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NCERT · Class 11 · Maths

Chapter 12: Introduction To 3 D Geometry Miscellaneous Exercise

Practice each question, then open the answer or worked solution.

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Questions

6 questions
Q1Written answer

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.

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Worked solution

Given:

ABCD is a parallelogram with vertices A (3, -1, 2), B (1, 2, -4), C (-1, 1, 2).

Where, x1 = 3, y1 = -1, z1 = 2

x2 = 1, y2 = 2, z2 = -4

x3 = -1, y3 = 1, z3 = 2

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 15

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

We also know that the diagonals of a parallelogram bisect each other, so the midpoints of AC and BD are equal, i.e., Midpoint of AC = Midpoint of BD ……….(1)

Now, by the Midpoint Formula, we know that the coordinates of the midpoint of the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) are [(x1+x2)/2, (y1+y2)/2, (z1+z2)/2]

So we have,

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 16

= (2/2, 0/2, 4/2)

= (1, 0, 2)

 

1 + x = 2, 2 + y = 0, -4 + z = 4

x = 1, y = -2, z = 8

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

Q2Written answer

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

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Worked solution

Given:

The vertices of the triangle are A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

x1 = 0, y1 = 0, z1 = 6

x2 = 0, y2 = 4, z2 = 0

x3 = 6, y3 = 0, z3 = 0

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 18

So, let the medians of this triangle be AD, BE and CF, corresponding to the vertices A, B and C, respectively.

D, E and F are the midpoints of the sides BC, AC and AB, respectively.

By the Midpoint Formula, we know that the coordinates of the mid-point of the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) are [(x1+x2)/2, (y1+y2)/2, (z1+z2)/2]

So we have,

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 19

By the Distance Formula, we know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 20

∴ The lengths of the medians of the given triangle are 7, √34 and 7.

Q3Written answer

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.

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Worked solution

ncert-solutions-class-11-maths-chapter-12-mis-q3

Given:

The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).

Where,

x1 = 2a, y1 = 2, z1 = 6;

x2 = -4, y2 = 3b, z2 = -10;

x3 = 8, y3 = 14, z3 = 2c

We know 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]

So, the coordinates of the centroid of the triangle PQR are

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 21

2a + 4 = 0, 3b + 16 = 0, 2c – 4 = 0

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

∴ The values of a, b and c are a = -2, b = -16/3, c = 2

Q4Written answer

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

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Worked solution

Let the point on the y-axis be A (0, y, 0).

Then, it is given that the distance between the points A (0, y, 0) and P (3, -2, 5) is 5√2.

Now, by using the distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

Distance of PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, the distance between the points A (0, y, 0) and P (3, -2, 5) is given by

Distance of AP = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

= √[(3-0)2 + (-2-y)2 + (5-0)2]

= √[32 + (-2-y)2 + 52]

= √[(-2-y)2 + 9 + 25]

5√2 = √[(-2-y)2 + 34]

Squaring on both sides, we get

(-2 -y)2 + 34 = 25 × 2

(-2 -y)2 = 50 – 34

4 + y2 + (2 × -2 × -y) = 16

y2 + 4y + 4 -16 = 0

y2 + 4y – 12 = 0

y2 + 6y – 2y – 12 = 0

y (y + 6) – 2 (y + 6) = 0

(y + 6) (y – 2) = 0

y = -6, y = 2

∴ The points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.

Q5Written answer

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
NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 22

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Worked solution

Given:

The coordinates of the points P (2, -3, 4) and Q (8, 0, 10).

x1 = 2, y1 = -3, z1 = 4

x2 = 8, y2 = 0, z2 = 10

Let the coordinates of the required point be (4, y, z).

So now, let the point R (4, y, z) divides the line segment joining the points P (2, -3, 4) and Q (8, 0, 10) in the ratio k: 1.

By using the Section Formula,

We know that the coordinates of the point R, which divides the line segment joining two points, P (x1, y1, z1) and Q (x2, y2, z2), internally in the ratio m: n, are given by

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 23

So, the coordinates of the point R are given by
NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 24

So, we have
NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 25

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 26

8k + 2 = 4 (k + 1)

8k + 2 = 4k + 4

8k – 4k = 4 – 2

4k = 2

k = 2/4

= 1/2

Now, let us substitute the values, and we get

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 27

= 6

∴ The coordinates of the required point are (4, -2, 6).

Q6Written answer

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.

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Worked solution

Given:

Points A (3, 4, 5) and B (-1, 3, -7)

x1 = 3, y1 = 4, z1 = 5

x2 = -1, y2 = 3, z2 = -7

PA2 + PB2 = k2 ……….(1)

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

Now by using the distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 28

So,

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 29

And

NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry image - 30

Now, substituting these values in (1), we have

22222222222222

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

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

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

2 (x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 109

(x2 + y2 + z2 – 2x – 7y + 2z) = (k2 – 109)/2

Hence, the required equation is (x2 + y2 + z2 – 2x – 7y + 2z) = (k2 – 109)/2

Access Other Exercise Solutions of Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry

Exercise 12.1 Solutions 4 Questions

Exercise 12.2 Solutions 5 Questions

Exercise 12.3 Solutions 5 Questions

Also explore – NCERT Class 11 Solutions