NCERT Solutions for Class 11 Maths Chapter 12 Introduction To Three Dimensional Geometry Miscellaneous Exercise

Access the NCERT Solutions for Class 11 Maths Chapter 12 Miscellaneous Exercise here. Chapter 12 Introduction to Three Dimensional Geometry of Class 11 Maths is categorized under the term – II CBSE Syllabus for 2021-22. Students can practice these questions and improve their reasoning skills. The Class 11 Maths NCERT Solutions can be considered as the best study material for improving the term – II exam score.

The Miscellaneous Exercise of NCERT Solutions for Class 11 Maths Chapter 12- Introduction to Three Dimensional Geometry is based on the topics listed above.

1. Coordinate Axes and Coordinate Planes in Three Dimensional Space
2. Coordinates of a Point in Space
3. Distance between Two Points
4. Section Formula

Download PDF of NCERT Solutions for Class 11 Maths Chapter 12- Introduction to Three Dimensional Geometry Miscellaneous Exercise

Solutions for Class 11 Maths Chapter 12 – Miscellaneous Exercise

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.

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

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 mid points of AC and BD are equal, i.e. Midpoint of AC = Midpoint of BD ……….(1)

Now, by 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,

= (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 is D (1, -2, 8).

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

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

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 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,

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

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

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.

Solution:

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

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

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).

Solution:

Let the point on 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 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 the 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.

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

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 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 is given by:

So, the coordinates of the point R are given by

So, we have

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, we get

= 6

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

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.

Solution:

Given:

The 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 distance formula,

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

So,

And

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

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