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

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 12
Chapter Name : Introduction to Three Dimensional Geometry

Exercise 12.1

Q1 A point is on the x-axis. What are its y-coordinate and z-coordinates?

Answer. It a point Is on the x-axis, then Its "y-coordinates and z-coordinates are zero.

Page : 271 , Block Name : Exercise 12.1

Q2 A point is in the XZ-plane. What can you say about its y-coordinate?

Answer. If a point is in the XZ plane, then its y-coordinate is zero.

Page : 271 , Block Name : Exercise 12.1

Q3 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).

Answer. The x-coordinate, y-coordinate, and z-coordinate Of point (I, 2, 3) are all positive,
Therefore, this point lies in octant I.
The x-coordinate, Y-coordinate, and z-coordinate of point (4, -2, 3) are positive, negative, and
positive respectively Therefore, this point lies in octant IV
The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, -5) are positive, negative, and
negative respectively. Therefore, this point lies in octant VIII.
The x-coordinate, y-coordinate. and z-coordinate of point (4, 2, -5) are positive, positive, and
negative respectively. Therefore, this point lies in octant V.
The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, -5) are negative, positive, and
negative respectively. Therefore, this point lies in octant VI.
The x-coordinate, y-coordinate, and z-coordinate or point (-4, 2, 5) are negative, positive, and
positive respectively. Therefore, this point lies in octant II.
The x-coordinate, y-coordinate, and z-coordinate Of point (-3, -1, 6) are negative, negative, and
positive respectively, Therefore, this point lies in octant Ill.
The x-coordinate, y-coordinate, and z-coordinate of point (2, -4, -7) are positive, negative, and
negative respectively. Therefore, this point lies in octant VIII.

Page : 271 , Block Name : Exercise 12.1

Q4 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.

Answer. (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.

Page : 271 , Block Name : Exercise 12.1

Exercise 12.2

Q1 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).

Answer. (i) The distance between points P (x , y z ) and P (x , y , z ) is given by
1, 1, 1 2 2, 2

2 2 2
PQ = √(xz − x1 ) + (y2 − y1 ) + (z2 − z1 )

(1) Distance between points (2, 3, 5) and (4, 3, 1)

2 2 2
= √(4 − 2) + (3 − 3) + (1 − 5)

= √4 + 16

= √20

= 2 √5

(ii) D Distance between points (−3, 7, 2) and (2, 4, −1)

2 2 2
= √(2 + 3) + (4 − 7) + (−1 − 2)

2 2 2
= √(5) + (−3) + (−3)

= √25 + 9 + 9

= √43

(iii) Distance between points (−1, 3, −4) and (1, −3, 4)

2 2 2
= √(1 + 1) + (−3 − 3) + (4 + 4)

2 3 2
= √(2) + (−6) + (8)

= √4 + 36 + 64 = √104 = 2√26

(iv) Distance between points (2, -1, 3) and (-2, 1, 3)
2 2 2
= √(−2 − 2) + (1 + 1) + (3 − 3)

2 2 2 2
= √(−4) + (2) + (0) + (3 − 3)

= √16 + 4

= √20

= 2 √5

Page : 273 , Block Name : Exercise 12.2

Q2 Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Answer. Let points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) be denoted by P, Q and R respectively.
Points P, Q, R are collinear if they lie on a line.
2 2 2
PQ = √(1 + 2) + (2 − 3) + (3 − 5)

2 2 2
= √(3) + (−1) + (−2)

= √9 + 1 + 4

= √14

2 2 2
QR = √(7 − 1) + (0 − 2) + (−1 − 3)

2 2 2
= √(6) + (−2) + (−4)

= √36 + 4 + 16

= √56

= 2√14

2 2 2
P R =√(7 + 2) + (0 − 3) + (−1 − 5)

2 2 2
= √(9) + (−3) + (−6)

= √81 + 9 + 36

= √126

= 3√14

Here, PQ + QR = √14 + 2√14 = 3√14 = PR
Hence, points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Page : 273 , Block Name : Exercise 12.2

Q3 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.

Answer. (i) Let points (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) be denoted by A, B, and C respectively.
2 2 2
AB = √(1 − 0) + (6 − 7) + (−6 + 10)

2 2 2
= √(1) + (−1) + (4)

= √1 + 1 + 16

= √18

= 3 √2

2 2 2
BC = √(4 − 1) + (9 − 6) + (−6 + 6)

2 2
= √(3) + (3)

= √9 + 9 = √18 = 3√2

2 2 2
CA = √(0 − 4) + (7 − 9) + (−10 + 6)

2 2 2
= √(−4) + (−2) + (−4)

= √16 + 4 + 16 = √36 = 6

Here, AB = BC ≠ CA

Thus, the Given points are the vertices of an isosceles triangle.
(ii) Let (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) be denoted by A, B, and C respectively.
2 2 2
AB = √(−1 − 0) + (6 − 7) + (6 − 10)

2 2 2
= √(−1) + (−1) + (−4)

= √1 + 1 + 16 = √18

= 3 √2

2 2 2
BC = √(−4 + 1) + (9 − 6) + (6 − 6)

2 2 2
= √(−3) + (3) + (0)

= √9 + 9 = √18

= 3 √2

2 2 2
CA = √(0 + 4) + (7 − 9) + (10 − 6)

2 2 2
= √(4) + (−2) + (4)

= √16 + 4 + 16

= √36

= 6

Now, AB 2
+ BC
2
= (3√2)
2
+ (3√2)
2
= 18 + 18 = 36 = AC
2

Therefore, the given points are the vertices of a right-angled triangle.
(iii) Let (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) be denoted by A, B, C and D respectively.
2 2 2
AB = √(1 + 1) + (−2 − 2) + (5 − 1)

= √4 + 16 + 16

= √36

= 6

2 2 2
BC = √(4 − 1) + (−7 + 2) + (8 − 5)

= √9 + 25 + 9 = √43

2 2 2
CD = √(2 − 4) + (−3 + 7) + (4 − 8)

= √4 + 16 + 16

= √36

= 6

2 2 2
DA = √(−1 − 2) + (2 + 3) + (1 − 4)

= √9 + 25 + 9 = √43

Here, AB = CD = 6, BC = AD = √43
Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.
Therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.

Page : 273 , Block Name : Exercise 12.2

Q4 Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2,
–1).

Answer. Let P(X, Y, Z) be the point that is equidistant from points A (1, 2, 3) amd B(3, 2, -1)
Accordingly, PA = PB
2 2
⇒ PA = PB

2 2 2 2 2 2
⇒ (x − 1) + (y − 2) + (z − 3) = (x − 3) + (y − 2) + (z + 1)

2 2 2 2 2 2
⇒ x − 2x + 1 + y − 4y + 4 + z − 6z + 9 = x − 6x + 9 + y − 4y + 4 + z + 2z + 1

⇒ −2x − 4y − 6z + 14 = −6x − 4y + 2z + 14

⇒ −2x − 6z + 6x − 2z = 0

⇒ 4x − 8z = 0

⇒ x − 2z = 0

Thus the required equation is x - 2z = 0.

Page : 273 , Block Name : Exercise 12.2

Q5 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.

Answer. Let the coordinates of P be (x, y, z).
The coordinates of points A and B are (4, 0, 0) and (-4, 0, 0) respectively.
It is given that PA + PB = 10.
2 2 2 2 2 2
⇒ √(x − 4) + y + z + √(x + 4) + y + z = 10

2 2 2 2 2 2
⇒ √(x − 4) + y + z = 10 − √(x + 4) + y + z

On squaring both sides, we obtain

2 2 2 2 2 2 2 2 2
⇒ (x − 4) + y + z = 100 − 20√(x + 4) + y + z + (x + 4) + y + z

2 2 2 2 2 2 2 2 2
⇒ x − 8x + 16 + y + z = 100 − 20√x + 8x + 16 + y + z + x + 8x + 16 + y + z

2 2 2
⇒ 20√x + 8x + 16 + y + z = 100 + 16x

2 2 2
⇒ 5√x + 8x + 16 + y + z = (25 + 4x)

On squaring both sides again, we obtain

2 2 2 2
25 (x + 8x + 16 + y + z ) = 625 + 16x + 200x

⇒ 25x
2
+ 200x + 400 + 25y
2
+ 25z
2
= 625 + 16x
2
+ 200x .
2 2 2
⇒ 9x + 25y + 25z − 225 = 0

2 2 2
Thus, the required equation is 9x + 25y + 25z − 225 = 0

Page : 273 , Block Name : Exercise 12.2

Exercise 12.3

Q1 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.

Answer. (i) The coordinates of point R that divides the line segment joining points P (x , y , z ) 1 1 1

Q (x2 , y2 , z2 ) internally in the ratio m : n

mx2 +nx1 my2 +my1 mz2 +nz1
( , , )
m+n m+n m+n

Let R (x, y, z) be point that divides the line segment joining points (-2, 3, 5) and (1, -4, 6)
internally in the ratio 2:3
2(1)+3(−2) 2(−4)+3(3) 2(6)+3(5)
x = ,y = , and z =
2+3 2+3 2+3

−4 1 27
Le. x = ,y = , and z =
5 5 5

Thus, the coordinates of the required point are (− 4

5
,
1

5
,
27

5
) .
(ii) The coordinates of point R that divides the line segment joining points P(x , y , z ) 1 1 1

Q (x , y , z ) internally in the ratio m : n are
2 2 2

mx2 −nx1 my2 −ny1 mz2 −nz1
( , , )
m−n m−n m−n

Let R (x, y, z) be the point that divides the line segment joining points (-2, 3, 5) and (1, -4, 6)
externally in the ratio 2:3
2(1)−3(−2) 2(−4)−3(3) 2(6)−3(5)
x = ,y = , and z =
2−3 2−3 2−3

i.e.. x = −8, y = 17, and z = 3

Thus, the coordinates of the required point are (-8, 17, 3).

Page : 277 , Block Name : Exercise 12.3

Q2 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.

Answer. Let point Q (5, 4, -6) divide the line segment Joining points P(3, 2, -4) and R (9, 8, -10)
in the ratio k:1.
Therefore, by section formula,
k(9)+3 k(8)+2 k(−10)−4
(5, 4, −6) = ( , , )
k+1 k+1 k+1

9k+3
⇒ = 5
k+1

⇒ 9k + 3 = 5k + 5

⇒ 4k = 2

2 1
⇒ k = =
4 2

Thus, point Q divides PR in ratio 1:2.

Page : 277 , Block Name : Exercise 12.3

Q3 Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–
2, 4, 7) and (3, –5, 8).

Answer. Let the YZ plane divide the line segment joining points (–2, 4, 7) and (3, –5, 8) in the
ratio k:1.
Hence, by section formula, the coordinates of point of intersection are given by
k(3)−2 k(−5)+4 k(8)+7
( , , )
k+1 k+1 k+1

On the YZ-plane, the x-coordinate of any point is zero.
3k−2
= 0
k+1

⇒ 3k − 2 = 0

2
⇒ k =
3

Thus, the YZ plane divides the line segment formed by joining the given points in the ratio 2:3.

Page : 277 , Block Name : Exercise 12.3

Q4 Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and c (0, are
1
, 2)
3

collinear.

Answer. The given points are A(2, -3, 4), B(-1, 2, 1) and c (0, 1

3
, 2)

Let P be a point that divides AB in the ratio k:1.
Hence, by section formula, the coordinates of P are given by
k(−1)+2 k(2)−3 k(1)+4
( , , )
k+1 k+1 k+1

Now, we nd the value of k at which point P coincides with point C.
By taking , we obtain k = 2.
−k+2
= 0
k+1

For k = 2, the coordinates of point P are (0, 1

3
, 2)

I.e., is a point that divides AB externally in the ratio 2:1 and is the same as point P.
Hence, points A, B, and C are collinear.

Page : 277 , Block Name : Exercise 12.3

Q5 Find the coordinates of the points which trisect the line segment joining the points P (4, 2, –
6) and Q (10, –16, 6).

Answer. Let A and B be the points that trisect the line segment joining points P(4, 2, -6) and
Q(10, -16, 6).

Point A divides PQ in the ratio 1:2. Therefore, by section formula, the coordinates of point A are
given by
1(10)+2(4) 1(−16)+2(2) 1(6)+2(−6)
( , , ) = (6, −4, −2)
1+2 1+2 1+2

Point B divides PQ in the ratio 2:1. Therefore, by section formula, the coordinates of Point B are
given by
2(10)+1(4) 2(−16)+1(2) 2(6)−1(6)
( , , ) = (8, −10, 2)
2+1 2+1 2+1

Thus, (6, -4, -2) and (8, -10, 2) are the points that intersect the line segment joining points P(4,
2, -6) and Q(10, -16, 6).

Page : 277 , Block Name : Exercise 12.3

Miscellaneous Exercise

Q1 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.

Answer. The three vertices of a parallelogram ABCD are given as A(3, -1, 2), B (1, 2, -4) and C(-1,

1, 2). Let the coordinates of the fourth vertex be D (x, y, z).

We know that the diagonals of a parallelogram bisect each other.
Therefore, in parallelogram ABCD, AC and BD bisect each other.
∴ Mid-point of AC = Mid-point of BD
3−1 −1+1 2+2 x+1 y+2 z−4
⇒ ( , , ) = ( , , )
2 2 2 2 2 2

x+1 y+2 z−4
⇒ (1, 0, 2) = ( , , )
2 2 2

x+1 y+2 z−4
⇒ = 1, = 0, and = 2
2 2 2

⇒ x = 1, y = −2, and z = 8

Thus, the coordinates of the fourth vertex are (1, -2, 8).

Page : 278 , Block Name : Miscellaneous Exercise

Q2 Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).

Answer. Let AD, BE, and CF be the medians of the given triangle ABC.

Since AD is the median, D is the mid-point of BC.
Coordinates of point D = = ( = (3, 2, 0)
0+6 4+0 0+0
∴ , ⋅ )
2 2 2

2 2 2
AD = √(0 − 3) + (0 − 2) + (6 − 0) = √9 + 4 + 36 = √49 = 7

Since BE is the median, E is the mid-point of AC.
0+6 0+0 6+0
∴ Coordinates of point E = ( , , ) = (3, 0, 3)
2 2 2

2 2 2
BE = √(3 − 0) + (0 − 4) + (3 − 0) = √9 + 16 + 9 = √34

Since CF is the median, F is the mid-point of AB.
0+0 0+4 6+0
∴ Coordinates of point F = ( , , ) = (0, 2, 3)
2 2 2

2 2 2
Length of CF = √(6 − 0) + (0 − 2) + (0 − 3) = √36 + 4 + 9 = √49 = 7

Thus, the lengths of the medians of ΔABC are 7, √34, and 7

Page : 278 , Block Name : Miscellaneous Exercise

Q3 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 nd the values of a, b and c.

Answer.

It is known that the coordinates of the centroid of the triangle, whose vertices are (x , y , z ), 1 1 1

y1 +y2 +y2
are ( .
x1 +x2 +x3 z1 +z2 +z3
(x2 , y2 , z2 ) and (x3 , y3 , z3 ) , , )
3 3 3

Therefore, coordinates of the centroid of △PQR
2a−4+8 2+3b+14 6−10+2c 2a+4 3b+16 2c−4
= ( , , ) = ( , , )
3 3 3 3 3 3

It is given that origin is the centroid of △PQR.
2a+4 3b+16 2c−4
∴ (0, 0, 0) = ( , , )
3 3 3

2a+4 3b+16 2c−4
⇒ = 0, = 0 and = 0
3 3 3

16
⇒ a = −2, b = − and c = 2
3

Thus, the respective values of a, b, c are −2, − 16

3
, and 2 .

Page : 278 , Block Name : Miscellaneous Exercise

Q4 Find the coordinates of a point on y-axis which are at a distance of 5√2 from the point P (3,
–2, 5).

Answer. If a point is on the y - axis, then x-coordinate and the z-coordinate of the point are zero.
Let A(0, b, 0) be the point on the y-axis at a distance of 5√2 from the point P(3, -2, 5).
Accordingly, AP = 5√2
2
∴ AP = 50

2 2 2
⇒ (3 − 0) + (−2 − b) + (5 − 0) = 50

2
⇒ 9 + 4 + b + 4b + 25 = 50

2
⇒ b + 4b − 12 = 0

2
⇒ b + 6b − 2b − 12 = 0

⇒ (b + 6)(b − 2) = 0

⇒ b = −6 or 2

Thus, the coordinates of the required points are (0, 2, 0) and (0, -6, 0).

Page : 278 , Block Name : Miscellaneous Exercise

Q5 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
]
8k+2 −3 10k+4
( , , )
k+1 k+1 k+1

Answer. The coordinates of points P and Q are given as P(2, -3, 4) and Q(8, 0, 10).
Let R divide line segment PQ in the ratio k:1.
Hence, by section formula, the coordinates of point R are given by
k(8)+2 k(0)−3 k(10)+4 8k+2 −3 10k+4
( , , ) = ( , , )
k+1 k+1 k+1 k+1 k+1 k+1

It is given that the x-coordinate of point R is 4.
8k+2
∴ = 4
k+1

⇒ 8k + 2 = 4k + 4

⇒ 4k = 2

1
⇒ k =
2

1 1
10( )+4 10( )+4

Therefore, the coordinates of point R are (4,
−3 2 2

, , ) = (4, −2, 6)
1 1
1 +1 +1
2 2

Page : 279 , Block Name : Miscellaneous Exercise

Q6 If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, nd the equation of the set of
points P such that PA , where k is a constant.
2 2 2
+ PB = k

Answer. The coordinates of points A and B are given as (3, 4, 5) and (-1, 3, -7) respectively.
Let the coordinates of formula, we obtain
2 2 2 2
PA = (x − 3) + (y − 4) + (z − 5)

2 2 2
= x + 9 − 6x + y + 16 − 8y + z + 25 − 10z

2 2 2
= x − 6x + y − 8y + z − 10z + 50

2 2 2 2
PB = (x + 1) + (y − 3) + (z + 7)

2 2 2
= x + 2x + y − 6y + z + 14z + 59
2 2 2
Now, if PA + PB = k , then

2 2 2 2 2 2 2
(x − 6x + y − 8y + z − 10z + 50) + (x + 2x + y − 6y + z + 14z + 59) = k

2 2 2 2
⇒ 2x + 2y + 2z − 4x − 14y + 4z + 109 = k

2 2 2 2
⇒ 2 (x + y + z − 2x − 7y + 2z) = k − 109

2
2 2 2 k −109
⇒ x + y + z − 2x − 7y + 2z =
2
2

Thus, the required equation is x .
2 2 2 k −109
+ y + z − 2x − 7y + 2z =
2

Page : 279 , Block Name : Miscellaneous Exercise