NCERT Solutions for Class 12 Maths Miscellaneous Exercise chapter 11 3 Dimensional Geometry

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.

Download PDF of NCERT Solutions for Class 12 Maths 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 of Maths NCERT Class 12 Chapter 11 Miscellaneous

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 be the line joining the origin (0, 0, 0) and the 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 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

NCERT Solutions for Class 12 Maths Chapter 11 image - 102

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

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

So, its direction cosines are

NCERT Solutions for Class 12 Maths Chapter 11 image - 103

NCERT Solutions for Class 12 Maths Chapter 11 image - 104

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 105

Given:

a1 = a, b1 = b, c1 = c

a2 = b – c, b2 = c – a, c2 = a – b

Let us substitute the values in the above equation we get,

NCERT Solutions for Class 12 Maths Chapter 11 image - 106

= 0

Cos θ = 0

So, θ = 90° [Since, cos 90 = 0]

Hence, 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

NCERT Solutions for Class 12 Maths Chapter 11 image - 107

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

x1 = 0, y1 = 0, z1 = 0

Since line is parallel to x – axis,

a = 1, b = 0, c = 0

∴ Equation of Line is given by

NCERT Solutions for Class 12 Maths Chapter 11 image - 108

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 109

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 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 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,

NCERT Solutions for Class 12 Maths Chapter 11 image - 110

6. If the lines 

NCERT Solutions for Class 12 Maths Chapter 11 image - 111 and 
NCERT Solutions for Class 12 Maths Chapter 11 image - 112 are perpendicular, find the value of k.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 113

NCERT Solutions for Class 12 Maths Chapter 11 image - 114

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

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 115

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 116

NCERT Solutions for Class 12 Maths Chapter 11 image - 117

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

 NCERT Solutions for Class 12 Maths Chapter 11 image - 118

Solution:

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 119

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

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

The Equation of 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 plane is x + y + z = a + b + c

9. Find the shortest distance between lines
NCERT Solutions for Class 12 Maths Chapter 11 image - 120 and NCERT Solutions for Class 12 Maths Chapter 11 image - 121

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 122

NCERT Solutions for Class 12 Maths Chapter 11 image - 123

NCERT Solutions for Class 12 Maths Chapter 11 image - 124

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 125

NCERT Solutions for Class 12 Maths Chapter 11 image - 126

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)

Substitute the value of λ in equation (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 is (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:

NCERT Solutions for Class 12 Maths Chapter 11 image - 127

NCERT Solutions for Class 12 Maths Chapter 11 image - 128

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)

Substitute the value of λ in equation (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 is (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

NCERT Solutions for Class 12 Maths Chapter 11 image - 129

It is given that the line passes through the 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 line is

NCERT Solutions for Class 12 Maths Chapter 11 image - 130

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 value of x, y, z in equation (1) in the equation of plane, we get

2x + y + z + 7 = 0

2(-k + 3) + (k – 4) + (6k – 5) = 7

5k – 3 = 7

5k = 10

k = 2

Now substitute the value of k in x, y, 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, C are the direction ratios of normal to the plane.

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

So, equation of 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 normal of both planes.

We know that

NCERT Solutions for Class 12 Maths Chapter 11 image - 131

So, required normal is cross product of normal of planes

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 132

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

 NCERT Solutions for Class 12 Maths Chapter 11 image - 133, then find the value of p.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 134

NCERT Solutions for Class 12 Maths Chapter 11 image - 135

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 NCERT Solutions for Class 12 Maths Chapter 11 image - 136 and NCERT Solutions for Class 12 Maths Chapter 11 image - 137 and parallel to x-axis.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 138

Since this plane is parallel to x-axis.

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

The direction ratios of Normal (a1, b1, c1) ≡ [(1 – 2λ), (1 – 3λ), (1 +)]

The direction ratios of x–axis (a2, b2, 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

NCERT Solutions for Class 12 Maths Chapter 11 image - 139

16. If O be 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, 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

Normal vector to plane is = 
NCERT Solutions for Class 12 Maths Chapter 11 image - 140

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

So, direction ratios of 
NCERT Solutions for Class 12 Maths Chapter 11 image - 141is = (1 – 0), (2 – 0), (-3 – 0)

= (1, 2, – 3)

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

Equation of 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

NCERT Solutions for Class 12 Maths Chapter 11 image - 142

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 143

Since this plane is perpendicular to the plane

NCERT Solutions for Class 12 Maths Chapter 11 image - 144

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

Direction ratios of Normal of plane (1) = (a1, b1, c1) ≡ [(1 – 2λ), (2 – λ), (3 + λ)]

Direction ratios of Normal of plane (2) = (a2, b2, 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

NCERT Solutions for Class 12 Maths Chapter 11 image - 145

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 146

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 147

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 148

NCERT Solutions for Class 12 Maths Chapter 11 image - 149

∴ The distance is 13 units.

NCERT Solutions for Class 12 Maths Chapter 11 image - 150

NCERT Solutions for Class 12 Maths Chapter 11 image - 151

NCERT Solutions for Class 12 Maths Chapter 11 image - 152

NCERT Solutions for Class 12 Maths Chapter 11 image - 153

20. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:
NCERT Solutions for Class 12 Maths Chapter 11 image - 154 and NCERT Solutions for Class 12 Maths Chapter 11 image - 155.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 156

NCERT Solutions for Class 12 Maths Chapter 11 image - 157

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 158

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 159

NCERT Solutions for Class 12 Maths Chapter 11 image - 160

NCERT Solutions for Class 12 Maths Chapter 11 image - 161

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 162

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

NCERT Solutions for Class 12 Maths Chapter 11 image - 163

= 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,

NCERT Solutions for Class 12 Maths Chapter 11 image - 164

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

∴ Both the given planes are parallel.

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