) from the point of intersection of the line 

Solution:

Where,

x = 2, y = -1, z = 2

So, the point of intersection is (2, -1, 2).

Now, the distance between points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by

∴ The distance is 13 units.

20. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:
 and .

Solution:

21. Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then

Solution:

22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
A. 2 units
B. 4 units
C. 8 units

D. 2/√29 units 

Solution:

We know that the distance between two parallel planes Ax + By + Cz = d₁ and Ax + By + Cz = d₂ is given as

It is given that:

First Plane:

2x + 3y + 4z = 4

Let us compare with Ax + By + Cz = d₁, we get

A = 2, B = 3, C = 4, d₁ = 4

Second Plane:

4x + 6y + 8z = 12 [Divide the equation by 2]

We get,

2x + 3y + 4z = 6

Now comparing with Ax + By + Cz = d₁, we get

A = 2, B = 3, C = 4, d₂ = 6

So,

Distance between two planes is given as

= 2/√29

∴ Option (D) is the correct option.

23. The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
A. Perpendicular
B. Parallel
C. intersect y–axis
D. passes through

Solution:

It is given that:

First Plane:

2x – y + 4z = 5 [Multiply both sides by 2.5]

We get,

5x – 2.5y + 10z = 12.5 … (1)

Given second Plane:

5x – 2.5y + 10z = 6 … (2)

So,

It is clear that the direction ratios of normal of both planes (1) and (2) are the same.

∴ Both the given planes are parallel.

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