NCERT Solutions for Class 11 Maths Chapter 12- Introduction to Three Dimensional Geometry Exercise 12.2

The Distance Between Two Points is the length of the line segment that connects the two points. The Exercise 12.2 of NCERT Solutions for Class 11 Maths Chapter 12- Introduction to Three Dimensional Geometry is based on the topic Distance between Two Points. The topic revolves around an equation that can be used to solve the problems present in this exercise. To know the solving process of these problems, understanding the concept is a must. The NCERT Solutions for Class 11 maths will help the students in getting thorough with the concepts and hence, scoring high marks in the exams.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 12- Introduction to Three Dimensional Geometry Exercise 12.2

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Solutions for Class 11 Maths Chapter 12 – Exercise 12.2

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


Access Other Exercise Solutions of Class 11 Maths Chapter 12- Introduction to Three Dimensional Geometry

Exercise 12.1 Solutions 4 Questions

Exercise 12.3 Solutions 5 Questions

Miscellaneous Exercise On Chapter 12 Solutions 6 Questions

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