NCERT Class 12 Mathematics Chapter 11 Three dimensional Geometry

NCERT Class 12 Mathematics Chapter 11 Three dimensional Geometry Solutions, NCERT Solutions For Class 12 Maths, NCERT Class 12 Mathematics Chapter 11 Three dimensional Geometry Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 12 Mathematics Chapter 11 Three dimensional Geometry Question Answer and select needs one.

NCERT Class 12 Mathematics Chapter 11 Three dimensional Geometry

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Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 12 Mathematics Chapter 11 Three dimensional Geometry Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 12 Mathematics Chapter 11 Three dimensional Geometry Solutions for All Subject, You can practice these here.

Three dimensional Geometry

Chapter – 11

1. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines. 

Ans: Let direction cosines of the line be I,m and n. 

Hence,

l = cos 90⁰ = 0

2. Find the direction cosines l,m and n of f a line which makes equal angles with the coordinate axes.

Ans: Let the direction cosines of the line make an angle α with each of the coordinates axes. 

Hence,

l = cos α 

m = cos α 

n = cos α 

Since, l² + m² + n² = 1 

Hence, 

⇒ cos² α + cos² α + cos² α = 1

⇒ 3cos² α = 1

⇒ cos² α = 1/3

⇒ cos α = ± 1/√3

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are

3. If a line has the direction ratios-18, 12, -4, then what are its direction cosines?

Ans: If a line has direction ratios -18,12, -4, then its direction cosines are:

4. Show that the points (2,3,4), (-1,-2,1), (5,8,7) are collinear.

Ans: Given points are A(2, 3, 4) B(- 1, – 2, 1) and C(5, 8, 7).

As we know that the direction cosines of points, 

(x₁, y₁, z₁) and (x₂, y₂, z₂) are given by (x₂ – x₁), (y₂ – y₁) and (z₂ – z₁).

Therefore, the direction ratios of AB are 

(- 1 – 2), (- 2 – 3) and (1 – 4) 

⇒ – 3 ,-5 and -3

The direction ratios of BC are 

[5-(-1)],[8-(-2)] and (7-1) 

⇒ 6 ,10 and 6

It can be seen that the direction ratios of BC are -2 times that AB i.e., they are proportional.

Hence, AB is parallel to BC. Since point B is common to both AB and BC, points A, B, and C are collinear.

5. Find the direction cosines of the sides of the triangle whose vertices are (3,5,-4), (-1,1,2) and (-5,-5,-2).

Ans: Vertices of the triangle are A(3, 5, – 4), B(- 1, 1, 2) and C(- 5, – 5, – 2) 

The direction ratios of the side AB are 

(- 1 – 3), (1 – 5) and [2-(- 4)] 

⇒- 4,- 4 and 6 

Hence, the direction cosines of AB are

The direction ratios of CA are

1. Show that the three lines with direction cosines are mutually perpendicular.

Ans: Two lines with direction cosines l₁, m₁, n₁ and l₂,m₂,n₂ are perpendicular to each other, if l₁ l₂ + m₁m₂ + n₁ n₂ = 0

For the lines with direction cosines, we get

Hence, the lines are perpendicular.

For the lines with direction cosines, we get

Hence, the lines are perpendicular.

For the lines with direction cosines, we get

Hence, the lines are perpendicular.

So, the all three lines are mutually perpendicular.

2. Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Ans: Let AB be the line joining the points (1,-1,2) and (3,4,-2); and CD be the line through the points (0,3,2) and (3,5,6)

Hence, 

a₁ = (3 – 1) = 2 

b₁ = [4 – (- 1)] = 5 

c₁ = (- 2 – 2) = – 4 

a₂ = (3 – 0) = 3 

b₂ = (5 – 3) = 2 

c₂ = (6 – 2) = 4

If, AB ⏊ CD ; ⇒ a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

Here,

a₁ a₂ + b₁ b₂ + c₁ c₂ = 2 × 3 + 5 × 2 + (- 4) × 4

= 6+10-16

= 0

Hence, AB and CD are perpendicular to each other.

3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (1, 2, 5).

Ans: Let AB be the line through the points (4,7,8) and (2,3,4); CD be the line through the points (-1,-2,1) and (1,2,5).

Hence,

a₁ = (2 – 4) = – 2 

b₁ = (3 – 7) = – 4 

c₁ = (4 – 8) = – 4

a₂ = [1 – (- 1)] = 2 

b₂ = [2 – (- 2)] = 4 

c₂ = (5 – 1) = 4

Hence, AB is parallel to CD.

4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3 î + 2 ĵ – 2 k̂.

Ans: It is given that the line passes through the point A(1,2,3). 

Therefore, the position vector through A(1,2,3) is

So, line passes through point A(1,2,3) and parallel to real number.

Hence,

This is the required equation of the line.

5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2 î – j + 4 k̂ and is in the direction î + 2 ĵ – k̂.

Ans: It is given that

Since, the vector equation of the line is given by where λ is some real number. 

Hence, 

x î – y ĵ + z k̂ = 2 î – ĵ + 4 k̂ + λ( î + 2 ĵ – k̂ ) 

= (2 + λ) î + (- 1 + 2λ) ĵ + (4 – λ) k̂

Eliminating [], we get the Cartesian form equation as

Thus, the equation cartesian form is of the line in vector form is and cartesian form is

6. Find the cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by

Ans: It is given that the required line passes through the point (-2,4,-5) and is parallel to

Therefore, its direction ratios are 3k,5k and 6k, where k ≠ 0

It is known that the equation of the line through the point (x₁, y₁, z₁) and with direction ratios a,b,c is given by

Hence, the equation of the required line is

Thus, the cartesian equation of the line is

7. The cartesian equation of a line is Write its vector form.

Ans: It is given that the Cartesian equation of the line is

Hence,

The given line passes through the point (5,-4,6)

Therefoe,

The position vector of the point is

Also, the direction ratios of the given line are 3,7 and 2 

This means that the line is in the direction of the vector,

As we known that the line through positive vector given by the equation,

Hence,

This is the required equation of the given line in vector form.

8. Find the angle between the following pairs of lines:

Ans: Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,

The given lines are parallel to the vectors,

respectively

= 3 × 1 + 2 × 2 + 6 × 2 

= 3 + 4 + 12

= 19

Ans: The given lines are parallel to the vectors,

respectively

9. Find the angle between the following pair of lines:

Ans: 

= 2 × (- 1) + 5 × 8 + (- 3) × 4 

= – 2 + 40 – 12 

= 26 

The angle [] between the given pair of lines is given by the relation,

Ans: 

= 2 × 4 + 2 × 1 + 1 × 8 

= 8 + 2 + 8 

= 18

If [] is the angle between the pair of lines, then

10. Find the values of p so that the lines and are at right angles.

Ans: The given equations can be written in the standard form as and

The direction ratios of the lines are given by

Since, both the lines are perpendicular to each other,

Therefore,

11. Show that the lines and are perpendicular to each other.

Ans: The equations of the given lines are and

Here,

a₁ = 7, b₁ = – 5 and c₁ = 1

a₂ = 1, b₂ = 2 and c₂ = 3

Two lines with direction ratios, a₁ b₁, c₁ and a₂ ,b₂, c₂ are perpendicular to each other, if a₁a₂ + b₁b₂ + c₁c₂ = 0

Since,

7 × 1 + (- 5) × 2 + 1 × 3 = 7 – 10 + 3

= 0

Hence, the given lines are perpendicular to each other.

12. Find the shortest distance between the lines

Ans: 

= √9 + 9

= √18

= 3√2

Putting all the values in equation (1), we get

13. Find the shortest distance between the lines

Ans: 

14. Find the shortest distance between the lines whose vector equations are

Ans: 

Putting all the values in equation (1), we get

15. Find the shortest distance between the lines whose vector equations are

Ans: 

Putting all the values in equation (1), we get

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

Ans: The angle θ between the lines with direction cosines a,b,c and (b – c), (c – a), (a – b) is given by,

= cos⁻¹ 0

= 90⁰

Thus, the required angle is 90°

2. Find the equation of a line parallel to x-axis and passing through the origin.

Ans: The line parallel to x-axis and passing through the origin is x-axis itself. 

Let A be a point on x-axis.

Therefore, the coordinates of A are given by (a,0,0), where a ∈ R 

Hence, the direction ratios of OA are a,0,0

The equation of OA is given by,

Hence, the equation of line parallel to x-axis and passing origin is

Ans: Here,

Two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂,c₂ are perpendicular, if a₁a₂ + b₁b₂ + c₁c₂ = 0

Therefore,

⇒ – 3(3k) + 2k × 1 + 2(- 5) = 0 

⇒ – 9k + 2k – 10 = 0 

⇒ 7k = – 10

Ans: The given lines are

Putting all these values in equation (1), we get

Hence, the shortest distance between the two given lines is 9 units.

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

Ans: Here,