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

## NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2 – CBSE Term II Free PDF Download

The 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 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 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.

Firstly 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.

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 the points (4, 7, 8), (2, 3, 4) is parallel to the line through the 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 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 passes 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, as 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 is 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 rationalizing the fraction by multiplying the numerator and denominator by √2, we get ∴ The shortest distance is 32/2

15. Find the shortest distance between the lines Solution:    ∴ The shortest distance is 229

16. Find the shortest distance between the lines whose vector equations are Solution: Here by comparing the equations we get,  ∴ The shortest distance is 319

17. Find the shortest distance between the lines whose vector equations are Solution:  And, ∴ The shortest distance is 829