RD Sharma Solutions for Class 11 Chapter 28 - Introduction to Three Dimensional Coordinate Geometry

Students can obtain a good score in their examination by using the solution module prepared by our subject experts at BYJU’s. Chapter 28 – Introduction to Three Dimensional Coordinate Geometry is based on the latest CBSE syllabus. The solutions here are developed in an interesting manner, which makes learning fun and easy. For further reference, students can make use of RD Sharma Class 11 Maths Solutions, which help students effortlessly prepare for their exams. Students can refer and easily download the pdf for free from the links given below.

Chapter 28 – Introduction to Three Dimensional Coordinate Geometry contains three exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Coordinates of a point in space.
  • Signs of coordinates of a point.
  • Distance formula.
  • Section formulae.

Download the pdf of RD Sharma Solutions for Class 11 Maths Chapter 28 – Introduction to Three Dimensional Coordinate Geometry

 

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Access answers to RD Sharma Solutions for Class 11 Maths Chapter 28 – Introduction to Three Dimensional Coordinate Geometry

ExERCISE 28.1 PAGE NO: 28.6

1. Name the octants in which the following points lie:
(i) (5, 2, 3)
(ii) (-5, 4, 3)
(iii) (4, -3, 5)
(iv) (7, 4, -3)
(v) (-5, -4, 7)
(vi) (-5, -3, -2)
(vii) (2, -5, -7)
(viii) (-7, 2, -5)

Solution:

(i) (5, 2, 3)

In this case, since x, y and z all three are positive then octant will be XOYZ


(ii) (-5, 4, 3)

In this case, since x is negative and y and z are positive then the octant will be X′OYZ


(iii) (4, -3, 5)

In this case, since y is negative and x and z are positive then the octant will be XOY′Z


(iv) (7, 4, -3)

In this case, since z is negative and x and y are positive then the octant will be XOYZ′


(v) (-5, -4, 7)

In this case, since x and y are negative and z is positive then the octant will be X′OY′Z


(vi) (-5, -3, -2)

In this case, since x, y and z all three are negative then octant will be X′OY′Z′


(vii) (2, -5, -7)

In this case, since z and y are negative and x is positive then the octant will be XOY′Z′


(viii) (-7, 2, -5)

In this case, since x and z are negative and x is positive then the octant will be X′OYZ′

2. Find the image of:
(i) (-2, 3, 4) in the yz-plane
(ii) (-5, 4, -3) in the xz-plane

(iii) (5, 2, -7) in the xy-plane
(iv) (-5, 0, 3) in the xz-plane
(v) (-4, 0, 0) in the xy-plane

Solution:

(i) (-2, 3, 4)

Since we need to find its image in yz-plane, a sign of its x-coordinate will change

So, Image of point (-2, 3, 4) is (2, 3, 4)


(ii)(-5, 4, -3)

Since we need to find its image in xz-plane, sign of its y-coordinate will change

So, Image of point (-5, 4, -3) is (-5, -4, -3)


(iii) (5, 2, -7)

Since we need to find its image in xy-plane, a sign of its z-coordinate will change

So, Image of point (5, 2, -7) is (5, 2, 7)


(iv) (-5, 0, 3)

Since we need to find its image in xz-plane, sign of its y-coordinate will change

So, Image of point (-5, 0, 3) is (-5, 0, 3)


(v) (-4, 0, 0)

Since we need to find its image in xy-plane, sign of its z-coordinate will change

So, Image of point (-4, 0, 0) is (-4, 0, 0)

3. A cube of side 5 has one vertex at the point (1, 0, 1), and the three edges from this vertex are, respectively, parallel to the negative x and y-axes and positive z-axis. Find the coordinates of the other vertices of the cube.

Solution:

Given: A cube has side 4 having one vertex at (1, 0, 1)

Side of cube = 5

We need to find the coordinates of the other vertices of the cube.

So let the Point A(1, 0, 1) and AB, AD and AE is parallel to –ve x-axis, -ve y-axis and +ve z-axis respectively.

Description: RD Sharma Solutions for Class 11 Maths Chapter 28 – image 1

Since side of cube = 5

Point B is (-4, 0, 1)

Point D is (1, -5, 1)

Point E is (1, 0, 6)

Now, EH is parallel to –ve y-axis

Point H is (1, -5, 6)

HG is parallel to –ve x-axis

Point G is (-4, -5, 6)


Now, again GC and GF is parallel to –ve z-axis and +ve y-axis respectively

Point C is (-4, -5, 1)

Point F is (-4, 0, 6)

4. Planes are drawn parallel to the coordinates planes through the points (3, 0, -1) and (-2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.

Solution:

Given: 

Points are (3, 0, -1) and (-2, 5, 4)

We need to find the lengths of the edges of the parallelepiped formed.


For point (3, 0, -1)

x1 = 3, y1 = 0 and z1 = -1


For point (-2, 5, 4)

x2 = -2, y2 = 5 and z2 = 4


Plane parallel to coordinate planes of x1 and x2 is yz-plane


Plane parallel to coordinate planes of y1 and y2 is xz-plane


Plane parallel to coordinate planes of z1 and z2 is xy-plane


Distance between planes x1 = 3 and x2 = -2 is 3 – (-2) = 3 + 2 = 5

Distance between planes x1 = 0 and y2 = 5 is 5 – 0 = 5

Distance between planes z1 = -1 and z2 = 4 is 4 – (-1) = 4 + 1 = 5


∴Theedges of parallelepiped is 5, 5, 5

5. Planes are drawn through the points (5, 0, 2) and (3, -2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed.

Solution:

Given: 

Points are (5, 0, 2) and (3, -2, 5)

We need to find the lengths of the edges of the parallelepiped formed


For point (5, 0, 2)

x1 = 5, y1 = 0 and z1 = 2


For point (3, -2, 5)

x2 = 3, y2 = -2 and z2 = 5


Plane parallel to coordinate planes of x1 and x2 is yz-plane

Plane parallel to coordinate planes of y1 and y2 is xz-plane


Plane parallel to coordinate planes of z1 and z2 is xy-plane


Distance between planes x1 = 5 and x2 = 3 is 5 – 3 = 2

Distance between planes x1 = 0 and y2 = -2 is 0 – (-2) = 0 + 2 = 2

Distance between planes z1 = 2 and z2 = 5 is 5 – 2 = 3


∴Theedges of parallelepiped is 2, 2, 3

6. Find the distances of the point P (-4, 3, 5) from the coordinate axes.

Solution:

Given:

The point P (-4, 3, 5)

The distance of the point from x-axis is given as:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 2

The distance of the point from y-axis is given as:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 3

The distance of the point from z-axis is given as:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 4

7. The coordinates of a point are (3, -2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.

Solution:

Given: 

Point (3, -2, 5)

The Absolute value of any point(x, y, z) is given by,

√(x2 + y2 + z2)

We need to make sure that absolute value to be the same for all points.

So let the point A(3, -2, 5)

Remaining 7 points are:

Point B(3, 2, 5) (By changing the sign of y coordinate)

Point C(-3, -2, 5) (By changing the sign of x coordinate)

Point D(3, -2, -5) (By changing the sign of z coordinate)

Point E(-3, 2, 5) (By changing the sign of x and y coordinate)

Point F(3, 2, -5) (By changing the sign of y and z coordinate)

Point G(-3, -2, -5) (By changing the sign of x and z coordinate)

Point H(-3, 2, -5) (By changing the sign of x, y and z coordinate)


ExERCISE 28.2 PAGE NO: 28.9

1. Find the distance between the following pairs of points:
(i) P(1, -1, 0) and Q (2, 1, 2)

(ii) A(3, 2, -1) and B (-1, -1, -1)

Solution:

(i) P(1, -1, 0) and Q (2, 1, 2)

Given:

The points P(1, -1, 0) and Q (2, 1, 2)

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 5

∴ The Distance between P and Q is 3 units.

(ii) A (3, 2, -1) and B (-1, -1, -1)

Given:

The points A (3, 2, -1) and B (-1, -1, -1)

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 6

= 5

∴ The Distance between A and B is 5 units.

2. Find the distance between the points P and Q having coordinates (-2, 3, 1) and (2, 1, 2).

Solution:

Given:

The points (-2, 3, 1) and (2, 1, 2)

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 7

∴ The Distance between the given two points is 21 units.

3. Using distance formula prove that the following points are collinear:
(i) A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

(ii) P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)

(iii) A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)
Solution:

(i) A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

Given:

The points A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

Points A, B and C are collinear if AB + BC = AC or AB + AC = BC or AC + BC = AB

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 8

= 66

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 9

∴The points A, B and C are collinear.

(ii) P (0, 7, -7), Q (1, 4, -5) and R (-1, 10, -9)

Given:

The points P (0, 7, -7), Q (1, 4, -5) and R (-1, 10, -9)

Points P, Q and R are collinear if PQ + QR = PR or PQ + PR = QR or PR + QR = PQ

By using the formula,

Distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 10

= 14

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 11

∴The points P, Q and R are collinear.

(iii) A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

Given:

The points A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

Points A, B and C are collinear if AB + BC = AC or AB + AC = BC or AC + BC = AB

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 12

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 13

Description: RD Sharma Solutions for Class 11 Maths Chapter 28 – image 14

∴The points A, B and C are collinear.

4. Determine the points in (i) xy-plane (ii) yz-plane and (iii) zx-plane which are equidistant from the points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1).

Solution:

Given:

The points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)

(i) xy-plane

We know z = 0 in xy-plane.

So let P(x, y, 0) be any point in xy-plane

According to the question:

PA = PB = PC

PA2 = PB2 = PC2

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 15

We know PA2 = PB2

So, (x – 1)2+ (y + 1)2 = (x – 2)2 + (y – 1)2 + 4

x2+ 1 – 2x + y2 + 1 + 2y = x2+ 4 – 4x + y2 + 1 – 2y + 4

– 2x + 2 + 2y = 9 – 4x – 2y

– 2x + 2 + 2y – 9 + 4x + 2y = 0

2x + 4y – 7 = 0

2x = – 4y + 7……………………(1)


Since, PA2 = PC2

So, (x – 1)2+ (y + 1)2 = (x – 3)2 + (y – 2)2 + 1

x2+ 1 – 2x + y2 + 1 + 2y = x2+ 9 – 6x + y2 + 4 – 4y + 1

– 2x + 2 + 2y = 14 – 6x – 4y

– 2x + 2 + 2y – 14 + 6x + 4y = 0

4x + 6y – 12 = 0

2(2x + 3y – 6) = 0


Now substitute the value of 2x (obtained in equation (1)), we get

7 – 4y + 3y – 6 = 0

– y + 1 = 0

y = 1


By substituting the value of y back in equation (1) we get,

2x = 7 – 4y

2x = 7 – 4(1)

2x = 3

x = 3/2

∴The point P (3/2, 1, 0) in xy-plane is equidistant from A, B and C.

(ii) yz-plane

We know x = 0 in yz-plane.

Let Q(0, y, z) any point in yz-plane

According to the question:

QA = QB = QC

QA2 = QB2 = QC2

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 16

We know, QA2 = QB2

So, 1 + z2+ (y + 1)2 = (z – 2)2 + (y – 1)2 + 4

z2+ 1 + y2 + 1 + 2y = z2+ 4 – 4z + y2 + 1 – 2y + 4

2 + 2y = 9 – 4z – 2y

2 + 2y – 9 + 4z + 2y = 0

4y + 4z – 7 = 0

4z = –4y + 7

z = [–4y + 7]/4 …. (1)

Since, QA2 = QC2

So, 1 + z2+ (y + 1)2 = (z + 1)2 + (y – 2)2 + 9

2+ 1 + y2 + 1 + 2y = z2+ 1 + 2z + y2 + 4 – 4y + 9

2 + 2y = 14 + 2z – 4y

2 + 2y – 14 – 2z + 4y = 0

–2z + 6y – 12 = 0

2(–z + 3y – 6) = 0


Now, substitute the value of z [obtained from (1)] we get

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 17

12y + 4y – 7 – 24 = 0

16y – 31 = 0

y = 31/16

Substitute the value of y back in equation (1), we get

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 18

= -3/16

∴The point Q (0, 31/16, -3/16) in yz-plane is equidistant from A, B and C.

(iii) zx-plane

We know y = 0 in xz-plane.

Let R(x, 0, z) any point in xz-plane

According to the question:

RA = RB = RC

RA2 = RB2 = RC2

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 19

We know, RA2 = RB2

So, 1 + z2+ (x – 1)2 = (z – 2)2 + (x – 2)2 + 1

z2+ 1 + x2 + 1 – 2x = z2+ 4 – 4z + x2 + 4 – 4x + 1

2 – 2x = 9 – 4z – 4x

2 + 4z – 9 + 4x – 2x = 0

2x + 4z – 7 = 0

2x = –4z + 7……………………………(1)


Since, RA2 = RC2

So, 1 + z2+ (x – 1)2 = (z + 1)2 + (x – 3)2 + 4

z2+ 1 + x2 + 1 – 2x = z2+ 1 + 2z + x2 + 9 – 6x + 4

2 – 2x = 14 + 2z – 6x

2 – 2x – 14 – 2z + 6x = 0

–2z + 4x – 12 = 0

2(2x) = 12 + 2z

Substitute the value of 2x [obtained from equation (1)] we get,

2(–4z + 7) = 12 + 2z

–8z + 14 = 12 + 2z

14 – 12 = 8z + 2z

10z = 2

z = 2/10

= 1/5

Now, substitute the value of z back in equation (1), we get

2x = -4z + 7

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 20

∴The point R (31/10, 0, 1/5) in xz-plane is equidistant from A, B and C.

5. Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, -4)

Solution:

Given:

The points (1, 5, 7) and (5, 1, -4)

We know x = 0 and y = 0 on z-axis

Let R(0, 0, z) any point on z-axis

According to the question:

RA = RB

RA2 = RB2

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 21

We know, RA2 = RB2

26+ (z – 7)2 = (z + 4)2 + 26

z2+ 49 – 14z + 26 = z2+ 16 + 8z + 26

49 – 14z = 16 + 8z

49 – 16 = 14z + 8z

22z = 33

z = 33/22

= 3/2

∴The point R (0, 0, 3/2) on z-axis is equidistant from (1, 5, 7) and (5, 1, -4).

6. Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).

Solution:

Given:

The points (3, 1, 2) and (5, 5, 2)

We know x = 0 and z = 0 on y-axis

Let R(0, y, 0) any point on the y-axis

According to the question:

RA = RB

RA2 = RB2

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 22

So,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 23

We know, RA2 = RB2

13+ (y – 1)2 = (y – 5)2 + 29

y2+ 1 – 2y + 13 = y2+ 25 – 10y + 29

10y – 2y = 54 – 14

8y = 40

y = 40/8

= 5

∴The point R (0, 5, 0) on y-axis is equidistant from (3, 1, 2) and (5, 5, 2).

7. Find the points on z-axis which are at a distance√21 from the point (1, 2, 3).

Solution:

Given:

The point (1, 2, 3)

Distance = √21 

We know x = 0 and y = 0 on z-axis

Let R(0, 0, z) any point on z-axis

According to question:

RA = √21 

RA2 = 21

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 24

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 25

We know, RA2 = 21

5 + (z – 3)2 = 21

z2+ 9 – 6z + 5 = 21

z2 – 6z = 21 – 14

z2– 6z – 7 = 0

z2– 7z + z – 7 = 0

z(z– 7) + 1(z – 7) = 0

(z– 7) (z + 1) = 0

(z– 7) = 0 or (z + 1) = 0

z= 7 or z = -1

∴The points (0, 0, 7) and (0, 0, -1) on z-axis is equidistant from (1, 2, 3).

8. Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.

Solution:

Given:

The points (1, 2, 3), (2, 3, 1) and (3, 1, 2)

An equilateral triangle is a triangle whose all sides are equal.

So let us prove AB = BC = AC

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 26

= 6

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 27

It is clear that,

AB = BC = AC

Δ ABC is a equilateral triangle

Hence Proved.

9. Show that the points (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an isosceles right-angled triangle.

Solution:

Given:

The points (0, 7, 10), (-1, 6, 6) and (-4, 9, 6)

Isosceles right-angled triangle is a triangle whose two sides are equal and also satisfies Pythagoras Theorem.

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 28

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 29

Since, AB = BC

So, AB2 + BC2

= (32)2 + (32)2

= 18 + 18

= 36

= AC2

We know that, AB = BC and AB2 + BC2 = AC2

So, Δ ABC is an isosceles-right angled triangle

Hence Proved.

10. Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of squares.

Solution:

Given:

The points A (3, 3, 3), B (0, 6, 3), C (1, 7, 7) and D (4, 4, 7)

We know that all sides of a square are equal.

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 30

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 31

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 32

It is clear that,

AB = BC = CD = AD

Quadrilateral formed by ABCD is a square. [Since all sides are equal]

Hence Proved.

11. Prove that the point A(1, 3, 0), B(-5, 5, 2), C(-9, -1, 2) and D(-3, -3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.

Solution:

Given:

The points A (1, 3, 0), B (-5, 5, 2), C (-9, -1, 2) and D (-3, -3, 0)

We know that, opposite sides of both parallelogram and rectangle are equal.

But diagonals of a parallelogram are not equal whereas they are equal for rectangle.

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 33

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 34

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 35

It is clear that,

AB = CD

BC = AD

Opposite sides are equal

Now, let us find the length of diagonals

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 36

It is clear that,

AC ≠ BD

The diagonals are not equal, but opposite sides are equal.

So we can say that quadrilateral formed by ABCD is a parallelogram but not a rectangle.

Hence Proved.

12.Show that the points A(1, 3, 4), B(-1, 6, 10), C(-7, 4, 7) and D(-5, 1, 1) are the vertices of a rhombus.

Solution:

Given:

The points A (1, 3, 4), B (-1, 6, 10), C (-7, 4, 7) and D (-5, 1, 1)

We know that, all sides of both square and rhombus are equal.

But diagonals of a rhombus are not equal whereas they are equal for square.

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 37

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 38

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 39

It is clear that,

AB = BC = CD = AD

So, all sides are equal

Now, let us find the length of diagonals

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 40

It is clear that,

AC ≠ BD

The diagonals are not equal but all sides are equal.

So we can say that quadrilateral formed by ABCD is a rhombus but not square.

Hence Proved.


ExERCISE 28.3 PAGE NO: 28.19

1. The vertices of the triangle are A(5, 4, 6), B(1, -1, 3) and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the length AD.

Solution:

Given:

The vertices of the triangle are A (5, 4, 6), B (1, -1, 3) and C (4, 3, 2).

By using the formulas let us find the coordinates of D and the length of AD

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 41

The section formula is given as

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 42

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 43

AB : AC = 5:3

BD: DC = 5:3

So, m = 5 and n = 3


B(1, -1, 3) and C(4, 3, 2)

Coordinates of D using section formula:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 44

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 45

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 46

2. A point C with z-coordinate 8 lies on the line segment joining the points A(2, -3, 4) and B(8, 0, 10). Find the coordinates.

Solution:

Given:

The points A (2, -3, 4) and B (8, 0, 10)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 47

Let Point C(x, y, 8), and C divides AB in ratio k: 1

So, m = k and n = 1

A(2, -3, 4) and B(8, 0, 10)

Coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 48

On comparing we get,

[10k + 4] / [k + 1] = 8

10k + 4 = 8(k + 1)

10k + 4 = 8k + 8

10k – 8k = 8 – 4

2k = 4

k = 4/2

= 2

Here C divides AB in ratio 2:1

x = [8k + 2] / [k + 1]

= [8(2) + 2] / [2 + 1]

= [16 + 2] / [3]

= 18/3

= 6

y = -3 / [k + 1]

= -3 / [2 + 1]

= -3 / 3

= -1

∴ The Coordinates of C are (6, -1, 8).

3. Show that the three points A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10) are collinear and find the ratio in which C divides AB.

Solution:

Given:

The points A (2, 3, 4), B (-1, 2, -3) and C (-4, 1, -10)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 49

Let C divides AB in ratio k: 1

Three points are collinear if the value of k is the same for x, y and z coordinates.

So, m = k and n = 1

A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)

Coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 50

On comparing we get,

[-k + 2] / [k + 1] = -4

-k + 2 = -4(k + 1)

-k + 2 = -4k – 4

4k – k = – 2 – 4

3k = -6

k = -6/3

= -2

[2k + 3] / [k + 1] = 1

2k + 3 = k + 1

2k – k = 1 – 3

k = – 2

[-3k + 4] / [k + 1] = -10

-3k + 4 = -10(k + 1)

-3k + 4 = -10k – 10

-3k + 10k = -10 – 4

7k = -14

k = -14/7

= -2

The value of k is the same in all three cases.

So, A, B and C are collinear [as k = -2]

∴We can say that, C divides AB externally in ratio 2:1

4. Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Solution:

Given:

The points (2, 4, 5) and (3, 5, 4)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 51

We know X coordinate is always 0 on yz-plane

So, let Point C(0, y, z), and let C divide AB in ratio k: 1

Then, m = k and n = 1

A(2, 4, 5) and B(3, 5, 4)

The coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 52

On comparing we get,

[3k + 2] / [k + 1] = 0

3k + 2 = 0(k + 1)

3k + 2 = 0

3k = – 2

k = -2/3

∴We can say that, C divides AB externally in ratio 2: 3

5. Find the ratio in which the line segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x + y + z = 5.

Solution:

Given:

The points(2, -1, 3) and (-1, 2, 1)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 53

Let C(x, y, z) be any point on the given plane and C divides AB in ratio k: 1

Then, m = k and n = 1

A(2, -1, 3) and B(-1, 2, 1)


Coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 54

On comparing we get,

[-k + 2] / [k + 1] = x

[2k – 1] / [k + 1] = y

[-k + 3] / [k + 1] = z

We know that x + y + z = 5

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 55

5(k + 1) = 4

5k + 5 = 4

5k = 4 – 5

5k = – 1

k = -1/5

∴We can say that, the plane divides AB externally in the ratio 1:5

6. If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.

Solution:

Given:

The points A (3, 2, -4), B (9, 8, -10) and C (5, 4, -6)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 56

Let C divides AB in ratio k: 1

Three points are collinear if the value of k is the same for x, y and z coordinates.

Then, m = k and n = 1

A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6)

Coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 57

On comparing we get,

[9k + 3] / [k + 1] = 5

9k + 3 = 5(k + 1)

9k + 3 = 5k + 5

9k – 5k = 5 – 3

4k = 2

k = 2/4

= ½

[8k + 2] / [k + 1] = 4

8k + 2 = 4(k + 1)

8k + 2 = 4k + 4

8k – 4k = 4 – 2

4k = 2

k = 2/4

= ½

[-10k – 4] / [k + 1] = -6

-10k – 4 = -6(k + 1)

-10k – 4 = -6k – 6

-10k + 6k = 4 – 6

-4k = -2

k = -2/-4

= ½

The value of k is the same in all three cases.

So, A, B and C are collinear [as, k = ½]

∴We can say that, C divides AB externally in ratio 1:2

7. The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

Solution:

Given:

The mid-points of the sides of a triangle ABC is given as (-2, 3, 5), (4, -1, 7) and (6, 5, 3).

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 58

We know the mid-point divides side in the ratio of 1:1.

The coordinates of C is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 59

P(-2, 3, 5) is mid-point of A(x1, y1, z1) and B(x2, y2, z2)

Then,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 60

Q(4, -1, 7) is mid-point of B(x2, y2, z2) and C(x3, y3, z3)

Then,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 61

R(6, 5, 3) is mid-point of A(x1, y1, z1) and C(x3, y3, z3)

Then,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 62

Now solving for ‘x’ terms

x1 + x2 = -4……………………(4)

x2 + x3 = 8………………………(5)

x1 + x3 = 12……………………(6)

By adding equation (4), (5), (6)

x1 + x2 + x2 + x3 + x1 + x3 = 8 + 12 – 4

2x1 + 2x2 + 2x3 = 16

2(x1 + x2 + x3) = 16

x1 + x2 + x3 = 8………………………(7)

Now, subtract equation (4), (5) and (6) from equation (7) separately:

x1 + x2 + x3 – x1 – x2 = 8 – (-4)

x3 = 12


x1 + x2 + x3 – x2 – x3 = 8 – 8

x1 = 0


x1 + x2 + x3 – x1 – x3 = 8 – 12

x2 = -4

Now solving for ‘y’ terms

y1 + y2 = 6……………………(8)

y2 + y3 = -2……………………(9)

y1 + y3 = 10……………………(10)


By adding equation (8), (9) and (10) we get,

y1 + y2 + y2 + y3 + y1 + y3 = 10 + 6 – 2

2y1 + 2y2 + 2y3 = 14

2(y1 + y2 + y3) = 14

y1 + y2 + y3 = 7………………………(11)


Now, subtract equation (8), (9) and (10) from equation (11) separately:

y1 + y2 + y3 – y1 – y2 = 7 – 6

y3 = 1


y1 + y2 + y3 – y2 – y3 = 7 – (-2)

y1 = 9


y1 + y2 + y3 – y1 – y3 = 7 – 10

y2 = -3

Now solving for ‘z’ terms

z1 + z2 = 10……………………(12)

z2 + z3 = 14……………………(13)

z1 + z3 = 6……………………(14)


By adding equation (12), (13) and (14) we get,

z1 + z2 + z2 + z3 + z1 + z3 = 6 + 14 + 10

2z1 + 2z2 + 2z3 = 30

2(z1 + z2 + z3) = 30

z1 + z2 + z3 = 15………………………(15)


Now, subtract equation (8), (9) and (10) from equation (11) separately:

z1 + z2 + z3 – z1 – z2 = 15 – 10

z3 = 5


z1 + z2 + z3 – z2 – z3 = 15 – 14

z1 = 1


z1 + z2 + z3 – z1 – z3 = 15 – 6

z2 = 9

∴Thevertices of sides of a triangle ABC are A(0, 9, 1) B(-4,-3, 9) and C(12, 1, 5).

8. A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

Solution:

Given:

The vertices of a triangle are A (1, 2, 3), B (0, 4, 1), C (-1, -1, -3)

By using the distance formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 63

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 64

So, AB/AC = 3/7

AB: AC = 3:7

BD: DC = 3:7

Then, m = 3 and n = 7

B(0, 4, 1) and C(-1, -1, -3)


Coordinates of D by using section formula is given as

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 65

∴The coordinates of D are (-3/10, 5/2, -1/5).

Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 28 – Introduction to Three Dimensional Coordinate Geometry

Exercise 28.1 Solutions

Exercise 28.2 Solutions

Exercise 28.3 Solutions

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