Class 11 - Maths - Introduction To 3D Geometry

Exercise 12.1

Question 1:

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

Answer:

If a point is on the x-axis, then its y-coordinates and z-coordinates are zero.

Question 2:

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.

Question 3:

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 (1, 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 of 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 III.

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.

Question 4:

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.

 

                                                                         Exercise 12.2

Question 1:

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:

The distance between points P(x1, y1, z1) and P(x2, y2, z2) is given by

PQ = √{(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2}

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

    = √{(4 – 2)2 + (3 – 3)2 + (1 – 5)2}

    = √{22 + 0 + (-4)2}

    = √(4 + 16)

    = √20

    = 2√5   

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

    = √{(2 + 3)2 + (4 – 7)2 + (-1 – 2)2}

    = √{52 + (-3)2 + (-3)2}

    = √(25 + 9 + 9)

    = √43

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

    = √{(1 + 1)2 + (-3 – 3)2 + (4 + 4)2}

    = √{22 + (-6)2 + 82}

    = √(4 + 36 + 64)

    = √104

    = 2√26

(iv) Distance between points (2, –1, 3) and (–2, 1, 3)

    = √{(-2 – 2)2 + (1 + 1)2 + (3 – 3)2}

    = √{(-4)2 + 22 + 0}

    = √(16 + 4)

    = √20

    = 2√5

Question 2:

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, and R are collinear if they lie on a line.

Now,

PQ = √{(1 + 2)2 + (2 - 3)2 + (3 – 5)2}

      = √{32 + (-1)2 +(-2)2}

      = √(9 + 1 + 4)

      = √14

QR = √{(7 - 1)2 + (0 - 2)2 + (-1 – 3)2}

      = √{62 + (-2)2 +(-4)2}

      = √(36 + 4 + 16)

      = √56

      = 2√14

PR = √{(7 + 2)2 + (0 - 3)2 + (-1 – 5)2}

      = √{92 + (-3)2 +(-6)2}

      = √(81 + 9 + 36)

      = √126

      = 3√14

Here, PQ + QR = √14 + 2√14 = 3√14 = PR

Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.

Question 3:

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.

Now,

AB = √{(1 - 0)2 + (6 - 7)2 + (-6 + 10)2}

      = √{12 + (-1)2 + 42}

      = √(1 + 1 + 16)

      = √18

      = 3√2

BC = √{(4 - 1)2 + (9 - 6)2 + (-6 + 6)2}

      = √{32 + 32 + 0}

      = √(9 + 9)

      = √18

      = 3√2

CA = √{(0 - 4)2 + (7 - 9)2 + (-10 + 6)2}

      = √{(-4)2 + (-2)2 +(-4)2}

      = √(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.

AB = √{(-1 - 0)2 + (6 - 7)2 + (6 - 10)2}

      = √{(-1)2 + (-1)2 + (-4)2}

      = √(1 + 1 + 16)

      = √18

      = 3√2

BC = √{(-4 + 1)2 + (9 - 6)2 + (6 - 6)2}

      = √{(-3)2 + 32 + 0}

      = √(9 + 9)

      = √18

      = 3√2

CA = √{(0 + 4)2 + (7 - 9)2 + (10 - 6)2}

      = √{42 + (-2)2 + 42}

      = √(16 + 4 + 16)

      = √36

      = 6

Now, AB2 + BC2 = (3√2)2 + (3√2)2 = 18 + 18 = 36 = AC2

Therefore, by Pythagoras theorem, ABC is a right triangle.

Hence, 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.

AB = √{(1 + 1)2 + (-2 + 2)2 + (5 - 1)2}

      = √{22 + (-4)2 + (4)2}

      = √(4 + 16 + 16)

      = √36

      = 6

BC = √{(4 - 1)2 + (-7 + 2)2 + (8 - 5)2}

      = √{32 + (-5)2 + 32}

      = √(9 + 25 + 9)

      = √43

CD = √{(2 - 4)2 + (-3 + 7)2 + (4 - 8)2}

      = √{(-2)2 + 42 + (-4)2}

      = √(4 + 16 + 16)

      = √36

      = 6

DA = √{(-1 - 2)2 + (2 + 3)2 + (1 - 4)2}

      = √{(-3)2 + 52 + (-3)2}

      = √(9 + 25 + 9)

      = √43

Here, AB = CD = 6 and 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.

Question 4:

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) and B(3, 2, –1).

Now, PA = PB

=> PA2 = PB2

=> (x - 1)2 + (y - 2)2 + (z - 3)2 = (x - 3)2 + (y - 2)2 + (z + 1)2

=> x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 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

Question 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.

 

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

=> √{(x - 4)2 + y2 + z2} + √{(x + 4)2 + y2 + z2} = 10

=> √{(x - 4)2 + y2 + z2} = 10 - √{(x + 4)2 + y2 + z2}

On squaring both sides, we obtain

=> (x - 4)2 + y2 + z2 = 102 + (x + 4)2 + y2 + z2 - 20√{(x + 4)2 + y2 + z2}

=> x2 + 16 – 8x + y2 + z2 = 100 + x2 + 16 + 8x + y2 + z2 - 20√{(x + 4)2 + y2 + z2}

=> 20√{(x + 4)2 + y2 + z2} = 100 + 16x

=> 5√{(x + 4)2 + y2 + z2} = 25 + 4x

On squaring both sides again, we obtain

=> [5√{(x + 4)2 + y2 + z2}]2 = (25 + 4x)2

=> 25(x2 + 8x + 16 + y2 + z2) = 625 + 16x2 + 200x

=> 25x2 + 200x + 400 + 25y2 + 25z2 = 625 + 16x2 + 200x

=> 9x2 + 25y2 + 25z2 – 225 = 0

Thus, the required equation is 9x2 + 25y2 + 25z2 – 225 = 0

 

                                                                    Exercise 12.3

Question 1:

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(x1, y1, z1) and

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

{(mx2 + nx1)/(m + n), (my2 + ny1)/(m + n), (mz2 + nz1)/(m + n)}

Let R (x, y, z) be the point that divides the line segment joining points (–2, 3, 5) and (1, –4, 6)

internally in the ratio 2 : 3

x = {2 * 1 + 3 * (-2)}/(2 + 3) = (2 - 6)/5 = -4/5

y = {2 * (-4) + 3 * 3}/(2 + 3) = (-8 + 9)/5 = 1/5

z = (2 * 6 + 3 * 5)/(2 + 3) = (12 + 15)/5 = 27/5

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(x1, y1, z1) and

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

{(mx2 - nx1)/(m - n), (my2 - ny1)/(m - n), (mz2 - nz1)/(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

x = {2 * 1 - 3 * (-2)}/(2 - 3) = (2 + 6)/(-1) = -8

y = {2 * (-4) - 3 * 3}/(2 - 3) = (-8 - 9)/(-1) = 17

z = (2 * 6 - 3 * 5)/(2 - 3) = (12 - 15)/(-1) = 3

Thus, the coordinates of the required point is (–8, 17, 3).

 

 

Question 2:

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,

     (5, 4, -6) = {(9 * k + 3)/(k + 1), (8 * k + 2)/(k + 1), (-10 * k - 4)/(k + 1)}

=> 5 = (9k + 3)/(k + 1)

=> 9k + 3 = 5(k + 1)

=> 9k + 3 = 5k + 5

=> 9k – 5k = 5 – 3

=> 4k = 2

=> k = 2/4

=> k = 1/2

Thus, point Q divides PR in the ratio = 1/2 : 1 = 1 : 2

Question 3:

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

{(3 * k - 2)/(k + 1), (-5 * k + 4)/(k + 1), (8 * k + 7)/(k + 1)}

On the YZ plane, the x-coordinate of any point is zero.

  • * k - 2)/(k + 1) = 0

=> 3k – 2 = 0

=> k = 2/3

Thus, the YZ plane divides the line segment formed by joining the given points in the ratio

2/3 : 1 = 2 : 3

Question 4:

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

{(-1 * k + 2)/(k + 1), (2 * k - 3)/(k + 1), (1 * k + 4)/(k + 1)}

Now, we find the value of k at which point P coincides with point C.

By taking (-k + 2)/(k + 1) = 0, we get k = 2

For k = 2, the coordinates of point P are (0, 1/3, 2)

i.e., C (0, 1/3, 2) is a point that divides AB externally in the ratio 2 : 1 and is the same as point

  1. Hence, points A, B, and C are collinear.

Question 5:

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 + 2), (-16 * 1 + 2 * 2)/(1 + 2), (1 * 6 + 2 * (-6))/(1 + 2)} = (6, -4, -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 + 1), (-16 * 2 + 2 * 1)/(2 + 1), (2 * 6 + 1 * (-6))/(2 + 1)} = (8, -10, 2)

Thus, (6, –4, –2) and (8, –10, 2) are the points that trisect the line segment joining points  

P (4, 2, –6) and Q (10, –16, 6).

 

 

                                                  Miscellaneous Exercise on Chapter 12

Question 1:

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).

                           Class_11_Maths_Intro_To_3D_Geometry_Figure_3                     

We know that the diagonals of a parallelogram bisect each other.

Therefore, in parallelogram ABCD, AC and BD bisect each other.

So, Mid-point of AC = Mid-point of BD

       {(3 - 1)/2, (-1 + 1)/2, (2 + 2)/2} = {(x + 1)/2, (y + 2)/2, (z - 4)/2}

=> (1, 0, 2) = {(x + 1)/2, (y + 2)/2, (z - 4)/2}

=> (x + 1)/2 = 1, (y + 2)/2 = 0, (z - 4)/2 = 2

=> x = 1, y = –2, and z = 8

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

Question 2:

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.

     Class_11_Maths_Intro_To_3D_Geometry_Figure_2                                                       

Since AD is the median, D is the mid-point of BC.

So, coordinates of point D = {(0 + 6)/2, (4 + 0)/2, (0 + 0)/2} = (3, 2, 0)

AD = √{(0 - 3)2 + (0 - 2)2 + (6 - 0)2} = √(9 + 4 + 36) = √49 = 7

Since BE is the median, E is the mid-point of AC.

So, coordinates of point E = {(0 + 6)/2, (0 + 0)/2, (6 + 0)/2} = (3, 0, 3)

BE = √{(3 - 0)2 + (0 - 4)2 + (3 - 0)2} = √(9 + 16 + 9) = √34

Since CF is the median, EF is the mid-point of AB.

So, coordinates of point F = {(0 + 0)/2, (0 + 4)/2, (6 + 0)/2} = (0, 2, 3)

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

Thus, the lengths of the medians of ∆ABC are 7, √34 and 7.

Question 3:

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 find the values of a, b and c.

Answer:

It is known that the coordinates of the centroid of the triangle, whose vertices are (x1, y1, z1),

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

Therefore, coordinates of the centroid of ∆PQR

Class_11_Maths_Intro_To_3D_Geometry_Figure_1

= {(2a - 4 + 8)/3, (2 + 3b + 14)/3, (6 - 10 + 2c)/3}

= {(2a + 4)/3, (3b + 16)/3, (2c - 4)/3}

It is given that origin is the centroid of ∆PQR.

So, (0, 0, 0) = {(2a + 4)/3, (3b + 16)/3, (2c - 4)/3}

=> (2a + 4)/3 = 0, (3b + 16)/3 = 0, (2c - 4)/3 = 0

=> a = -2, b = -16/3, c = 2

Thus, the respective values of a, b, and c are -2, -16/3 and 2.

Question 4:

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 point P (3, –2, 5).

Given, AP = 5√2

=> AP2 = 50

=> (3 - 0)2 + (-2 - b)2 + (5 - 0)2 = 50

=> 9 + 4 + b2 + 4b + 25 = 50

=> b2 + 4b + 38 – 50 = 0

=> b2 + 4b – 12 = 0

=> (b + 6)(b - 2) = 0

=> b = -6, 2

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

Question 5:

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)/(k + 1), -3/(k + 1), (10k + 4)/(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

{(8 * k + 2)/(k + 1), (k * 0 - 3)/(k + 1), (10 * k + 4)/(k + 1)}]

= {(8k + 2)/(k + 1), -3/(k + 1), (10k + 4)/(k + 1)}]

It is given that the x-coordinate of point R is 4.

=> (8k + 2)/(k + 1) = 4

=> (8k + 2) = 4(k + 1)

=> 8k + 2 = 4k + 4

=> 8k – 4k = 4 – 2

=> 4k = 2

=> k = 1/2

Therefore, the coordinates of point R are

{4, -3/(1/2 + 1), (10 * 1/2 + 4)/(1/2 + 1)}] = (4, -2, 6)

Question 6:

If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively,

find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.

Answer:

The coordinates of points A and B are given as (3, 4, 5) and (–1, 3, –7) respectively.

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

Given, PA2 + PB2 = k2

=> {(x - 3)2 + (y - 4)2 + (z - 5)2} + (x + 1)2 + (y - 3)2 + (z + 7)2 = k2

=> {x2 + 9 – 6x + y2 + 16 – 8y + z2 + 25 – 10z} + {x2 + 1 + 2x + y2 + 9 – 6y + z2 + 49 + 14z} = k2

=> x2 – 6x + y2 – 8y + z2 – 10z + 50 + x2 + 2x + y2 – 6y + z2 + 14z + 59 = k2

=> 2x2 – 4x + 2y2 – 14y + 2z2 + 4z + 109 = k2

=> 2x2 + 2y2 + 2z2 – 4x – 14y + 4z = k2 – 190

=> 2(x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 190

=> x2 + y2 + z2 – 2x – 7y + 2z = (k2 – 190)/2

Thus, the required equation is: x2 + y2 + z2 – 2x – 7y + 2z = (k2 – 190)/2

 

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