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

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

NCERT Solutions for Class 11 Maths Chapter 12 Introduction To Three Dimensional Geometry empowers the students to solve the problems in a dynamic way of consuming less time. The NCERT Solutions are written by highly experienced teachers according to the latest update on term-wise CBSE Syllabus 2021-22 making the clarification of every problem straightforward and precise. The students can refer to this Chapter’s NCERT Solutions for Class 11 Maths PDF from the link provided below for optimum term – II exam preparations.

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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 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 I octant.

(ii) (4, -2, 3)

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

So it lies in IV octant.

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

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

So it lies in VIII octant.

(iv) (4, 2, -5)

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

So it lies in 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 II octant.

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

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

So it lies in III octant.

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

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

So it lies in 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 a line.

Firstly let us calculate 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 be

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

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

So firstly 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)

Firstly 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 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 be: 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

Firstly 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)

So, 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

Firstly 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 the 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]

Now it is given that:

PA + PB = 10

PA = 10 – PB

Square on both the 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 the 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 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 divides 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 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 divides 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 co-ordinates 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 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 now, 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 section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2) are collinear.

Solution:

Let the 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 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 the 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 same as P.

Hence, A, B, 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), 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 also know that the diagonals of a parallelogram bisect each other, so the mid points of AC and BD are equal, i.e. Midpoint of AC = Midpoint of BD ……….(1)

Now, by 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 is 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 (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 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, By Distance Formula, we know that 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, c = 2

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

Solution:

Let the point on 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 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 the 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 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 now, 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 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, 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:

The 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 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

 Also Access NCERT Exemplar for Class 11 Maths Chapter 12 CBSE Notes for Class 11 Maths Chapter 12

## 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 categorized under the second term CBSE Class 11 Maths Syllabus 2021-22.

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, 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, naming of a coordinate plane and different notations in coordinate planes.

Rahul was riding his bicycle back home from a basketball game from 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 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 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 horizontal distance be x, and the vertical distance to x be y, and 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 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 is 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)(iii) Any point on the z-axis is represented as (0, 0, z).
• The coordinates of the 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 would appear in the second term exam. These solutions provide alternative methods and explanations to solve problems that make the students feel confident while facing the term – II 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 second term exam.

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

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

3D Geometry is a branch of Mathematics which 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 basic theory which has its applications in other sciences and higher Mathematics. Trigonometric ratios concept has application in 3D Geometry.

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

The topics covered in the Chapter 12 of NCERT Solutions for Class 11 Maths are –
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

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

NCERT Solutions by BYJU’S is the best reference guide for the students to score well in the term-wise exams. These solutions contain detailed explanations for each and every concept covered in the chapter. Students who find it difficult in solving 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 experts at BYJU’S with the aim of helping students, irrespective of their intelligence quotient.