# Class 11 NCERT Solutions- Chapter 12 Introduction to three dimensional Geometry – Exercise 12.2

### Problem 1: Find the distance between the following pairs of points:

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

**Solution:**

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

Now, by using the distance formula,

Length of 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

Length of distance PQ = √[(4 – 2)^{2}+ (3 – 3)^{2}+ (1 – 5)^{2}]= √[(2)

^{2}+ (0)^{2}+ (-4)^{2}]= √[4 + 0 + 16]

= √20

= 2√5

∴ The length of distance PQ is 2√5 units.

### (ii) (–3, 7, 2) and (2, 4, –1)

**Solution:**

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

Now, by using the distance formula,

Length of 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

Length of distance PQ = √[(2 – (-3))^{2}+ (4 – 7)^{2}+ (-1 – 2)^{2}]= √[(5)

^{2}+ (-3)^{2}+ (-3)^{2}]= √[25 + 9 + 9]

= √43

∴ The length of distance PQ is √43 units.

### (iii) (–1, 3, – 4) and (1, –3, 4)

**Solution:**

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

Now, by using the distance formula,

Length of 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

Length of distance PQ = √[(1 – (-1))^{2}+ (-3 – 3)^{2}+ (4 – (-4))^{2}]= √[(2)

^{2}+ (-6)^{2}+ (8)^{2}]= √[4 + 36 + 64]

= √104

= 2√26

∴ The length of distance PQ is 2√26 units.

### (iv) (2, –1, 3) and (–2, 1, 3)

**Solution:**

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

Now, by using the distance formula,

Length of 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

Length of distance PQ = √[(-2 – 2)^{2}+ (1 – (-1))^{2}+ (3 – 3)^{2}]= √[(-4)

^{2}+ (2)^{2}+ (0)^{2}]= √[16 + 4 + 0]

= √20

= 2√5

∴ The required distance is 2√5 units.

### Problem 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

P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)

Now, by using the distance formula,

Length of 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

Length of distance PQ = √[(1 – (-2))^{2}+ (2 – 3)^{2}+ (3 – 5)^{2}]= √[(3)2 + (-1)2 + (-2)2]

= √[9 + 1 + 4]

= √14

Length of distance PQ is √14Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)

Now, by using the distance formula,

Length of 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

Length of distance QR = √[(7 – 1)^{2}+ (0 – 2)^{2}+ (-1 – 3)^{2}]= √[(6)

^{2}+ (-2)^{2 }+ (-4)^{2}]= √[36 + 4 + 16]

= √56

= 2√14

Length of distance QR is 2√14P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)

Now, by using the distance formula,

Length of 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

Length of distance PR = √[(7 – (-2))^{2}+ (0 – 3)^{2}+ (-1 – 5)^{2}]= √[(9)

^{2}+ (-3)^{2}+ (-6)^{2}]= √[81 + 9 + 36]

= √126

= 3√14

Length of distance PR is 3√14Thus, PQ = √14, QR = 2√14 and PR = 3√14

So, PQ + QR = √14 + 2√14

= 3√14

= PR

∴ The points P, Q and R are collinear.

### Problem 3: Verify the following:

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

**Solution:**

(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

P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)

Now, by using the distance formula,

Length of 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

Length of 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)

Now, by using the distance formula,

Length of 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

Length of distance QR = √[(4 – 1)^{2}+ (9 – 6)^{2}+ (-6 – (-6))^{2}]= √[(3)

^{2}+ (3)^{2}+ (-6+6)^{2}]= √[9 + 9 + 0]

= √18

Hence,

Length of distance PQ = Length of distance QR i.e

√18 = √18∴ Length of 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.

**Solution:**

(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)

Now, by using the distance formula,

Length of 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

Length of distance PQ = √[(-1 – 0)^{2}+ (6 – 7)^{2}+ (6 – 10)^{2}]= √[(-1)

^{2}+ (-1)^{2}+ (-4)^{2}]= √[1 + 1 + 16]

= √18

Length of distance PQ is √18cmQ ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

Now, by using the distance formula,

Length of 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

Length of distance QR = √[(4 – 1)^{2}+ (9 – 6)^{2}+ (-6 – (-6))^{2}]= √[(3)

^{2}+ (3)^{2}+ (-6+6)^{2}]= √[9 + 9 + 0]

= √18

Length of distance QR is √18cmP ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

Now, by using the distance formula,

Length of 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

Length of distance PR = √[(-4 – 0)^{2}+ (9 – 7)^{2}+ (6 – 10)^{2}]= √[(-4)

^{2}+ (2)^{2}+ (-4)^{2}]= √[16 + 4 + 16]

= √36

Length of distance PR is √36cmNow,

PQ

^{2}+ QR^{2}= 18 + 18= 36

= PR

^{2}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.

**Solution :**

(–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

A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)

Now, by using the distance formula,

Length of 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

Length of distance AB = √[(1 – (-1))2 + (-2 – 2)^{2}+ (5 – 1)^{2}]= √[(2)

^{2}+ (-4)^{2}+ (4)^{2}]= √[4 + 16 + 16]

= √36

= 6

Length of distance AB is 6cmB ≡ (1, – 2, 5) and C ≡ (4, – 7, 8)

Now, by using the distance formula,

Length of 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

Length of distance BC = √[(4 – 1)^{2}+ (-7 – (-2))^{2}+ (8 – 5)^{2}]= √[(3)

^{2}+ (-5)

^{2}+ (3)^{2}]= √[9 + 25 + 9]

= √43

Length of distance BC is √43cmC ≡ (4, – 7, 8) and D ≡ (2, – 3, 4)

Now, by using the distance formula,

Length of 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

Length of distance CD = √[(2 – 4)^{2}+ (-3 – (-7))^{2}+ (4 – 8)^{2}]= √[(-2)

^{2}+ (4)^{2}+ (-4)^{2}]= √[4 + 16 + 16]

= √36

= 6

Length of distance CD is 6cmD ≡ (2, – 3, 4) and A ≡ (– 1, 2, 1)

By using the formula,

Length of 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

Length of distance DA = √[(-1 – 2)^{2}+ (2 – (-3))^{2}+ (1 – 4)^{2}]= √[(-3)

^{2}+ (5)^{2}+ (-3)^{2}]= √[9 + 25 + 9]

= √43

Length of distance DA is √43cmSince AB = CD and BC = DA (given)

So, In ABCD both pairs of opposite sides are equal

∴ ABCD is a parallelogram

### Problem 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

P ≡ (x, y, z) and A ≡ (1, 2, 3)

Now, by using the distance formula,

Now, by using the distance formula, PA = √[(x2 – x1)

^{2}+ (y2 – y1)^{2}+ (z2 – z1)^{2}]So here,

x1 = x, y1 = y, z1 = z

x2 = 1, y2 = 2, z2 = 3

Length of distance PA = √[(1 – x)^{2}+ (2 – y)^{2}+ (3 – z)^{2}]P ≡ (x, y, z) and B ≡ (3, 2, – 1)

Now, by using the distance formula,

Length of 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

Length of distance PB = √[(3 – x)^{2}+ (2 – y)^{2}+ (-1 – z)^{2}]Since PA = PB

Square on both the sides, we get

PA

^{2}= PB^{2}(1 – x)

^{2}+ (2 – y)^{2}+ (3 – z)^{2}= (3 – x)^{2}+ (2 – y)^{2}+ (– 1 – z)^{2}(1 + x

^{2}– 2x) + (4 + y^{2}– 4y) + (9 + z^{2}– 6z)(9 + x

^{2}– 6x) + (4 + y^{2}– 4y) + (1 + z^{2}+ 2z)– 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14

4x – 8z = 0

x – 2z = 0

∴ The required equation is x – 2z = 0

### Problem 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)

Now, by using the distance formula,

Length of 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

Length of distancePA = √[(4– x)^{2}+ (0 – y)^{2}+ (0 – z)^{2}]Calculating PB

P ≡ (x, y, z) and B ≡ (– 4, 0, 0)

Now, by using the distance formula,

Length of 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

Length of 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 getPA

^{2}= (10 – PB)^{2}PA

^{2}= 100 + PB^{2}– 20 PB(4 – x)

^{2}+ (0 – y)^{2}+ (0 – z)^{2}100 + (– 4 – x)

^{2}+ (0 – y)^{2}+ (0 – z)^{2}– 20 PB(16 + x

^{2}– 8x) + (y^{2}) + (z^{2})100 + (16 + x

^{2}+ 8x) + (y^{2}) + (z^{2}) – 20 PB20 PB = 16x + 100

5 PB = (4x + 25)

Square on both the sides again, we get25 PB

^{2}= 16x^{2}+ 200x + 62525 [(– 4 – x)

^{2}+ (0 – y)^{2}+ (0 – z)^{2}] = 16x^{2}+ 200x + 62525 [x

^{2}+ y^{2}+ z^{2}+ 8x + 16] = 16x^{2}+ 200x + 62525x

^{2}+ 25y^{2}+ 25z^{2}+ 200x + 400 = 16x^{2}+ 200x + 6259x

^{2}+ 25y^{2}+ 25z^{2}– 225 = 0

∴ The required equation is 9x^{2}+ 25y^{2}+ 25z^{2}– 225 = 0