NCERT Solutions Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry are available here to help the students understand the ways in which different questions should be solved. Looking around, we can observe that everything in the real world is in a three-dimensional shape. In this chapter, students learn about Three-Dimensional Geometry in detail. Students can easily score full marks in the questions from this chapter by solving all the questions present in the NCERT textbook.

The Class 12 NCERT solutions for Three Dimensional Geometry are very easy to understand. These solutions cover all the exercise questions included in the book and are according to the latest guidelines of CBSE. Here, the PDF of the Class 12 Maths Chapter 11 NCERT solutions is available which can be downloaded and referred in online mode.

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

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Access Exercises of Class 12 Maths Chapter 11 – Three Dimensional Geometry

Exercise 11.1 Solutions 5 Questions

Exercise 11.2 Solutions 17 Questions

Exercise 11.3 Solutions 14 Questions

Miscellaneous Exercise On Chapter 11 Solutions 23 Questions

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

EXERCISE 11.1 PAGE NO: 467

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

Solution:

Let the direction cosines of the line be l, m and n.

Here let α = 90°, β = 135° and γ = 45°

So,

l = cos α, m = cos β and n = cos γ

So direction cosines are

l = cos 90° = 0

m = cos 135°= cos (180° – 45°) = -cos 45° = -1/2

n = cos 45° = 1/2

∴ The direction cosines of the line are 0, -1/2, 1/2

2. Find the direction cosines of a line which makes equal angles with the coordinate axes.

Solution:

Given:

Angles are equal.

So let the angles be α, β, γ

Let the direction cosines of the line be l, m and n

l = cos α, m = cos β and n = cos γ

Here given α = β = γ (Since, line makes equal angles with the coordinate axes) … (1)

The direction cosines are

l = cos α, m = cos β and n = cos γ

We have,

l2 + m 2 + n2 = 1

cos2 α + cos2β + cos2γ = 1

From (1) we have,

cos2 α + cos2 α + cos2 α = 1

3 cos2 α = 1

Cos α = ± √(1/3)

∴ The direction cosines are

l = ± √(1/3), m = ± √(1/3), n = ± √(1/3)

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

Solution:

Given

Direction ratios as -18, 12, -4

Where, a = -18, b = 12, c = -4

Let us consider the direction ratios of the line as a, b and c

Then the direction cosines are

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 1

∴ The direction cosines are

-18/22, 12/22, -4/22 => -9/11, 6/11, -2/11

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

Solution:

If the direction ratios of two lines segments are proportional, then the lines are collinear.

Given:

A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7)

Direction ratio of line joining A (2, 3, 4) and B (−1, −2, 1), are

(−1−2), (−2−3), (1−4) = (−3, −5, −3)

Where, a1 = -3, b1 = -5, c1 = -3

Direction ratio of line joining B (−1, −2, 1) and C (5, 8, 7) are

(5− (−1)), (8− (−2)), (7−1) = (6, 10, 6)

Where, a2 = 6, b2 = 10 and c2 =6

Now,

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 2

∴ A, B, 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).

Solution:

Given:

The vertices are (3, 5, –4), (-1, 1, 2) and (–5, –5, –2).

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 3

The direction cosines of the two points passing through A(x1, y1, z1) and B(x2, y2, z2) is given by (x2 – x1), (y2-y1), (z2-z1)

Firstly let us find the direction ratios of AB

Where, A = (3, 5, -4) and B = (-1, 1, 2)

Ratio of AB = [(x2 – x1)2, (y2 – y1)2, (z2 – z1)2]

= (-1-3), (1-5), (2-(-4)) = -4, -4, 6

Then by using the formula,

√[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

√[(-4)2 + (-4)2 + (6)2] = √(16+16+36)

= √68

= 2√17

Now let us find the direction cosines of the line AB

By using the formula,

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 4

-4/2√17 , -4/2√17, 6/2√17

Or -2/√17, -2/√17, 3/√17

Similarly,

Let us find the direction ratios of BC

Where, B = (-1, 1, 2) and C = (-5, -5, -2)

Ratio of AB = [(x2 – x1)2, (y2 – y1)2, (z2 – z1)2]

= (-5+1), (-5-1), (-2-2) = -4, -6, -4

Then by using the formula,

√[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

√[(-4)2 + (-6)2 + (-4)2] = √(16+36+16)

= √68

= 2√17

Now let us find the direction cosines of the line AB

By using the formula,

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 5

-4/2√17, -6/2√17, -4/2√17

Or -2/√17, -3/√17, -2/√17

Similarly,

Let us find the direction ratios of CA

Where, C = (-5, -5, -2) and A = (3, 5, -4)

Ratio of AB = [(x2 – x1)2, (y2 – y1)2, (z2 – z1)2]

= (3+5), (5+5), (-4+2) = 8, 10, -2

Then by using the formula,

√[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

√[(8)2 + (10)2 + (-2)2] = √(64+100+4)

= √168

= 2√42

Now let us find the direction cosines of the line AB

By using the formula,

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 6

8/2√42, 10/2√42, -2/2√42

Or 4/√42, 5/√42, -1/√42


EXERCISE 11.2 PAGE NO: 477

1. Show that the three lines with direction cosines

 NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 7 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 |

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 8
NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 9

So,  L1⊥ L2 …… (1)

Similarly,

Let us compute, | l2l3 + m2m3 + n2n3 |

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 10

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 11

So, L2⊥ L3 ….. (2)

Similarly,

Let us compute, | l3l1 + m3m1 + n3n1 |

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 12

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 13

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

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 14

So, a1/a2 = -2/2 = -1

b1/b2 = -4/4 = -1

c1/c2 = -4/4 = -1

∴ We can say that,

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 15

-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 NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 16.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 17

5. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 18and NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 19 is in the direction 

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 20

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

 NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 21

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

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 22

7. The Cartesian equation of a line is

 NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 23. Write its vector form.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 24

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

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 25

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 26

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 27

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 28

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 29

10. Find the angle between the following pairs of lines:
NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 30

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 31

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 32

So,

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 33

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 34

By (3), we have

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 35

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 36

11.  Find the angle between the following pair of lines:
NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 37

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 38

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 39

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 40

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 41

12. Find the values of p so that the lines

 NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 42 are at right angles.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 43

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

 NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 44 are perpendicular to each other.

Solution:

The equations of the given lines are

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 45

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
NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 46

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 47

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 48

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 49

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

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 50

∴ The shortest distance is 32/2

15. Find the shortest distance between the lines
NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 51

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 52

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 53

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 54

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 55

∴ The shortest distance is 229

16. Find the shortest distance between the lines whose vector equations are
NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 56

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 57

Here by comparing the equations we get,

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 58

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 59

∴ The shortest distance is 319

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

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 60

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 61

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 62

And,

NCERT Solutions for Class 12 Maths Chapter 11 –Three Dimensional Geometry image - 63

∴ The shortest distance is 829


EXERCISE 11.3 PAGE NO: 493

1. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z = 2

(b) x + y + z = 1

(c) 2x + 3y – z = 5

(d) 5y + 8 = 0

Solution:

(a) z = 2

Given:

The equation of the plane, z = 2 or 0x + 0y + z = 2 …. (1)

Direction ratio of the normal (0, 0, 1)

By using the formula,

[(0)2 + (0)2 + (1)2] = 1

= 1

Now,

Divide both the sides of equation (1) by 1, we get

0x/(1) + 0y/(1) + z/1 = 2

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

∴ The direction cosines are 0, 0, 1

Distance (d) from the origin is 2 units

(b) x + y + z = 1

Given:

The equation of the plane, x + y + z = 1…. (1)

Direction ratio of the normal (1, 1, 1)

By using the formula,

[(1)2 + (1)2 + (1)2] = 3

Now,

Divide both the sides of equation (1) by 3, we get

x/(3) + y/(3) + z/(3) = 1/3

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

∴ The direction cosines are 1/3, 1/3, 1/3

Distance (d) from the origin is 1/3 units

(c) 2x + 3y – z = 5

Given:

The equation of the plane, 2x + 3y – z = 5…. (1)

Direction ratio of the normal (2, 3, -1)

By using the formula,

[(2)2 + (3)2 + (-1)2] = 14

Now,

Divide both the sides of equation (1) by 14, we get

2x/(14) + 3y/(14) – z/(14) = 5/14

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

∴ The direction cosines are 2/14, 3/14, -1/14

Distance (d) from the origin is 5/14 units

(d) 5y + 8 = 0

Given:

The equation of the plane, 5y + 8 = 0

-5y = 8 or

0x – 5y + 0z = 8…. (1)

Direction ratio of the normal (0, -5, 0)

By using the formula,

[(0)2 + (-5)2 + (0)2] = 25

= 5

Now,

Divide both the sides of equation (1) by 5, we get

0x/(5) – 5y/(5) – 0z/(5) = 8/5

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

∴ The direction cosines are 0, -1, 0

Distance (d) from the origin is 8/5 units

2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector

NCERT Solutions for Class 12 Maths Chapter 11 image - 64

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 65

NCERT Solutions for Class 12 Maths Chapter 11 image - 66

3. Find the Cartesian equation of the following planes:
(a) NCERT Solutions for Class 12 Maths Chapter 11 image - 67

Solution:

Given:

The equation of the plane.

NCERT Solutions for Class 12 Maths Chapter 11 image - 68

NCERT Solutions for Class 12 Maths Chapter 11 image - 69

NCERT Solutions for Class 12 Maths Chapter 11 image - 70

NCERT Solutions for Class 12 Maths Chapter 11 image - 71

4. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
(a) 2x + 3y + 4z – 12 = 0

(b) 3y + 4z – 6 = 0

(c) x + y + z = 1

(d) 5y + 8 = 0

Solution:

(a) 2x + 3y + 4z – 12 = 0

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x, y, z).

2x + 3y + 4z = 12 …. (1)

Direction ratio are (2, 3, 4)

[(2)2 + (3)2 + (4)2] = (4 + 9 + 16)

= 29

Now,

Divide both the sides of equation (1) by 29, we get

2x/(29) + 3y/(29) + 4z/(29) = 12/29

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

∴ The direction cosines are 2/29, 3/29, 4/29

Coordinate of the foot (ld, md, nd) =

= [(2/29) (12/29), (3/29) (12/29), (4/29) (12/29)]

= 24/29, 36/29, 48/29

(b) 3y + 4z – 6 = 0

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x, y, z).

0x + 3y + 4z = 6 …. (1)

Direction ratio are (0, 3, 4)

[(0)2 + (3)2 + (4)2] = (0 + 9 + 16)

= 25

= 5

Now,

Divide both the sides of equation (1) by 5, we get

0x/(5) + 3y/(5) + 4z/(5) = 6/5

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

∴ The direction cosines are 0/5, 3/5, 4/5

Coordinate of the foot (ld, md, nd) =

= [(0/5) (6/5), (3/5) (6/5), (4/5) (6/5)]

= 0, 18/25, 24/25

(c) x + y + z = 1

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x, y, z).

x + y + z = 1 …. (1)

Direction ratio are (1, 1, 1)

[(1)2 + (1)2 + (1)2] = (1 + 1 + 1)

= 3

Now,

Divide both the sides of equation (1) by 3, we get

1x/(3) + 1y/(3) + 1z/(3) = 1/3

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

∴ The direction cosines are 1/3, 1/3, 1/3

Coordinate of the foot (ld, md, nd) =

= [(1/3) (1/3), (1/3) (1/3), (1/3) (1/3)]

= 1/3, 1/3, 1/3

(d) 5y + 8 = 0

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x, y, z).

0x – 5y + 0z = 8 …. (1)

Direction ratio are (0, -5, 0)

[(0)2 + (-5)2 + (0)2] = (0 + 25 + 0)

= 25

= 5

Now,

Divide both the sides of equation (1) by 5, we get

0x/(5) – 5y/(5) + 0z/(5) = 8/5

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

∴ The direction cosines are 0, -1, 0

Coordinate of the foot (ld, md, nd) =

= [(0/5) (8/5), (-5/5) (8/5), (0/5) (8/5)]

= 0, -8/5, 0

5. Find the vector and Cartesian equations of the planes

(a) that passes through the point (1, 0, –2) and the normal to the plane is

NCERT Solutions for Class 12 Maths Chapter 11 image - 72

(b) that passes through the point (1,4, 6) and the normal vector to the plane is

 NCERT Solutions for Class 12 Maths Chapter 11 image - 73

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 74

x – 1 – 2y + 8 + z – 6 = 0

x – 2y + z + 1 = 0

x – 2y + z = -1

∴ The required Cartesian equation of the plane is x – 2y + z = -1

NCERT Solutions for Class 12 Maths Chapter 11 image - 75

NCERT Solutions for Class 12 Maths Chapter 11 image - 76

x – 1 – 2y + 8 + z – 6 = 0

x – 2y + z + 1 = 0

x – 2y + z = -1

∴ The required Cartesian equation of the plane is x – 2y + z = -1

6. Find the equations of the planes that passes through three points.
(a) (1, 1, –1), (6, 4, –5), (–4, –2, 3)

(b) (1, 1, 0), (1, 2, 1), (–2, 2, –1)

Solution:

Given:

The points are (1, 1, -1), (6, 4, -5), (-4, -2, 3).

Let,

NCERT Solutions for Class 12 Maths Chapter 11 image - 77

= 1(12 – 10) – 1(18 – 20) -1 (-12 + 16)

= 2 + 2 – 4

= 0

Since, the value of determinant is 0.

∴ The points are collinear as there will be infinite planes passing through the given 3 points.

(b) (1, 1, 0), (1, 2, 1), (–2, 2, –1)

NCERT Solutions for Class 12 Maths Chapter 11 image - 78

NCERT Solutions for Class 12 Maths Chapter 11 image - 79

7. Find the intercepts cut off by the plane 2x + y – z = 5.

Solution:

Given:

The plane 2x + y – z = 5

Let us express the equation of the plane in intercept form

x/a + y/b + z/c = 1

Where a, b, c are the intercepts cut-off by the plane at x, y and z axes respectively.

2x + y – z = 5 …. (1)

Now divide both the sides of equation (1) by 5, we get

2x/5 + y/5 – z/5 = 5/5

2x/5 + y/5 – z/5 = 1

x/(5/2) + y/5 + z/(-5) = 1

Here, a = 5/2, b = 5 and c = -5

∴ The intercepts cut-off by the plane are 5/2, 5 and -5.

8. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Solution:

We know that the equation of the plane ZOX is y = 0

So, the equation of plane parallel to ZOX is of the form, y = a

Since the y-intercept of the plane is 3, a = 3

∴ The required equation of the plane is y = 3

9. Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

Solution:

Given:

Equation of the plane passes through the intersection of the plane is given by

(3x – y + 2z – 4) + λ (x + y + z – 2) = 0 and the plane passes through the points (2, 2, 1).

So, (3 × 2 – 2 + 2 × 1 – 4) + λ (2 + 2 + 1 – 2) = 0

2 + 3λ = 0

3λ = -2

λ = -2/3 …. (1)

Upon simplification, the required equation of the plane is given as

(3x – y + 2z – 4) – 2/3 (x + y + z – 2) = 0

(9x – 3y + 6z – 12 – 2x – 2y – 2z + 4)/3 = 0

7x – 5y + 4z – 8 = 0

∴ The required equation of the plane is 7x – 5y + 4z – 8 = 0

10. Find the vector equation of the plane passing through the intersection of the planes NCERT Solutions for Class 12 Maths Chapter 11 image - 80 and through the point (2, 1, 3).

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 81

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

NCERT Solutions for Class 12 Maths Chapter 11 image - 82

NCERT Solutions for Class 12 Maths Chapter 11 image - 83

11. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

Solution:

Let the equation of the plane that passes through the two-given planes

x + y + z = 1 and 2x + 3y + 4z = 5 is

(x + y + z – 1) + λ (2x + 3y + 4z – 5) = 0

(2λ + 1) x + (3λ + 1) y + (4λ + 1) z -1 – 5λ = 0…… (1)

So the direction ratio of the plane is (2λ + 1, 3λ + 1, 4λ + 1)

And direction ratio of another plane is (1, -1, 1)

Since, both the planes are ⊥

So by substituting in a1a2 + b1b2 + c1c2 = 0

(2λ + 1 × 1) + (3λ + 1 × (-1)) + (4λ + 1 × 1) = 0

2λ + 1 – 3λ – 1 + 4λ + 1 = 0

3λ + 1 = 0

λ = -1/3

Substitute the value of λ in equation (1) we get,

NCERT Solutions for Class 12 Maths Chapter 11 image - 84

x – z + 2 = 0

∴ The required equation of the plane is x – z + 2 = 0

12. Find the angle between the planes whose vector equations are

NCERT Solutions for Class 12 Maths Chapter 11 image - 85

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 86

NCERT Solutions for Class 12 Maths Chapter 11 image - 87

13. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

(d) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0

Solution:

(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

Given:

The equation of the given planes are

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

Two planes are ⊥ if the direction ratio of the normal to the plane is

a1a2 + b1b2 + c1c2 = 0

21 – 5 – 60

-44 ≠ 0

Both the planes are not ⊥ to each other.

Now, two planes are || to each other if the direction ratio of the normal to the plane is

NCERT Solutions for Class 12 Maths Chapter 11 image - 88

NCERT Solutions for Class 12 Maths Chapter 11 image - 89

∴ The angle is cos-1 (2/5)

(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

Given:

The equation of the given planes are

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

Two planes are ⊥ if the direction ratio of the normal to the plane is

a1a2 + b1b2 + c1c2 = 0

2 × 1 + 1 × (-2) + 3 × 0

= 0

∴ The given planes are ⊥ to each other.

(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

Given:

The equation of the given planes are

2x – 2y + 4z + 5 =0 and x – 2y + 5 = 0

We know that, two planes are ⊥ if the direction ratio of the normal to the plane is

a1a2 + b1b2 + c1c2 = 0

6 + 6 + 24

36 ≠ 0

∴ Both the planes are not ⊥ to each other.

Now let us check, both planes are || to each other if the direction ratio of the normal to the plane is

NCERT Solutions for Class 12 Maths Chapter 11 image - 90

∴ The given planes are || to each other.

(d) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

Given:

The equation of the given planes are

2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

We know that, two planes are ⊥ if the direction ratio of the normal to the plane is

a1a2 + b1b2 + c1c2 = 0

2 × 2 + (-1) × (-1) + 3 × 3

14 ≠ 0

∴ Both the planes are not ⊥ to each other.

Now, let us check two planes are || to each other if the direction ratio of the normal to the plane is

NCERT Solutions for Class 12 Maths Chapter 11 image - 91

∴ The given planes are || to each other.

(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0

Given:

The equation of the given planes are

4x + 8y + z – 8 = 0 and y + z – 4 = 0

We know that, two planes are ⊥ if the direction ratio of the normal to the plane is

a1a2 + b1b2 + c1c2 = 0

0 + 8 + 1

9 ≠ 0

∴ Both the planes are not ⊥ to each other.

Now let us check, two planes are || to each other if the direction ratio of the normal to the plane is

NCERT Solutions for Class 12 Maths Chapter 11 image - 92

∴ Both the planes are not || to each other.

Now let us find the angle between them which is given as

NCERT Solutions for Class 12 Maths Chapter 11 image - 93

∴ The angle is 45o.

14. In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0, 0, 0) 3x – 4y + 12 z = 3

(b) (3, -2, 1) 2x – y + 2z + 3 = 0

(c) (2, 3, -5) x + 2y – 2z = 9

(d) (-6, 0, 0) 2x – 3y + 6z – 2 = 0

Solution:

(a) Point Plane

(0, 0, 0) 3x – 4y + 12 z = 3

We know that, distance of point P(x1, y1, z1) from the plane Ax + By + Cz – D = 0 is given as:

NCERT Solutions for Class 12 Maths Chapter 11 image - 94

Given point is (0, 0, 0) and the plane is 3x – 4y + 12z = 3

NCERT Solutions for Class 12 Maths Chapter 11 image - 95

= |3/169|

= 3/13

∴ The distance is 3/13.

(b) Point Plane

(3, -2, 1) 2x – y + 2z + 3 = 0

We know that, distance of point P(x1, y1, z1) from the plane Ax + By + Cz – D = 0 is given as:

NCERT Solutions for Class 12 Maths Chapter 11 image - 96

Given point is (3, -2, 1) and the plane is 2x – y + 2z + 3 = 0

NCERT Solutions for Class 12 Maths Chapter 11 image - 97

= |13/9|

= 13/3

∴ The distance is 13/3.

(c) Point Plane

(2, 3, -5) x + 2y – 2z = 9

We know that, distance of point P(x1, y1, z1) from the plane Ax + By + Cz – D = 0 is given as:

NCERT Solutions for Class 12 Maths Chapter 11 image - 98

Given point is (2, 3, -5) and the plane is x + 2y – 2z = 9

NCERT Solutions for Class 12 Maths Chapter 11 image - 99

= |9/9|

= 9/3

= 3

∴ The distance is 3.

(d) Point Plane

(-6, 0, 0) 2x – 3y + 6z – 2 = 0

We know that, distance of point P(x1, y1, z1) from the plane Ax + By + Cz – D = 0 is given as:

NCERT Solutions for Class 12 Maths Chapter 11 image - 100

Given point is (-6, 0, 0) and the plane is 2x – 3y + 6z – 2 = 0

NCERT Solutions for Class 12 Maths Chapter 11 image - 101

= |14/49|

= 14/7

= 2

∴ The distance is 2.


Miscellaneous EXERCISE PAGE NO: 497

1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, –1), (4, 3, –1).

Solution:

Let us consider OA be the line joining the origin (0, 0, 0) and the point A (2, 1, 1).

And let BC be the line joining the points B (3, 5, −1) and C (4, 3, −1)

So the direction ratios of OA = (a1, b1, c1) ≡ [(2 – 0), (1 – 0), (1 – 0)] ≡ (2, 1, 1)

And the direction ratios of BC = (a2, b2, c2) ≡ [(4 – 3), (3 – 5), (-1 + 1)] ≡ (1, -2, 0)

Given:

OA is ⊥ to BC

Now we have to prove that:

a1a2 + b1b2 + c1c2 = 0

Let us consider LHS: a1a2 + b1b2 + c1c2

a1a2 + b1b2 + c1c2 = 2 × 1 + 1 × (−2) + 1 × 0

= 2 – 2

= 0

We know that R.H.S is 0

So LHS = RHS

∴ OA is ⊥ to BC

Hence proved.

2. If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are (m1n2 – m2n1), (n1l2 – n2l1), (l1m2 – l2m1)

Solution:

Let us consider l, m, n be the direction cosines of the line perpendicular to each of the given lines.

Then, ll1 + mm1 + nn1 = 0 … (1)

And ll2 + mm2 + nn2 = 0 … (2)

Upon solving (1) and (2) by using cross – multiplication, we get

NCERT Solutions for Class 12 Maths Chapter 11 image - 102

Thus, the direction cosines of the given line are proportional to

(m1n2 – m2n1), (n1l2 – n2l1), (l1m2 – l2m1)

So, its direction cosines are

NCERT Solutions for Class 12 Maths Chapter 11 image - 103

NCERT Solutions for Class 12 Maths Chapter 11 image - 104

We know that

(l12 + m12 + n12) (l22 + m22 + n22) – (l1l2 + m1m2 + n1n2)2

= (m1n2 – m2n1)2 + (n1l2 – n2l1)2 + (l1m2 – l2m1)2 … (3)

It is given that the given lines are perpendicular to each other.

So, l1l2 + m1m2 + n1n2 = 0

Also, we have

l12 + m12 + n12 = 1

And, l22 + m22 + n22 = 1

Substituting these values in equation (3), we get

(m1n2 – m2n1)2 + (n1l2 – n2l1)2 + (l1m2 – l2m1)2 = 1

λ = 1

Hence, the direction cosines of the given line are (m1n2 – m2n1), (n1l2 – n2l1), (l1m2 – l2m1)

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

Solution:

Angle between the lines with direction ratios a1, b1, c1 and a2, b2, c2 is given by

NCERT Solutions for Class 12 Maths Chapter 11 image - 105

Given:

a1 = a, b1 = b, c1 = c

a2 = b – c, b2 = c – a, c2 = a – b

Let us substitute the values in the above equation we get,

NCERT Solutions for Class 12 Maths Chapter 11 image - 106

= 0

Cos θ = 0

So, θ = 90° [Since, cos 90 = 0]

Hence, Angle between the given pair of lines is 90°.

4. Find the equation of a line parallel to x – axis and passing through the origin.

Solution:

We know that, equation of a line passing through (x1, y1, z1) and parallel to a line with direction ratios a, b, c is

NCERT Solutions for Class 12 Maths Chapter 11 image - 107

Given: the line passes through origin i.e. (0, 0, 0)

x1 = 0, y1 = 0, z1 = 0

Since line is parallel to x – axis,

a = 1, b = 0, c = 0

∴ Equation of Line is given by

NCERT Solutions for Class 12 Maths Chapter 11 image - 108

5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (–4, 3, –6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Solution:

We know that the angle between the lines with direction ratios a1, b1, c1 and a2, b2, c2 is given by

NCERT Solutions for Class 12 Maths Chapter 11 image - 109

So now, a line passing through A (x1, y1, z1) and B (x2, y2, z2) has direction ratios (x1 – x2), (y1 – y2), (z1 – z2)

The direction ratios of line joining the points A (1, 2, 3) and B (4, 5, 7)

= (4 – 1), (5 – 2), (7 – 3)

= (3, 3, 4)

∴ a1 = 3, b1 = 3, c1 = 4

The direction ratios of line joining the points C (-4, 3, -6) and B (2, 9, 2)

= (2 – (-4)), (9 – 3), (2-(-6))

= (6, 6, 8)

∴ a2 = 6, b2 = 6, c2 = 8

Now let us substitute the values in the above equation we get,

NCERT Solutions for Class 12 Maths Chapter 11 image - 110

6. If the lines 

NCERT Solutions for Class 12 Maths Chapter 11 image - 111 and 
NCERT Solutions for Class 12 Maths Chapter 11 image - 112 are perpendicular, find the value of k.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 113

NCERT Solutions for Class 12 Maths Chapter 11 image - 114

We get –

x2 = 1, y2 = 2, z2 = 3

And a2 = 3k, b2 = 1, c2 = -5

Since the two lines are perpendicular,

a1a2 + b1b2 + c1c2 = 0

(-3) × 3k + 2k × 1 + 2 × (-5) = 0

-9k + 2k – 10 = 0

-7k = 10

k = -10/7

∴ The value of k is -10/7.

7. Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane

NCERT Solutions for Class 12 Maths Chapter 11 image - 115

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 116

NCERT Solutions for Class 12 Maths Chapter 11 image - 117

8. Find the equation of the plane passing through (a, b, c) and parallel to the plane

 NCERT Solutions for Class 12 Maths Chapter 11 image - 118

Solution:

The equation of a plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is given as

A (x – x1) + B (y – y1) + C (z – z1) = 0

It is given that, the plane passes through (a, b, c)

So, x1 = a, y1 = b, z1 = c

Since both planes are parallel to each other, their normal will be parallel

NCERT Solutions for Class 12 Maths Chapter 11 image - 119

Direction ratios of normal = (1, 1, 1)

So, A = 1, B =1, C = 1

The Equation of plane in Cartesian form is given as

A (x – x1) + B (y – y1) + C (z – z1) = 0

1(x – a) + 1(y – b) + 1(z – c) = 0

x + y + z – (a + b + c) = 0

x + y + z = a + b + c

∴ The required equation of plane is x + y + z = a + b + c

9. Find the shortest distance between lines
NCERT Solutions for Class 12 Maths Chapter 11 image - 120 and NCERT Solutions for Class 12 Maths Chapter 11 image - 121

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 122

NCERT Solutions for Class 12 Maths Chapter 11 image - 123

NCERT Solutions for Class 12 Maths Chapter 11 image - 124

10. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ – plane.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 125

NCERT Solutions for Class 12 Maths Chapter 11 image - 126

We know that, two vectors are equal if their corresponding components are equal

So,

0 = 5 – 2λ

5 = 2λ

λ = 5/2

y = 1 + 3λ … (5)

And,

z = 6 – 5λ … (6)

Substitute the value of λ in equation (5) and (6), we get –

y = 1 + 3λ

= 1 + 3 × (5/2)

= 1 + (15/2)

= 17/2

And

z = 6 – 5λ

= 6 – 5 × (5/2)

= 6 – (25/2)

= – 13/2

∴ The coordinates of the required point is (0, 17/2, -13/2).

11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX – plane.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 127

NCERT Solutions for Class 12 Maths Chapter 11 image - 128

We know that, two vectors are equal if their corresponding components are equal

So,

x = 5 – 2λ … (5)

0 = 1 + 3λ

-1 = 3λ

λ = -1/3

And,

z = 6 – 5λ … (6)

Substitute the value of λ in equation (5) and (6), we get –

x = 5 – 2λ

= 5 – 2 × (-1/3)

= 5 + (2/3)

= 17/3

And

z = 6 – 5λ

= 6 – 5 × (-1/3)

= 6 + (5/3)

= 23/3

∴ The coordinates of the required point is (17/3, 0, 23/3).

12. Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane 2x + y + z = 7.

Solution:

We know that the equation of a line passing through two points A (x1, y1, z1) and B (x2, y2, z2) is given as

NCERT Solutions for Class 12 Maths Chapter 11 image - 129

It is given that the line passes through the points A (3, –4, –5) and B (2, –3, 1)

So, x1 = 3, y1 = -4, z1 = -5

And, x2 = 2, y2 = -3, z2 = 1

Then the equation of line is

NCERT Solutions for Class 12 Maths Chapter 11 image - 130

So, x = -k + 3 |, y = k – 4 |, z = 6k – 5 … (1)

Now let (x, y, z) be the coordinates of the point where the line crosses the given plane 2x + y + z + 7 = 0

By substituting the value of x, y, z in equation (1) in the equation of plane, we get

2x + y + z + 7 = 0

2(-k + 3) + (k – 4) + (6k – 5) = 7

5k – 3 = 7

5k = 10

k = 2

Now substitute the value of k in x, y, z we get,

x = – k + 3 = – 2 + 3 = 1

y = k – 4 = 2 – 4 = – 2

z = 6k – 5 = 12 – 5 = 7

∴ The coordinates of the required point are (1, -2, 7).

13. Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Solution:

We know that the equation of a plane passing through (x1, y1, z1) is given by

A (x – x1) + B (y – y1) + C (z – z1) = 0

Where, A, B, C are the direction ratios of normal to the plane.

It is given that the plane passes through (-1, 3, 2)

So, equation of plane is given by

A (x + 1) + B (y – 3) + C (z – 2) = 0 ……… (1)

Since this plane is perpendicular to the given two planes. So, their normal to the plane would be perpendicular to normal of both planes.

We know that

NCERT Solutions for Class 12 Maths Chapter 11 image - 131

So, required normal is cross product of normal of planes

x + 2y + 3z = 5 and 3x + 3y + z = 0

NCERT Solutions for Class 12 Maths Chapter 11 image - 132

Hence, the direction ratios are = -7, 8, -3

∴ A = -7, B = 8, C = -3

Substituting the obtained values in equation (1), we get

A (x + 1) + B (y – 3) + C (z – 2) = 0

-7(x + 1) + 8(y – 3) + (-3) (z – 2) = 0

-7x – 7 + 8y – 24 – 3z + 6 = 0

-7x + 8y – 3z – 25 = 0

7x – 8y + 3z + 25 = 0

∴ The equation of the required plane is 7x – 8y + 3z + 25 = 0.

14. If the points (1, 1, p) and (–3, 0, 1) be equidistant from the plane

 NCERT Solutions for Class 12 Maths Chapter 11 image - 133, then find the value of p.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 134

NCERT Solutions for Class 12 Maths Chapter 11 image - 135

20 – 12p = ± 8

20 – 12p = 8 or, 20 – 12p = -8

12p = 12 or, 12p = 28

p = 1 or, p = 7/3

∴ The possible values of p are 1 and 7/3.

15. Find the equation of the plane passing through the line of intersection of the planes NCERT Solutions for Class 12 Maths Chapter 11 image - 136 and NCERT Solutions for Class 12 Maths Chapter 11 image - 137 and parallel to x-axis.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 138

Since this plane is parallel to x-axis.

So, the normal vector of the plane (1) will be perpendicular to x-axis.

The direction ratios of Normal (a1, b1, c1) ≡ [(1 – 2λ), (1 – 3λ), (1 +)]

The direction ratios of x–axis (a2, b2, c2) ≡ (1, 0, 0)

Since the two lines are perpendicular,

a1a2 + b1b2 + c1c2 = 0

(1 – 2λ) × 1 + (1 – 3λ) × 0 + (1 + λ) × 0 = 0

(1 – 2λ) = 0

λ = 1/2

Substituting the value of λ in equation (1), we get

NCERT Solutions for Class 12 Maths Chapter 11 image - 139

16. If O be the origin and the coordinates of P be (1, 2, –3), then find the equation of the plane passing through P and perpendicular to OP.

Solution:

We know that the equation of a plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is given as

A(x – x1) + B(y – y1) + C (z – z1) = 0

It is given that the plane passes through P (1, 2, 3)

So, x1 = 1, y1 = 2, z1 = – 3

Normal vector to plane is =
NCERT Solutions for Class 12 Maths Chapter 11 image - 140

Where O (0, 0, 0), P (1, 2, -3)

So, direction ratios of
NCERT Solutions for Class 12 Maths Chapter 11 image - 141is = (1 – 0), (2 – 0), (-3 – 0)

= (1, 2, – 3)

Where, A = 1, B = 2, C = -3

Equation of plane in Cartesian form is given as

1(x – 1) + 2(y – 2) – 3(z – (-3)) = 0

x – 1 + 2y – 4 – 3z – 9 = 0

x + 2y – 3z – 14 = 0

∴ The equation of the required plane is x + 2y – 3z – 14 = 0

NCERT Solutions for Class 12 Maths Chapter 11 image - 142

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 143

Since this plane is perpendicular to the plane

NCERT Solutions for Class 12 Maths Chapter 11 image - 144

So, the normal vector of the plane (1) will be perpendicular to the normal vector of plane (2).

Direction ratios of Normal of plane (1) = (a1, b1, c1) ≡ [(1 – 2λ), (2 – λ), (3 + λ)]

Direction ratios of Normal of plane (2) = (a2, b2, c2) ≡ (-5, -3, 6)

Since the two lines are perpendicular,

a1a2 + b1b2 + c1c2 = 0

(1 – 2λ) × (-5) + (2 – λ) × (-3) + (3 + λ) × 6 = 0

-5 + 10λ – 6 + 3λ + 18 + 6λ = 0

19λ + 7 = 0

λ = -7/19

By substituting the value of λ in equation (1), we get

NCERT Solutions for Class 12 Maths Chapter 11 image - 145

18. Find the distance of the point (–1, –5, –10) from the point of intersection of the line 

NCERT Solutions for Class 12 Maths Chapter 11 image - 146

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 147

Where,

x = 2, y = -1, z = 2

So, the point of intersection is (2, -1, 2).

Now, the distance between points (x1, y1, z1) and (x2, y2, z2) is given by

NCERT Solutions for Class 12 Maths Chapter 11 image - 148

NCERT Solutions for Class 12 Maths Chapter 11 image - 149

∴ The distance is 13 units.

NCERT Solutions for Class 12 Maths Chapter 11 image - 150

NCERT Solutions for Class 12 Maths Chapter 11 image - 151

NCERT Solutions for Class 12 Maths Chapter 11 image - 152

NCERT Solutions for Class 12 Maths Chapter 11 image - 153

20. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:
NCERT Solutions for Class 12 Maths Chapter 11 image - 154 and NCERT Solutions for Class 12 Maths Chapter 11 image - 155.

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 156

NCERT Solutions for Class 12 Maths Chapter 11 image - 157

21. Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then

NCERT Solutions for Class 12 Maths Chapter 11 image - 158

Solution:

NCERT Solutions for Class 12 Maths Chapter 11 image - 159

NCERT Solutions for Class 12 Maths Chapter 11 image - 160

NCERT Solutions for Class 12 Maths Chapter 11 image - 161

22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
A. 2 units
B. 4 units
C. 8 units

D. 2/√29 units 

Solution:

We know that the distance between two parallel planes Ax + By + Cz = d1 and Ax + By + Cz = d2 is given as

NCERT Solutions for Class 12 Maths Chapter 11 image - 162

It is given that:

First Plane:

2x + 3y + 4z = 4

Let us compare with Ax + By + Cz = d1, we get

A = 2, B = 3, C = 4, d1 = 4

Second Plane:

4x + 6y + 8z = 12 [Divide the equation by 2]

We get,

2x + 3y + 4z = 6

Now comparing with Ax + By + Cz = d1, we get

A = 2, B = 3, C = 4, d2 = 6

So,

Distance between two planes is given as

NCERT Solutions for Class 12 Maths Chapter 11 image - 163

= 2/√29

∴ Option (D) is the correct option.

23. The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
A. Perpendicular
B. Parallel
C. intersect y–axis
D. passes through

Solution:

It is given that:

First Plane:

2x – y + 4z = 5 [Multiply both sides by 2.5]

We get,

5x – 2.5y + 10z = 12.5 … (1)

Given second Plane:

5x – 2.5y + 10z = 6 … (2)

So,

NCERT Solutions for Class 12 Maths Chapter 11 image - 164

It is clear that the direction ratios of normal of both the plane (1) and (2) are same.

∴ Both the given planes are parallel.


The major concepts of Maths covered in Chapter 11- Three Dimensional Geometry of NCERT Solutions for Class 12 includes:

11.1 Introduction

11.2 Direction Cosines and Direction Ratios of a Line

11.2.1 Relation between the direction cosines of a line

11.2.2 Direction cosines of a line passing through two points

11.3 Equation of a Line in Space

11.3.1 Equation of a line through a given point and parallel to a given vector b.

11.3.2 Equation of a line passing through two given points

11.4 Angle between Two Lines

11.5 Shortest Distance between Two Lines

11.5.1 Distance between two skew lines

11.5.2 Distance between parallel lines

11.6 Plane

11.6.1 Equation of a plane in normal form

11.6.2 Equation of a plane perpendicular to a given vector, passing through a given point

11.6.3 Equation of a plane passing through three non collinear points

11.6.4 Intercept form of the equation of a plane

11.6.5 Plane passing through the intersection of two given planes

11.7 Coplanarity of Two Lines

11.8 Angle between Two Planes

11.9 Distance of a Point from a Plane

11.10 Angle between a Line and a Plane

NCERT Solutions for Class 12 Maths Chapter 11- Three Dimensional Geometry

The chapter Three Dimensional Geometry belongs to the unit Vectors and Three – Dimensional Geometry, that adds up to 14 marks of the total marks. There are 3 exercises along with a miscellaneous exercise in this chapter to help students understand the concepts related to Three Dimensional Geometry clearly. Some of the topics discussed in Chapter 11 of NCERT Solutions for Class 12 Maths are as follows:

  1. Direction cosines of a line are the cosines of the angles made by the line with the positive directions of the coordinate axes.
  2. If l, m, n are the direction cosines of a line, then l2 + m2 + n2 = 1
  3. Direction ratios of a line are the numbers which are proportional to the direction cosines of a line.
  4. Skew lines are lines in space which are neither parallel nor intersecting. They lie in different planes.
  5. Angle between skew lines is the angle between two intersecting lines drawn from any point (preferably through the origin) parallel to each of the skew lines.
  6. 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 cosθ = |l1l2 + m1m2 + n1n2|

These are very few topics that are discussed in the chapter three dimensional geometry. To know more about the chapter, refer the NCERT Textbook of Class 12 Maths.

Key Features of NCERT Solutions for Class 12 Maths Chapter 11- Three Dimensional Geometry

Studying the Three Dimensional Geometry of Class 12 enables the students to understand the following:

Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a plane. Angle between (i) two lines, (ii) two planes, (iii) a line and a plane. Distance of a point from a plane.

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