NCERT Solution Class 12 Chapter 11 - Three Dimensional Geometry Exercise 11.2

NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2 – Free PDF Download

Exercise 11.2 of NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry is based on the following topics:

1. Equation of a Line in Space
   1. Equation of a line through a given point and parallel to a given vector b.
   2. Equation of a line passing through two given points
2. Angle between Two Lines
3. Shortest Distance between Two Lines
   1. Distance between two skew lines
   2. Distance between parallel lines

The exercise contains problems that are based on the topics mentioned above. Hence, solving this exercise helps the students understand the concepts covered in the chapter thoroughly.

NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry Exercise 11.2

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Access Other Exercises of Class 12 Maths Chapter 11

Exercise 11.1 Solutions 5 Questions

Exercise 11.3 Solutions 14 Questions

Miscellaneous Exercise on Chapter 11 Solutions 23 Questions

Access Answers to NCERT Class 12 Maths Chapter 11 Exercise 11.2

1. Show that the three lines with direction cosines

  are mutually perpendicular.

Solution:

Let us consider the direction cosines of L₁, L₂ and L₃ be l₁, m₁, n₁; l₂, m₂, n₂ and l₃, m₃, n₃.

We know that

If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two lines,

And θ is the acute angle between the two lines,

Then cos θ = |l₁l₂ + m₁m₂ + n₁n₂|

If two lines are perpendicular, then the angle between the two is θ = 90°

For perpendicular lines, | l₁l₂ + m₁m₂ + n₁n₂ | = cos 90° = 0, i.e. | l₁l₂ + m₁m₂ + n₁n₂ | = 0

So, in order to check if the three lines are mutually perpendicular, we compute | l₁l₂ + m₁m₂ + n₁n₂ | for all the pairs of the three lines.

First, let us compute, | l₁l₂ + m₁m₂ + n₁n₂ |

So,  L₁⊥ L₂ …… (1)

Similarly,

Let us compute, | l₂l₃ + m₂m₃ + n₂n₃ |

So, L₂⊥ L₃ ….. (2)

Similarly,

Let us compute, | l₃l₁ + m₃m₁ + n₃n₁ |

So, L₁⊥ L₃ ….. (3)

∴ By (1), (2) and (3), the lines are perpendicular.

Hence, L₁, L₂ and L₃ 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).

Solution:

Given:

The points (1, –1, 2), (3, 4, –2) and (0, 3, 2), (3, 5, 6).

Let us consider 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).

Now,

The direction ratios, a₁, b₁, c₁ of AB are

(3 – 1), (4 – (-1)), (-2 – 2) = 2, 5, -4.

Similarly,

The direction ratios, a₂, b₂, c₂ of CD are

(3 – 0), (5 – 3), (6 – 2) = 3, 2, 4.

Then, AB and CD will be perpendicular to each other, if a₁a₂ + b₁b₂ + c₁c₂ = 0

a₁a₂ + b₁b₂ + c₁c₂ = 2(3) + 5(2) + 4(-4)

= 6 + 10 – 16

= 0

∴ AB and CD are perpendicular to each other.

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

Solution:

Given:

The points (4, 7, 8), (2, 3, 4) and (–1, –2, 1), (1, 2, 5).

Let us consider AB be the line joining the points (4, 7, 8), (2, 3, 4), and CD be the line through the points (-1, -2, 1), (1, 2, 5).

Now,

The direction ratios, a₁, b₁, c₁ of AB are

(2 -4), (3 -7), (4 -8) = -2, -4, -4.

The direction ratios, a₂, b₂, c₂ of CD are

(1 – (-1)), (2 – (-2)), (5 – 1) = 2, 4, 4.

Then, AB will be parallel to CD, if

So, a₁/a₂ = -2/2 = -1

b₁/b₂ = -4/4 = -1

c₁/c₂ = -4/4 = -1

∴ We can say that,

-1 = -1 = -1

Hence, AB is parallel to CD, where 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).

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

Solution:

5. Find the equation of the line in vector and in a Cartesian form that passes through the point with position vector and is in the direction. 

Solution:

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

Solution:

Given:

The points (-2, 4, -5)

We know that

The Cartesian equation of a line through a point (x₁, y₁, z₁) and having direction ratios a, b, c is

7. The Cartesian equation of a line is

 . Write its vector form.

Solution:

So when comparing this standard form with the given equation, we get

x₁ = 5, y₁ = -4, z₁ = 6, and

l = 3, m = 7, n = 2

8. Find the vector and the Cartesian equations of the lines that pass through the origin and (5, -2, 3).

Solution:

9. Find the vector and the Cartesian equations of the line that passes through the points (3, -2, -5), (3, –2, 6).

Solution:

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

Solution:

So,

By (3), we have

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

Solution:

12. Find the values of p so that the lines

  are at right angles.

Solution:

So, the direction ratios of the lines are

-3, 2p/7, 2 and -3p/7, 1, -5

Now, both the lines are at right angles,

So, a₁a₂ + b₁b₂ + c₁c₂ = 0

(-3) (-3p/7) + (2p/7) (1) + 2 (-5) = 0

9p/7 + 2p/7 – 10 = 0

(9p+2p)/7 = 10

11p/7 = 10

11p = 70

p = 70/11

∴ The value of p is 70/11.

13. Show that the lines

  are perpendicular to each other.

Solution:

The equations of the given lines are

Two lines with direction ratios are given as

a₁a₂ + b₁b₂ + c₁c₂ = 0

So, the direction ratios of the given lines are 7, -5, 1 and 1, 2, 3

i.e., a₁ = 7, b₁ = -5, c₁ = 1, and

a₂ = 1, b₂ = 2, c₂ = 3

Now, considering

a₁a₂ + b₁b₂ + c₁c₂ = 7 × 1 + (-5) × 2 + 1 × 3

= 7 -10 + 3

= – 3 + 3

= 0

∴ The two lines are perpendicular to each other.

14. Find the shortest distance between the lines

Solution:

Let us rationalize the fraction by multiplying the numerator and denominator by √2, and we get

∴ The shortest distance is 3√2/2.

15. Find the shortest distance between the lines

Solution:

∴ The shortest distance is 2√29.

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

Solution:

Here, by comparing the equations, we get

∴ The shortest distance is 3√19.

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

Solution:

And,

∴ The shortest distance is 8√29.

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NCERT Solutions for Class 12 Maths

NCERT Solutions for Class 12

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