# 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. 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.

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### Solutions for Class 11 Maths Chapter 12 â€“ Exercise 12.2

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

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

(ii) (â€“3, 7, 2) and (2, 4, â€“1)

(iii) (â€“1, 3, â€“ 4) and (1, â€“3, 4)

(iv) (2, â€“1, 3) and (â€“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 = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 2, y1Â = 3, z1Â = 5

x2Â = 4, y2Â = 3, z2Â = 1

Distance PQ = âˆš[(4 â€“ 2)2 + (3 â€“ 3)2 + (1 â€“ 5)2]

= âˆš[(2)2 + 02 + (-4)2]

= âˆš[4 + 0 + 16]

= âˆš20

= 2âˆš5

âˆ´ The required distance is 2âˆš5 units.

(ii) (â€“3, 7, 2) and (2, 4, â€“1)

Let P be (â€“ 3, 7, 2) and Q be (2, 4, â€“ 1)

By using the formula,

Distance PQ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = â€“ 3, y1Â = 7, z1Â = 2

x2Â = 2, y2Â = 4, z2Â = â€“ 1

Distance PQ = âˆš[(2 â€“ (-3))2 + (4 â€“ 7)2 + (-1 â€“ 2)2]

= âˆš[(5)2 + (-3)2 + (-3)2]

= âˆš[25 + 9 + 9]

= âˆš43

âˆ´ The required distance is âˆš43 units.

(iii) (â€“1, 3, â€“ 4) and (1, â€“3, 4)

Let P be (â€“ 1, 3, â€“ 4) and Q be (1, â€“ 3, 4)

By using the formula,

Distance PQ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = â€“ 1, y1Â = 3, z1Â = â€“ 4

x2Â = 1, y2Â = â€“ 3, z2Â = 4

Distance PQ = âˆš[(1 â€“ (-1))2 + (-3 â€“ 3)2 + (4 â€“ (-4))2]

= âˆš[(2)2 + (-6)2 + (8)2]

= âˆš[4 + 36 + 64]

= âˆš104

= 2âˆš26

âˆ´ The required distance is 2âˆš26 units.

(iv) (2, â€“1, 3) and (â€“2, 1, 3)

Let P be (2, â€“ 1, 3) and Q be (â€“ 2, 1, 3)

By using the formula,

Distance PQ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 2, y1Â = â€“ 1, z1Â = 3

x2Â = â€“ 2, y2Â = 1, z2Â = 3

Distance PQ = âˆš[(-2 â€“ 2)2 + (1 â€“ (-1))2 + (3 â€“ 3)2]

= âˆš[(-4)2 + (2)2 + (0)2]

= âˆš[16 + 4 + 0]

= âˆš20

= 2âˆš5

âˆ´ The required distance is 2âˆš5 units.

2. Show that the points (â€“2, 3, 5), (1, 2, 3) and (7, 0, â€“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 â‰¡ (â€“ 2, 3, 5) and Q â‰¡ (1, 2, 3)

By using the formula,

Distance PQ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = â€“ 2, y1Â = 3, z1Â = 5

x2Â = 1, y2Â = 2, z2Â = 3

Distance PQ = âˆš[(1 â€“ (-2))2 + (2 â€“ 3)2 + (3 â€“ 5)2]

= âˆš[(3)2 + (-1)2 + (-2)2]

= âˆš[9 + 1 + 4]

= âˆš14

Calculating QR

Q â‰¡ (1, 2, 3) and R â‰¡ (7, 0, â€“ 1)

By using the formula,

Distance QR = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 1, y1Â = 2, z1Â = 3

x2Â = 7, y2Â = 0, z2Â = â€“ 1

Distance QR = âˆš[(7 â€“ 1)2 + (0 â€“ 2)2 + (-1 â€“ 3)2]

= âˆš[(6)2 + (-2)2 + (-4)2]

= âˆš[36 + 4 + 16]

= âˆš56

= 2âˆš14

Calculating PR

P â‰¡ (â€“ 2, 3, 5) and R â‰¡ (7, 0, â€“ 1)

By using the formula,

Distance PR = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = â€“ 2, y1Â = 3, z1Â = 5

x2Â = 7, y2Â = 0, z2Â = â€“ 1

Distance PR = âˆš[(7 â€“ (-2))2 + (0 â€“ 3)2 + (-1 â€“ 5)2]

= âˆš[(9)2 + (-3)2 + (-6)2]

= âˆš[81 + 9 + 36]

= âˆš126

= 3âˆš14

Thus, PQ =Â âˆš14, QR =Â 2âˆš14Â and PR =Â 3âˆš14

So, PQ + QR =Â âˆš14Â +Â 2âˆš14

=Â 3âˆš14

= PR

âˆ´ The points P, Q and R are collinear.

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

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

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

Solution:

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

Let us consider the points be

P(0, 7, â€“10), Q(1, 6, â€“ 6) and R(4, 9, â€“ 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 â‰¡ (0, 7, â€“ 10) and Q â‰¡ (1, 6, â€“ 6)

By using the formula,

Distance PQ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 0, y1Â = 7, z1Â = â€“ 10

x2Â = 1, y2Â = 6, z2Â = â€“ 6

Distance PQ = âˆš[(1 â€“ 0)2 + (6 â€“ 7)2 + (-6 â€“ (-10))2]

= âˆš[(1)2 + (-1)2 + (4)2]

= âˆš[1 + 1 + 16]

= âˆš18

Calculating QR

Q â‰¡ (1, 6, â€“ 6) and R â‰¡ (4, 9, â€“ 6)

By using the formula,

Distance QR = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 1, y1Â = 6, z1Â = â€“ 6

x2Â = 4, y2Â = 9, z2Â = â€“ 6

Distance QR = âˆš[(4 â€“ 1)2 + (9 â€“ 6)2 + (-6 â€“ (-6))2]

= âˆš[(3)2 + (3)2 + (-6+6)2]

= âˆš[9 + 9 + 0]

= âˆš18

Hence, PQ = QR

18 = 18

2 sides are equal

âˆ´ PQR is an isosceles triangle.

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

Let the points be

P(0, 7, 10), Q(â€“ 1, 6, 6) & R(â€“ 4, 9, 6)

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

Calculating PQ

P â‰¡ (0, 7, 10) and Q â‰¡ (â€“ 1, 6, 6)

By using the formula,

Distance PQ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 0, y1Â = 7, z1Â = 10

x2Â = â€“ 1, y2Â = 6, z2Â = 6

Distance PQ = âˆš[(-1 â€“ 0)2 + (6 â€“ 7)2 + (6 â€“ 10)2]

= âˆš[(-1)2 + (-1)2 + (-4)2]

= âˆš[1 + 1 + 16]

= âˆš18

Calculating QR

Q â‰¡ (1, 6, â€“ 6) and R â‰¡ (4, 9, â€“ 6)

By using the formula,

Distance QR = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 1, y1Â = 6, z1Â = â€“ 6

x2Â = 4, y2Â = 9, z2Â = â€“ 6

Distance QR = âˆš[(4 â€“ 1)2 + (9 â€“ 6)2 + (-6 â€“ (-6))2]

= âˆš[(3)2 + (3)2 + (-6+6)2]

= âˆš[9 + 9 + 0]

= âˆš18

Calculating PR

P â‰¡ (0, 7, 10) and R â‰¡ (â€“ 4, 9, 6)

By using the formula,

Distance PR = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 0, y1Â = 7, z1Â = 10

x2Â = â€“ 4, y2Â = 9, z2Â = 6

Distance PR = âˆš[(-4 â€“ 0)2 + (9 â€“ 7)2 + (6 â€“ 10)2]

= âˆš[(-4)2 + (2)2 + (-4)2]

= âˆš[16 + 4 + 16]

= âˆš36

Now,

PQ2Â + QR2Â = 18 + 18

= 36

= PR2

By using converse of Pythagoras theorem,

âˆ´ The given vertices P, Q & R are the vertices of a right â€“ angled triangle at Q.

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

Let the points be: A(â€“1, 2, 1), B(1, â€“2, 5), C(4, â€“7, 8) & D(2, â€“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 â‰¡ (â€“ 1, 2, 1) and B â‰¡ (1, â€“ 2, 5)

By using the formula,

Distance AB = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = â€“ 1, y1Â = 2, z1Â = 1

x2Â = 1, y2Â = â€“ 2, z2Â = 5

Distance AB = âˆš[(1 â€“ (-1))2 + (-2 â€“ 2)2 + (5 â€“ 1)2]

= âˆš[(2)2 + (-4)2 + (4)2]

= âˆš[4 + 16 + 16]

= âˆš36

= 6

Calculating BC

B â‰¡ (1, â€“ 2, 5) and C â‰¡ (4, â€“ 7, 8)

By using the formula,

Distance BCÂ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 1, y1Â = â€“ 2, z1Â = 5

x2Â = 4, y2Â = â€“ 7, z2Â = 8

Distance BCÂ = âˆš[(4 â€“ 1)2 + (-7 â€“ (-2))2 + (8 â€“ 5)2]

= âˆš[(3)2 + (-5)2 + (3)2]

= âˆš[9 + 25 + 9]

= âˆš43

Calculating CD

C â‰¡ (4, â€“ 7, 8) and D â‰¡ (2, â€“ 3, 4)

By using the formula,

Distance CD =Â âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 4, y1Â = â€“ 7, z1Â = 8

x2Â = 2, y2Â = â€“ 3, z2Â = 4

Distance CDÂ = âˆš[(2 â€“ 4)2 + (-3 â€“ (-7))2 + (4 â€“ 8)2]

= âˆš[(-2)2 + (4)2 + (-4)2]

= âˆš[4 + 16 + 16]

= âˆš36

= 6

Calculating DA

D â‰¡ (2, â€“ 3, 4) and A â‰¡ (â€“ 1, 2, 1)

By using the formula,

Distance DAÂ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = 2, y1Â = â€“ 3, z1Â = 4

x2Â = â€“ 1, y2Â = 2, z2Â = 1

Distance DA = âˆš[(-1 â€“ 2)2 + (2 â€“ (-3))2 + (1 â€“ 4)2]

= âˆš[(-3)2 + (5)2 + (-3)2]

= âˆš[9 + 25 + 9]

= âˆš43

Since AB = CD and BC = DA (given)

So, In ABCD both pairs of opposite sides are equal.

âˆ´ 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, â€“1).

Solution:

Let A (1, 2, 3) & B (3, 2, â€“ 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, â€“ 1)

i.e. PA = PB

Firstly let us calculate

Calculating PA

P â‰¡ (x, y, z) and A â‰¡ (1, 2, 3)

By using the formula,

Distance PA = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = x, y1Â = y, z1Â = z

x2Â = 1, y2Â = 2, z2Â = 3

Distance PA = âˆš[(1 â€“ x)2 + (2 â€“ y)2 + (3 â€“ z)2]

Calculating PB

P â‰¡ (x, y, z) and B â‰¡ (3, 2, â€“ 1)

By using the formula,

Distance PBÂ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = x, y1Â = y, z1Â = z

x2Â = 3, y2Â = 2, z2Â = â€“ 1

Distance PBÂ = âˆš[(3 â€“ x)2 + (2 â€“ y)2 + (-1 â€“ z)2]

Since PA = PB

Square on both the sides, we get

PA2Â = PB2

(1 â€“ x)2Â + (2 â€“ y)2Â + (3 â€“ z)2Â = (3 â€“ x)2Â + (2 â€“ y)2Â + (â€“ 1 â€“ z)2

(1 + x2Â â€“ 2x) + (4 + y2Â â€“ 4y) + (9 + z2Â â€“ 6z)

(9 + x2Â â€“ 6x) + (4 + y2Â â€“ 4y) + (1 + z2Â + 2z)

â€“ 2x â€“ 4y â€“ 6z + 14 = â€“ 6x â€“ 4y + 2z + 14

4x â€“ 8z = 0

x â€“ 2z = 0

âˆ´ The required equation is x â€“ 2z = 0

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

Solution:

Let A (4, 0, 0) & B (â€“ 4, 0, 0)

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

Calculating PA

P â‰¡ (x, y, z) and A â‰¡ (4, 0, 0)

By using the formula,

Distance PAÂ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = x, y1Â = y, z1Â = z

x2Â = 4, y2Â = 0, z2Â = 0

Distance PA = âˆš[(4â€“ x)2 + (0 â€“ y)2 + (0 â€“ z)2]

Calculating PB

P â‰¡ (x, y, z) and B â‰¡ (â€“ 4, 0, 0)

By using the formula,

Distance PBÂ = âˆš[(x2 â€“ x1)2 + (y2 â€“ y1)2 + (z2 â€“ z1)2]

So here,

x1Â = x, y1Â = y, z1Â = z

x2Â = â€“ 4, y2Â = 0, z2Â = 0

Distance PB = âˆš[(-4â€“ x)2 + (0 â€“ y)2 + (0 â€“ z)2]

Now it is given that:

PA + PB = 10

PA = 10 â€“ PB

Square on both the sides, we get

PA2Â = (10 â€“ PB)2

PA2Â = 100 + PB2Â â€“ 20 PB

(4 â€“ x)2Â + (0 â€“ y)2Â + (0 â€“ z)2

100 + (â€“ 4 â€“ x)2Â + (0 â€“ y)2Â + (0 â€“ z)2Â â€“ 20 PB

(16 + x2Â â€“ 8x) + (y2) + (z2)

100 + (16 + x2Â + 8x) + (y2) + (z2) â€“ 20 PB

20 PB = 16x + 100

5 PB = (4x + 25)

Square on both the sides again, we get

25 PB2Â = 16x2Â + 200x + 625

25 [(â€“ 4 â€“ x)2Â + (0 â€“ y)2Â + (0 â€“ z)2] = 16x2Â + 200x + 625

25 [x2Â + y2Â + z2Â + 8x + 16] = 16x2Â + 200x + 625

25x2Â + 25y2Â + 25z2Â + 200x + 400 = 16x2Â + 200x + 625

9x2Â + 25y2Â + 25z2Â â€“ 225 = 0

âˆ´ The required equation is 9x2Â + 25y2Â + 25z2Â â€“ 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