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

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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry enables the students to solve problems in a dynamic way in less time. The NCERT Solutions are designed by highly experienced teachers according to the latest CBSE Syllabus 2023-24, making the students solve every problem easily. The students can refer to the NCERT Solutions for Class 11 Maths Chapter 12 PDF from the link below for optimum preparation for the annual exam.

### Access Answers to NCERT Solutions Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry

Exercise 12.1, Page No. 271

1. A point is on the x-axis. What are its y-coordinate and z-coordinates?

Solution:

If a point is on the x-axis, then the coordinates of y and z are 0.

So the point is (x, 0, 0).

2. A point is in the XZ-plane. What can you say about its y-coordinate?

Solution:

If a point is in the XZ plane, then its y-co-ordinate is 0.

3. Name the octants in which the following points lie:
(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (2, –4, –7).

Solution:

Here is the table which represents the octants:

 Octants I II III IV V VI VII VIII x + – – + + – – + y + + – – + + – – z + + + + – – – –

(i) (1, 2, 3)

Here, x is positive, y is positive, and z is positive.

So, it lies in the I octant.

(ii) (4, -2, 3)

Here, x is positive, y is negative, and z is positive.

So, it lies in the IV octant.

(iii) (4, -2, -5)

Here, x is positive, y is negative, and z is negative.

So, it lies in the VIII octant.

(iv) (4, 2, -5)

Here, x is positive, y is positive, and z is negative.

So, it lies in the V octant.

(v) (-4, 2, -5)

Here, x is negative, y is positive, and z is negative.

So, it lies in VI octant.

(vi) (-4, 2, 5)

Here, x is negative, y is positive, and z is positive.

So, it lies in the II octant.

(vii) (-3, -1, 6)

Here, x is negative, y is negative, and z is positive.

So, it lies in the III octant.

(viii) (2, -4, -7)

Here, x is positive, y is negative, and z is negative.

So, it lies in the VIII octant.

4. Fill in the blanks:
(i) The x-axis and y-axis, taken together, determine a plane known as _______.
(ii) The coordinates of points in the XY-plane are of the form _______.
(iii) Coordinate planes divide the space into ______ octants.

Solution:

(i) The x-axis and y-axis, taken together, determine a plane known as XY Plane.

(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).

(iii) Coordinate planes divide the space into eight octants.

Exercise 12.2 Page No. 273

1. Find the distance between the following pairs of points:

(i) (2, 3, 5) and (4, 3, 1)

(ii) (–3, 7, 2) and (2, 4, –1)

(iii) (–1, 3, –4) and (1, –3, 4)

(iv) (2, –1, 3) and (–2, 1, 3)

Solution:

(i) (2, 3, 5) and (4, 3, 1)

Let P be (2, 3, 5) and Q be (4, 3, 1)

By using the formula,

Distance PQ = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 2, y1 = 3, z1 = 5

x2 = 4, y2 = 3, z2 = 1

Distance PQ = [(4 – 2)2 + (3 – 3)2 + (1 – 5)2]

= [(2)2 + 02 + (-4)2]

= [4 + 0 + 16]

= √20

= 25

∴ The required distance is 25 units.

(ii) (–3, 7, 2) and (2, 4, –1)

Let P be (– 3, 7, 2) and Q be (2, 4, –1)

By using the formula,

Distance PQ = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = – 3, y1 = 7, z1 = 2

x2 = 2, y2 = 4, z2 = – 1

Distance PQ = [(2 – (-3))2 + (4 – 7)2 + (-1 – 2)2]

= [(5)2 + (-3)2 + (-3)2]

= [25 + 9 + 9]

= √43

∴ The required distance is 43 units.

(iii) (–1, 3, –4) and (1, –3, 4)

Let P be (– 1, 3, – 4) and Q be (1, – 3, 4)

By using the formula,

Distance PQ = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = – 1, y1 = 3, z1 = – 4

x2 = 1, y2 = – 3, z2 = 4

Distance PQ = [(1 – (-1))2 + (-3 – 3)2 + (4 – (-4))2]

= [(2)2 + (-6)2 + (8)2]

= [4 + 36 + 64]

= √104

= 226

∴ The required distance is 226 units.

(iv) (2, –1, 3) and (–2, 1, 3)

Let P be (2, – 1, 3) and Q be (– 2, 1, 3)

By using the formula,

Distance PQ = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 2, y1 = – 1, z1 = 3

x2 = – 2, y2 = 1, z2 = 3

Distance PQ = [(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]

= [(-4)2 + (2)2 + (0)2]

= [16 + 4 + 0]

= √20

= 25

∴ The required distance is 25 units.

2. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Solution:

If three points are collinear, then they lie on the same line.

First, let us calculate the distance between the 3 points

i.e., PQ, QR and PR

Calculating PQ

P ≡ (–2, 3, 5) and Q ≡ (1, 2, 3)

By using the formula,

Distance PQ = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = – 2, y1 = 3, z1 = 5

x2 = 1, y2 = 2, z2 = 3

Distance PQ = [(1 – (-2))2 + (2 – 3)2 + (3 – 5)2]

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

= [9 + 1 + 4]

= √14

Calculating QR

Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)

By using the formula,

Distance QR = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 1, y1 = 2, z1 = 3

x2 = 7, y2 = 0, z2 = – 1

Distance QR = [(7 – 1)2 + (0 – 2)2 + (-1 – 3)2]

= [(6)2 + (-2)2 + (-4)2]

= [36 + 4 + 16]

= √56

= 214

Calculating PR

P ≡ (–2, 3, 5) and R ≡ (7, 0, – 1)

By using the formula,

Distance PR = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = –2, y1 = 3, z1 = 5

x2 = 7, y2 = 0, z2 = –1

Distance PR = [(7 – (-2))2 + (0 – 3)2 + (-1 – 5)2]

= [(9)2 + (-3)2 + (-6)2]

= [81 + 9 + 36]

= √126

= 314

Thus, PQ = 14, QR = 214 and PR = 314

So, PQ + QR = 14 + 214

= 314

= PR

∴ The points P, Q and R are collinear.

3. Verify the following:
(i) (0, 7, –10), (1, 6, –6), and (4, 9, –6) are the vertices of an isosceles triangle.

(ii) (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) are the vertices of a right-angled triangle.

(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) are the vertices of a parallelogram.

Solution:

(i) (0, 7, –10), (1, 6, –6), and (4, 9, – 6) are the vertices of an isosceles triangle.

Let us consider the points,

P(0, 7, –10), Q(1, 6, –6) and R(4, 9, –6)

If any 2 sides are equal, it will be an isosceles triangle

So, first, let us calculate the distance of PQ, QR

Calculating PQ

P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)

By using the formula,

Distance PQ = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here,

x1 = 0, y1 = 7, z1 = – 10

x2 = 1, y2 = 6, z2 = – 6

Distance PQ = [(1 – 0)2 + (6 – 7)2 + (-6 – (-10))2]

= [(1)2 + (-1)2 + (4)2]

= [1 + 1 + 16]

= √18

Calculating QR

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

By using the formula,

Distance QR = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 1, y1 = 6, z1 = – 6

x2 = 4, y2 = 9, z2 = – 6

Distance QR = [(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]

= [(3)2 + (3)2 + (-6+6)2]

= [9 + 9 + 0]

= √18

Hence, PQ = QR

18 = 18

2 sides are equal

∴ PQR is an isosceles triangle.

(ii) (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) are the vertices of a right-angled triangle.

Let the points be

P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)

First, let us calculate the distance of PQ, OR and PR

Calculating PQ

P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)

By using the formula,

Distance PQ = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 0, y1 = 7, z1 = 10

x2 = –1, y2 = 6, z2 = 6

Distance PQ = [(-1 – 0)2 + (6 – 7)2 + (6 – 10)2]

= [(-1)2 + (-1)2 + (-4)2]

= [1 + 1 + 16]

= √18

Calculating QR

Q ≡ (1, 6, –6) and R ≡ (4, 9, –6)

By using the formula,

Distance QR = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 1, y1 = 6, z1 = –6

x2 = 4, y2 = 9, z2 = –6

Distance QR = [(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]

= [(3)2 + (3)2 + (-6+6)2]

= [9 + 9 + 0]

= √18

Calculating PR

P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

By using the formula,

Distance PR = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 0, y1 = 7, z1 = 10

x2 = – 4, y2 = 9, z2 = 6

Distance PR = [(-4 – 0)2 + (9 – 7)2 + (6 – 10)2]

= [(-4)2 + (2)2 + (-4)2]

= [16 + 4 + 16]

= √36

Now,

PQ2 + QR2 = 18 + 18

= 36

= PR2

By using the converse of Pythagoras theorem,

∴ The given vertices P, Q & R are the vertices of a right–angled triangle at Q.

(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) are the vertices of a parallelogram.

Let the points: A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) & D(2, –3, 4)

ABCD can be vertices of parallelogram only if opposite sides are equal.

i.e., AB = CD and BC = AD

First, let us calculate the distance

Calculating AB

A ≡ (–1, 2, 1) and B ≡ (1, –2, 5)

By using the formula,

Distance AB = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = – 1, y1 = 2, z1 = 1

x2 = 1, y2 = – 2, z2 = 5

Distance AB = [(1 – (-1))2 + (-2 – 2)2 + (5 – 1)2]

= [(2)2 + (-4)2 + (4)2]

= [4 + 16 + 16]

= √36

= 6

Calculating BC

B ≡ (1, –2, 5) and C ≡ (4, –7, 8)

By using the formula,

Distance BC = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 1, y1 = – 2, z1 = 5

x2 = 4, y2 = – 7, z2 = 8

Distance BC = [(4 – 1)2 + (-7 – (-2))2 + (8 – 5)2]

= [(3)2 + (-5)2 + (3)2]

= [9 + 25 + 9]

= √43

Calculating CD

C ≡ (4, –7, 8) and D ≡ (2, –3, 4)

By using the formula,

Distance CD = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 4, y1 = –7, z1 = 8

x2 = 2, y2 = –3, z2 = 4

Distance CD = [(2 – 4)2 + (-3 – (-7))2 + (4 – 8)2]

= [(-2)2 + (4)2 + (-4)2]

= [4 + 16 + 16]

= √36

= 6

Calculating DA

D ≡ (2, –3, 4) and A ≡ (–1, 2, 1)

By using the formula,

Distance DA = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 2, y1 = – 3, z1 = 4

x2 = –1, y2 = 2, z2 = 1

Distance DA = [(-1 – 2)2 + (2 – (-3))2 + (1 – 4)2]

= [(-3)2 + (5)2 + (-3)2]

= [9 + 25 + 9]

= √43

Since AB = CD and BC = DA (given),

In ABCD, both pairs of opposite sides are equal.

∴ ABCD is a parallelogram.

4. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Solution:

Let A (1, 2, 3) & B (3, 2, –1)

Let point P be (x, y, z)

Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, 2, –1)

i.e. PA = PB

First, let us calculate

Calculating PA

P ≡ (x, y, z) and A ≡ (1, 2, 3)

By using the formula,

Distance PA = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = x, y1 = y, z1 = z

x2 = 1, y2 = 2, z2 = 3

Distance PA = [(1 – x)2 + (2 – y)2 + (3 – z)2]

Calculating PB

P ≡ (x, y, z) and B ≡ (3, 2, –1)

By using the formula,

Distance PB = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = x, y1 = y, z1 = z

x2 = 3, y2 = 2, z2 = –1

Distance PB = [(3 – x)2 + (2 – y)2 + (-1 – z)2]

Since PA = PB

Square on both sides, we get

PA2 = PB2

(1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (2 – y)2 + (– 1 – z)2

(1 + x2 – 2x) + (4 + y2 – 4y) + (9 + z2 – 6z)

(9 + x2 – 6x) + (4 + y2 – 4y) + (1 + z2 + 2z)

– 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14

4x – 8z = 0

x – 2z = 0

∴ The required equation is x – 2z = 0.

5. Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(–4, 0, 0) is equal to 10.

Solution:

Let A(4, 0, 0) & B(– 4, 0, 0)

Let the coordinates of point P be (x, y, z)

Calculating PA

P ≡ (x, y, z) and A ≡ (4, 0, 0)

By using the formula,

Distance PA = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = x, y1 = y, z1 = z

x2 = 4, y2 = 0, z2 = 0

Distance PA = [(4– x)2 + (0 – y)2 + (0 – z)2]

Calculating PB,

P ≡ (x, y, z) and B ≡ (– 4, 0, 0)

By using the formula,

Distance PB = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = x, y1 = y, z1 = z

x2 = – 4, y2 = 0, z2 = 0

Distance PB = [(-4– x)2 + (0 – y)2 + (0 – z)2]

It is given that,

PA + PB = 10

PA = 10 – PB

Square on both sides, we get

PA2 = (10 – PB)2

PA2 = 100 + PB2 – 20 PB

(4 – x)2 + (0 – y)2 + (0 – z)2

100 + (– 4 – x)2 + (0 – y)2 + (0 – z)2 – 20 PB

(16 + x2 – 8x) + (y2) + (z2)

100 + (16 + x2 + 8x) + (y2) + (z2) – 20 PB

20 PB = 16x + 100

5 PB = (4x + 25)

Square on both sides again, we get

25 PB2 = 16x2 + 200x + 625

25 [(– 4 – x)2 + (0 – y)2 + (0 – z)2] = 16x2 + 200x + 625

25 [x2 + y2 + z2 + 8x + 16] = 16x2 + 200x + 625

25x2 + 25y2 + 25z2 + 200x + 400 = 16x2 + 200x + 625

9x2 + 25y2 + 25z2 – 225 = 0

∴ The required equation is 9x2 + 25y2 + 25z2 – 225 = 0.

Exercise 12.3 Page No. 277

1. Find the coordinates of the point which divides the line segment joining the points (–2, 3, 5) and (1, –4, 6) in the ratio (i) 2:3 internally, (ii) 2:3 externally.

Solution:

Let the line segment joining the points P (-2, 3, 5) and Q (1, -4, 6) be PQ.

(i) 2:3 internally

By using the section formula,

We know that the coordinates of the point R, which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m:n, is given by:

Upon comparing, we have

x1 = -2, y1 = 3, z1 = 5;

x2 = 1, y2 = -4, z2 = 6 and

m = 2, n = 3

So, the coordinates of the point which divide the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2:3 internally is given by:

Hence, the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-4/5, 1/5, 27/5).

(ii) 2:3 externally

By using the section formula,

We know that the coordinates of the point R, which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) externally in the ratio m:n, is given by:

Upon comparing, we have

x1 = -2, y1 = 3, z1 = 5;

x2 = 1, y2 = -4, z2 = 6 and

m = 2, n = 3

So, the coordinates of the point which divide the line segment joining the points P (–2, 3, 5) and Q (1, –4, 6) in the ratio 2:3 externally is given by:

∴ The coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-8, 17, 3).

2. Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Solution:

Let us consider Q divides PR in the ratio k:1.

By using the section formula,

We know that the coordinates of the point R, which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m:n, is given by:

Upon comparing, we have,

x1 = 3, y1 = 2, z1 = -4;

x2 = 9, y2 = 8, z2 = -10 and

m = k, n = 1

So, we have

9k + 3 = 5 (k+1)

9k + 3 = 5k + 5

9k – 5k = 5 – 3

4k = 2

k = 2/4

= ½

Hence, the ratio in which Q divides PR is 1:2.

3. Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).

Solution:

Let the line segment formed by joining the points P (-2, 4, 7) and Q (3, -5, 8) be PQ.

We know that any point on the YZ-plane is of the form (0, y, z).

So, let R (0, y, z) divides the line segment PQ in the ratio k:1.

Then,

Upon comparing, we have,

x1 = -2, y1 = 4, z1 = 7;

x2 = 3, y2 = -5, z2 = 8 and

m = k, n = 1

By using the section formula,

We know that the coordinates of the point R, which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m:n, is given by:

So we have,

3k – 2 = 0

3k = 2

k = 2/3

Hence, the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8) is 2:3.

4. Using the section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2) are collinear.

Solution:

Let point P divides AB in the ratio k:1.

Upon comparing, we have,

x1 = 2, y1 = -3, z1 = 4;

x2 = -1, y2 = 2, z2 = 1 and

m = k, n = 1

By using the section formula,

We know that the coordinates of the point R, which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m:n, is given by:

So we have,

Now, we check if, for some value of k, the point coincides with point C.

Put (-k+2)/(k+1) = 0

-k + 2 = 0

k = 2

When k = 2, then (2k-3)/(k+1) = (2(2)-3)/(2+1)

= (4-3)/3

= 1/3

And, (k+4)/(k+1) = (2+4)/(2+1)

= 6/3

= 2

∴ C (0, 1/3, 2) is a point which divides AB in the ratio 2:1 and is the same as P.

Hence, A, B, and C are collinear.

5. Find the coordinates of the points which trisect the line segment joining the points P (4, 2, –6) and Q (10, –16, 6).

Solution:

Let A (x1, y1, z1) and B (x2, y2, z2) trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6).

A divides the line segment PQ in the ratio 1:2.

Upon comparing, we have,

x1 = 4, y1 = 2, z1 = -6;

x2 = 10, y2 = -16, z2 = 6 and

m = 1, n = 2

By using the section formula,

We know that the coordinates of the point R, which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m:n, is given by:

So, we have

Similarly, we know that B divides the line segment PQ in the ratio 2:1.

Upon comparing, we have,

x1 = 4, y1 = 2, z1 = -6;

x2 = 10, y2 = -16, z2 = 6 and

m = 2, n = 1

By using the section formula,

We know that the coordinates of the point R, which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m:n, is given by:

So, we have

∴ The coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6) are (6, -4, -2) and (8, -10, 2).

Miscellaneous Exercise Page No. 278

1. Three vertices of a parallelogram ABCD are A(3, – 1, 2), B(1, 2, – 4) and C(– 1, 1, 2). Find the coordinates of the fourth vertex.

Solution:

Given:

ABCD is a parallelogram with vertices A (3, -1, 2), B (1, 2, -4), and C (-1, 1, 2).

Where, x1 = 3, y1 = -1, z1 = 2;

x2 = 1, y2 = 2, z2 = -4;

x3 = -1, y3 = 1, z3 = 2

Let the coordinates of the fourth vertex be D (x, y, z).

We know that the diagonals of a parallelogram bisect each other, so the midpoints of AC and BD are equal, i.e., Midpoint of AC = Midpoint of BD ……….(1)

Now, by the midpoint formula, we know that the coordinates of the mid-point of the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) are [(x1+x2)/2, (y1+y2)/2, (z1+z2)/2]

So we have,

= (2/2, 0/2, 4/2)

= (1, 0, 2)

1 + x = 2, 2 + y = 0, -4 + z = 4

x = 1, y = -2, z = 8

Hence, the coordinates of the fourth vertex are D (1, -2, 8).

2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

Solution:

Given:

The vertices of the triangle are A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

x1 = 0, y1 = 0, z1 = 6;

x2 = 0, y2 = 4, z2 = 0;

x3 = 6, y3 = 0, z3 = 0

So, let the medians of this triangle be AD, BE and CF, corresponding to the vertices A, B and C, respectively.

D, E and F are the midpoints of the sides BC, AC and AB, respectively.

By the midpoint formula, the coordinates of the midpoint of the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) are [(x1+x2)/2, (y1+y2)/2, (z1+z2)/2]

So, we have

By the distance formula, the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

∴ The lengths of the medians of the given triangle are 7, 34 and 7.

3. If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.

Solution:

Given:

The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).

Where,

x1 = 2a, y1 = 2, z1 = 6;

x2 = -4, y2 = 3b, z2 = -10;

x3 = 8, y3 = 14, z3 = 2c

We know that the coordinates of the centroid of the triangle, whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), are [(x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3]

So, the coordinates of the centroid of the triangle PQR are

2a + 4 = 0, 3b + 16 = 0, 2c – 4 = 0

a = -2, b = -16/3, c = 2

∴ The values of a, b and c are a = -2, b = -16/3, and c = 2.

4. Find the coordinates of a point on the y-axis, which are at a distance of 5√2 from the point P (3, –2, 5).

Solution:

Let the point on the y-axis be A (0, y, 0).

Then, it is given that the distance between the points A (0, y, 0) and P (3, -2, 5) is 5√2.

Now, by using the distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

Distance of PQ = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, the distance between the points A (0, y, 0) and P (3, -2, 5) is given by

Distance of AP = [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

= [(3-0)2 + (-2-y)2 + (5-0)2]

= [32 + (-2-y)2 + 52]

= [(-2-y)2 + 9 + 25]

52 = [(-2-y)2 + 34]

Squaring on both sides, we get

(-2 -y)2 + 34 = 25 × 2

(-2 -y)2 = 50 – 34

4 + y2 + (2 × -2 × -y) = 16

y2 + 4y + 4 -16 = 0

y2 + 4y – 12 = 0

y2 + 6y – 2y – 12 = 0

y (y + 6) – 2 (y + 6) = 0

(y + 6) (y – 2) = 0

y = -6, y = 2

∴ The points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.

5. A point R with x-coordinate 4 lies on the line segment joining the points P (2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.
[Hint: Suppose R divides PQ in the ratio k:1. The coordinates of the point R are given by

Solution:

Given:

The coordinates of the points are P (2, -3, 4) and Q (8, 0, 10).

x1 = 2, y1 = -3, z1 = 4;

x2 = 8, y2 = 0, z2 = 10

Let the coordinates of the required point be (4, y, z).

So, let the point R (4, y, z) divides the line segment joining the points P (2, -3, 4) and Q (8, 0, 10) in the ratio k:1.

By using the section formula,

We know that the coordinates of the point R, which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m:n, is given by:

So, the coordinates of the point R are given by

So, we have

8k + 2 = 4 (k + 1)

8k + 2 = 4k + 4

8k – 4k = 4 – 2

4k = 2

k = 2/4

= 1/2

Now, let us substitute the values, and we get

= 6

∴ The coordinates of the required point are (4, -2, 6).

6. If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.

Solution:

Given:

Points A (3, 4, 5) and B (-1, 3, -7)

x1 = 3, y1 = 4, z1 = 5;

x2 = -1, y2 = 3, z2 = -7;

PA2 + PB2 = k2 ……….(1)

Let the point be P (x, y, z).

Now, by using the distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

So,

And

Now, substituting these values in (1), we have

[(3 – x)2 + (4 – y)2 + (5 – z)2] + [(-1 – x)2 + (3 – y)2 + (-7 – z)2] = k2 [(9 + x2 – 6x) + (16 + y2 – 8y) + (25 + z2 – 10z)] + [(1 + x2 + 2x) + (9 + y2 – 6y) + (49 + z2 + 14z)] = k2

9 + x2 – 6x + 16 + y2 – 8y + 25 + z2 – 10z + 1 + x2 + 2x + 9 + y2 – 6y + 49 + z2 + 14z = k2

2x2 + 2y2 + 2z2 – 4x – 14y + 4z + 109 = k2

2x2 + 2y2 + 2z2 – 4x – 14y + 4z = k2 – 109

2 (x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 109

(x2 + y2 + z2 – 2x – 7y + 2z) = (k2 – 109)/2

Hence, the required equation is (x2 + y2 + z2 – 2x – 7y + 2z) = (k2 – 109)/2.

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

Given below are the topics of Class 11 Maths Chapter 12, Introduction to Three Dimensional Geometry, which is categorised under the CBSE Class 11 Maths Syllabus 2023-24.

12.1 Introduction

This section introduces the concept of coordinate axes, coordinate planes in real life, coordinates of the point concerning the three coordinate planes, and the basics of geometry in three-dimensional space.

12.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space

This section defines the rectangular coordinate system, the naming of a coordinate plane and different notations in coordinate planes.

Rahul was riding his bicycle back home from a basketball game at a nearby stadium when he hit the divider to avoid a dog that had run onto the road. Unfortunately, his bicycle was stuck to the divider. When he couldn’t remove the bicycle by himself, he decided to take the help of his friend who stayed nearby. Rahul later looked at a topographical map and identified his friend’s house. He travelled 200 metres south and 550 metres west from where he left his bicycle. The map showed that he had also walked uphill from an altitude of 600 metres to an altitude of 650 metres above sea level. If we treat the location of Rahul’s bicycle as the origin of the coordinates, what is the position vector of the Kapur farm?

12.3 Coordinates of a Point in Space

This section explains the coordinate system in space, coordinates [x, y and z] with a few examples.

If a student is planning to place different pieces of furniture in a drawing room, a two-dimensional grid representing the room can be drawn, and an appropriate unit of measurement should be used. Let the horizontal distance be x, the vertical distance to x be y, and the origin is the starting point. If the width of the room is 10 meters, any point in the room can be defined as (x,y,z).

12.4 Distance between Two Points

This section covers the distance between three points in a three dimensional coordinate system using the distance formula along with a few solved problems.

If an object P is placed in the coordinate plane, the distance between the point P from the three axes [x, y and z] can be calculated using the distance formula.

12.5 Section Formula

This section talks about the section formula for a three dimensional geometry and its different cases. A few examples are solved for better understanding.

Exercise 12.1 Solutions: 4 Questions

Exercise 12.2 Solutions: 5 Questions

Exercise 12.3 Solutions: 5 Questions

Miscellaneous Exercise on Chapter 12 Solutions: 6 Questions

## A Few Points on Chapter 12 Introduction to Three Dimensional Geometry

• In three dimensional geometry, a cartesian coordinate system consists of three mutually perpendicular lines, namely x, y and z-axes. They are measured in the same unit of length.
• The three planes, XY-plane, YZ-plane and ZX-plane, are determined by the pair of axes called the axes of the coordinate planes.
• The three coordinate planes divide the space into eight parts known as octants.
• The coordinates of a point P (x, y, z) in three dimensional geometry are written in the form of an ordered triplet. Here, x, y and z are the distances from the YZ, ZX and XY-planes.
• (i) Any point on the x-axis is represented as (x, 0, 0), (ii) Any point on the y-axis is represented as (0, y, 0), and (iii) Any point on the z-axis is represented as (0, 0, z).
• The coordinates of point R divide the line segment joining two points P (x1,y1,z1) and Q (x2,y2,z2) internally and externally in the ratio m:n.

Subject experts at BYJU’S, who have prepared the NCERT Solutions for Class 11 Maths, have years of understanding about the question paper setting and types of questions that appear in the exam. These solutions provide alternative methods and explanations to solve problems that make the students feel confident to face the exam. Also, solving many complicated problems enhances the Mathematical ability of the students. The solutions cover all the necessary questions, which a student must and should have mastered to appear for the exam.

Disclaimer –

Dropped Topics –

12.5 Section Formula
Exercise 12.3
Ques. 4 and 5 (Miscellaneous Exercise)
Last Three Points in the Summary

## Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 12

Q1

### Explain the concept of 3D Geometry covered in Chapter 12 of NCERT Solutions for Class 11 Maths.

3D Geometry is a branch of Mathematics that deals with the study of points, lines or solid shapes in three dimensional coordinate systems. It will introduce the students to the concept of z-coordinate along with x and y to determine the exact location of a point in the 3D coordinate plane. It is a basic theory which has applications in other science streams and higher Mathematics. The trigonometric ratios concept has applications in 3D Geometry.
Q2

### Discuss the topics covered in Chapter 12 of NCERT Solutions for Class 11 Maths.

The topics covered in Chapter 12 of NCERT Solutions for Class 11 Maths are listed below:
1. Introduction
2. Coordinate Axes and Coordinate Planes in Three Dimensional Space
3. Coordinates of a Point in Space
4. Distance between Two Points
5. Section Formula
Q3

### Where can I get the NCERT Solutions for Class 11 Maths Chapter 12?

The NCERT Solutions, available at BYJU’S, is the best reference guide for students to score well in the exams. These solutions contain detailed explanations for each and every concept covered in the chapter. Students who find it difficult to solve exercise-wise problems can access the solutions, which are available online, to get their doubts cleared instantly. The chapter-wise and exercise-wise solutions are created by the subject-matter experts at BYJU’S with the aim of helping students, irrespective of their intelligence quotient.