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

## NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2 â€“ 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

Â Â Are mutually perpendicular.

Solution:

Let us consider the direction cosines of L1, L2Â and L3Â be l1, m1, n1; l2, m2, n2Â and l3, m3, n3.

We know that

If l1, m1, n1Â and l2, m2, n2Â are the direction cosines of two lines;

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

Then cos Î¸ = |l1l2Â + m1m2Â + n1n2|

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

For perpendicular lines, | l1l2Â + m1m2Â + n1n2Â | = cos 90Â° = 0, i.e. | l1l2Â + m1m2Â + n1n2Â | = 0

So, in order to check if the three lines are mutually perpendicular, we compute | l1l2Â + m1m2Â + n1n2Â | for all the pairs of the three lines.

Firstly let us compute, | l1l2Â + m1m2Â + n1n2Â |

So, Â L1âŠ¥Â L2Â â€¦â€¦ (1)

Similarly,

Let us compute, | l2l3Â + m2m3Â + n2n3Â |

So, L2âŠ¥Â L3Â â€¦.. (2)

Similarly,

Let us compute, | l3l1Â + m3m1Â + n3n1Â |

So, L1âŠ¥Â L3Â â€¦.. (3)

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

L1, L2Â and L3Â 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, a1, b1, c1 of AB are

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

Similarly,

The direction ratios, a2, b2, c2 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 a1a2 + b1b2 + c1c2 = 0

a1a2 + b1b2 + c1c2 = 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, a1, b1, c1 of AB are

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

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

(1 â€“ (-1)), (2 â€“ (-2)), (5 â€“ 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

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 (x1, y1, z1) 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

x1Â = 5, y1Â = -4, z1Â = 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, a1a2Â + b1b2Â + c1c2Â = 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

a1a2Â + b1b2Â + c1c2Â = 0

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

i.e., a1Â = 7, b1Â = -5, c1Â = 1 and

a2Â = 1, b2Â = 2, c2Â = 3

Now, Considering

a1a2Â + b1b2Â + c1c2Â = 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 3âˆš2/2

15. Find the shortest distance between the lines

Solution:

âˆ´ The shortest distance is 2âˆš29

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 3âˆš19

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

Solution:

And,

âˆ´ The shortest distance is 8âˆš29