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

- Equation of a Line in Space
- Equation of a line through a given point and parallel to a given vector b.
- Equation of a line passing through two given points

- Angle between Two Lines
- Shortest Distance between Two Lines
- Distance between two skew lines
- Distance between parallel lines

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

## Download PDF of NCERT Solutions for Class 12 Maths Chapter 11- Three Dimensional Geometry Exercise 11.2

### 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 of Maths NCERT Class 12 Chapter 11.2

**1. Show that the three lines with direction cosines**

**Â Â Are mutually perpendicular.**

**Solution:**

Let us consider the direction cosines of L_{1}, L_{2}Â and L_{3}Â be l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2}Â and l_{3}, m_{3}, n_{3}.

We know that

If l_{1}, m_{1}, n_{1}Â and l_{2}, m_{2}, n_{2}Â are the direction cosines of two lines;

And Î¸ is the acute angle between the two lines;

Then cos Î¸ = |l_{1}l_{2}Â + m_{1}m_{2}Â + n_{1}n_{2}|

If two lines are perpendicular, then the angle between the two is Î¸ = 90Â°

For perpendicular lines, | l_{1}l_{2}Â + m_{1}m_{2}Â + n_{1}n_{2}Â | = cos 90Â° = 0, i.e. | l_{1}l_{2}Â + m_{1}m_{2}Â + n_{1}n_{2}Â | = 0

So, in order to check if the three lines are mutually perpendicular, we compute | l_{1}l_{2}Â + m_{1}m_{2}Â + n_{1}n_{2}Â | for all the pairs of the three lines.

Firstly let us compute, | l_{1}l_{2}Â + m_{1}m_{2}Â + n_{1}n_{2}Â |

Â

So, Â L_{1}âŠ¥Â L_{2}Â â€¦â€¦ (1)

Similarly,

Let us compute, | l_{2}l_{3}Â + m_{2}m_{3}Â + n_{2}n_{3}Â |

Â

So, L_{2}âŠ¥Â L_{3}Â â€¦.. (2)

Similarly,

Let us compute, | l_{3}l_{1}Â + m_{3}m_{1}Â + n_{3}n_{1}Â |

So, L_{1}âŠ¥Â L_{3}Â â€¦.. (3)

âˆ´Â By (1), (2) and (3), the lines are perpendicular.

L_{1}, L_{2}Â and L_{3}Â are mutually perpendicular.

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

**Solution:**

Given:

The points (1, â€“1, 2), (3, 4, â€“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, a_{1}, b_{1}, c_{1} of AB are

(3 â€“ 1), (4 â€“ (-1)), (-2 â€“ 2) = 2, 5, -4.

Similarly,

The direction ratios, a_{2}, b_{2}, c_{2} of CD are

(3 â€“ 0), (5 â€“ 3), (6 â€“ 2) = 3, 2, 4.

Then, AB and CD will be perpendicular to each other, if a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 2(3) + 5(2) + 4(-4)

= 6 + 10 â€“ 16

= 0

âˆ´ 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 (â€“1, â€“2, 1), (1, 2, 5).**

**Solution:**

Given:

The points (4, 7, 8), (2, 3, 4) and (â€“1, â€“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 (â€“1, â€“2, 1), (1, 2, 5).

Now,

The direction ratios, a_{1}, b_{1}, c_{1} of AB are

(2 â€“ 4), (3 â€“ 7), (4 â€“ 8) = -2, -4, -4.

The direction ratios, a_{2}, b_{2}, c_{2} of CD are

(1 â€“ (-1)), (2 â€“ (-2)), (5 â€“ 1) = 2, 4, 4.

Then AB will be parallel to CD, if

So, a_{1}/a_{2} = -2/2 = -1

b_{1}/b_{2} = -4/4 = -1

c_{1}/c_{2} = -4/4 = -1

âˆ´ 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 (â€“1, â€“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 Cartesian form that passes through the point with position vector and is in the directionÂ **

**Solution:**

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

**Â **

**Solution:**

Given:

The points (-2, 4, -5)

We know that

The Cartesian equation of a line through a point (x_{1}, y_{1}, z_{1}) and having direction ratios a, b, c is

**7. The Cartesian equation of a line is**

**Â .Â Write its vector form.**

**Solution:**

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

x_{1}Â = 5, y_{1}Â = -4, z_{1}Â = 6 and

l = 3, m = 7, n = 2

**8. Find the vector and the Cartesian equations of the lines that passes through the origin and (5, â€“2, 3).**

**Solution:**

**9. Find the vector and the Cartesian equations of the line that passes through the points (3, â€“2, â€“5), (3, â€“2, 6).**

**Solution:**

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

**Solution:**

So,

By (3), we have

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

**Solution:**

**12.** **Find the values of p so that the lines**

**Â Â are at right angles.**

**Solution:**

So the direction ratios of the lines are

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

Now, as both the lines are at right angles,

So, a_{1}a_{2}Â + b_{1}b_{2}Â + c_{1}c_{2}Â = 0

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

9p/7 + 2p/7 â€“ 10 = 0

(9p+2p)/7 = 10

11p/7 = 10

11p = 70

p = 70/11

âˆ´ The value of p is 70/11

**13. Show that the lines**

**Â Â are perpendicular to each other.**

**Solution:**

The equations of the given lines are

Two lines with direction ratios is given as

a_{1}a_{2}Â + b_{1}b_{2}Â + c_{1}c_{2}Â = 0

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

i.e., a_{1}Â = 7, b_{1}Â = -5, c_{1}Â = 1 and

a_{2}Â = 1, b_{2}Â = 2, c_{2}Â = 3

Now, Considering

a_{1}a_{2}Â + b_{1}b_{2}Â + c_{1}c_{2}Â = 7 Ã— 1 + (-5) Ã— 2 + 1 Ã— 3

= 7 -10 + 3

= â€“ 3 + 3

= 0

âˆ´Â The two lines are perpendicular to each other.

**14. Find the shortest distance between the lines**

**Solution:**

Let us rationalizing the fraction by multiplying the numerator and denominator by âˆš2, we get

âˆ´ The shortest distance is 3**âˆš**2/2

**15.** **Find the shortest distance between the lines**

**Solution:**

âˆ´ The shortest distance is 2**âˆš**29

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

**Solution:**

Here by comparing the equations we get,

âˆ´ The shortest distance is 3**âˆš**19

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

**Solution:**

And,

âˆ´ The shortest distance is 8**âˆš**29