# 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 = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 2, y1 = 3, z1 = 5

x2 = 4, y2 = 3, z2 = 1

Length of distance PQ = âˆš[(4 â€“ 2)^{2}+ (3 â€“ 3)^{2}+ (1 â€“ 5)^{2}]= âˆš[(2)

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

= âˆš20

= 2âˆš5

âˆ´ The length of distance PQ is 2âˆš5 units.

### (ii) (â€“3, 7, 2) and (2, 4, â€“1)

**Solution:**

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

Now, by using the distance formula,

Length of distance PQ = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = â€“ 3, y1 = 7, z1 = 2

x2 = 2, y2 = 4, z2 = â€“ 1

Length of 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 length of distance PQ is âˆš43 units.

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

**Solution:**

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

Now, by using the distance formula,

Length of distance PQ = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = â€“ 1, y1 = 3, z1 = â€“ 4

x2 = 1, y2 = â€“ 3, z2 = 4

Length of 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 length of distance PQ is 2âˆš26 units.

### (iv) (2, â€“1, 3) and (â€“2, 1, 3)

**Solution:**

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

Now, by using the distance formula,

Length of distance PQ = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 2, y1 = â€“ 1, z1 = 3

x2 = â€“ 2, y2 = 1, z2 = 3

Length of 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.

### Problem 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

P â‰¡ (â€“ 2, 3, 5) and Q â‰¡ (1, 2, 3)

Now, by using the distance formula,

Length of distance PQ = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = â€“ 2, y1 = 3, z1 = 5

x2 = 1, y2 = 2, z2 = 3

Length of 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

Length of distance PQ is âˆš14Q â‰¡ (1, 2, 3) and R â‰¡ (7, 0, â€“ 1)

Now, by using the distance formula,

Length of distance QR = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 1, y1 = 2, z1 = 3

x2 = 7, y2 = 0, z2 = â€“ 1

Length of 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

Length of distance QR is 2âˆš14P â‰¡ (â€“ 2, 3, 5) and R â‰¡ (7, 0, â€“ 1)

Now, by using the distance formula,

Length of distance PR= âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = â€“ 2, y1 = 3, z1 = 5

x2 = 7, y2 = 0, z2 = â€“ 1

Length of 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

Length of distance PR is 3âˆš14Thus, PQ = âˆš14, QR = 2âˆš14 and PR = 3âˆš14

So, PQ + QR = âˆš14 + 2âˆš14

= 3âˆš14

= PR

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

### Problem 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.**

**Solution:**

(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

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

Now, by using the distance formula,

Length of distance PQ = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 0, y1 = 7, z1 = â€“ 10

x2 = 1, y2 = 6, z2 = â€“ 6

Length of 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)

Now, by using the distance formula,

Length of distance QR = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 1, y1 = 6, z1 = â€“ 6

x2 = 4, y2 = 9, z2 = â€“ 6

Length of 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,

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

âˆš18 = âˆš18âˆ´ Length of 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.

**Solution:**

(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)

Now, by using the distance formula,

Length of distance PQ = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 0, y1 = 7, z1 = 10

x2 = â€“ 1, y2 = 6, z2 = 6

Length of 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

Length of distance PQ is âˆš18cmQ â‰¡ (1, 6, â€“ 6) and R â‰¡ (4, 9, â€“ 6)

Now, by using the distance formula,

Length of distance QR = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 1, y1 = 6, z1 = â€“ 6

x2 = 4, y2 = 9, z2 = â€“ 6

Length of 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

Length of distance QR is âˆš18cmP â‰¡ (0, 7, 10) and R â‰¡ (â€“ 4, 9, 6)

Now, by using the distance formula,

Length of distance PR = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 0, y1 = 7, z1 = 10

x2 = â€“ 4, y2 = 9, z2 = 6

Length of 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

Length of distance PR is âˆš36cmNow,

PQ

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

= PR

^{2}By using converse of Pythagoras theorem,

âˆ´ The given vertices P, Q & R are the vertices of a right-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.

**Solution :**

(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

A â‰¡ (â€“ 1, 2, 1) and B â‰¡ (1, â€“ 2, 5)

Now, by using the distance formula,

Length of distance AB = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = â€“ 1, y1 = 2, z1 = 1

x2 = 1, y2 = â€“ 2, z2 = 5

Length of 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

Length of distance AB is 6cmB â‰¡ (1, â€“ 2, 5) and C â‰¡ (4, â€“ 7, 8)

Now, by using the distance formula,

Length of distance BC = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 1, y1 = â€“ 2, z1 = 5

x2 = 4, y2 = â€“ 7, z2 = 8

Length of distance BC = 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

Length of distance BC is âˆš43cmC â‰¡ (4, â€“ 7, 8) and D â‰¡ (2, â€“ 3, 4)

Now, by using the distance formula,

Length of distance CD = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 4, y1 = â€“ 7, z1 = 8

x2 = 2, y2 = â€“ 3, z2 = 4

Length of distance CD = 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

Length of distance CD is 6cmD â‰¡ (2, â€“ 3, 4) and A â‰¡ (â€“ 1, 2, 1)

By using the formula,

Length of distance DA = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = 2, y1 = â€“ 3, z1 = 4

x2 = â€“ 1, y2 = 2, z2 = 1

Length of 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

Length of distance DA is âˆš43cmSince AB = CD and BC = DA (given)

So, In ABCD both pairs of opposite sides are equal

âˆ´ 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, 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

P â‰¡ (x, y, z) and A â‰¡ (1, 2, 3)

Now, by using the distance formula,

Now, by using the distance formula, PA = âˆš[(x2 â€“ x1)

^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = x, y1 = y, z1 = z

x2 = 1, y2 = 2, z2 = 3

Length of distance PA = âˆš[(1 â€“ x)^{2}+ (2 â€“ y)^{2}+ (3 â€“ z)^{2}]P â‰¡ (x, y, z) and B â‰¡ (3, 2, â€“ 1)

Now, by using the distance formula,

Length of distance PB = âˆš[(x2 â€“ x1)

^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = x, y1 = y, z1 = z

x2 = 3, y2 = 2, z2 = â€“ 1

Length of distance PB = âˆš[(3 â€“ x)^{2}+ (2 â€“ y)^{2}+ (-1 â€“ z)^{2}]Since PA = PB

Square on both the sides, we get

PA

^{2}= PB^{2}(1 â€“ x)

^{2}+ (2 â€“ y)^{2}+ (3 â€“ z)^{2}= (3 â€“ x)^{2}+ (2 â€“ y)^{2}+ (â€“ 1 â€“ z)^{2}(1 + x

^{2}â€“ 2x) + (4 + y^{2}â€“ 4y) + (9 + z^{2}â€“ 6z)(9 + x

^{2}â€“ 6x) + (4 + y^{2}â€“ 4y) + (1 + z^{2}+ 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

### Problem 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)

Now, by using the distance formula,

Length of distance PA = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = x, y1 = y, z1 = z

x2 = 4, y2 = 0, z2 = 0

Length of distancePA = 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)

Now, by using the distance formula,

Length of distance PB = âˆš[(x2 â€“ x1)^{2}+ (y2 â€“ y1)^{2}+ (z2 â€“ z1)^{2}]So here,

x1 = x, y1 = y, z1 = z

x2 = â€“ 4, y2 = 0, z2 = 0

Length of 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 getPA

^{2}= (10 â€“ PB)^{2}PA

^{2}= 100 + PB^{2}â€“ 20 PB(4 â€“ x)

^{2}+ (0 â€“ y)^{2}+ (0 â€“ z)^{2}100 + (â€“ 4 â€“ x)

^{2}+ (0 â€“ y)^{2}+ (0 â€“ z)^{2}â€“ 20 PB(16 + x

^{2}â€“ 8x) + (y^{2}) + (z^{2})100 + (16 + x

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

5 PB = (4x + 25)

Square on both the sides again, we get25 PB

^{2}= 16x^{2}+ 200x + 62525 [(â€“ 4 â€“ x)

^{2}+ (0 â€“ y)^{2}+ (0 â€“ 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}â€“ 225 = 0

âˆ´ The required equation is 9x^{2}+ 25y^{2}+ 25z^{2}â€“ 225 = 0

## Please

Loginto comment...