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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

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 categorised under the CBSE Syllabus for the academic year 2023-24. Students can practise these questions and improve their reasoning skills. The Class 11 Maths NCERT Solutions can be considered the best study materials for improving the board 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 below.

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

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

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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 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,

= (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).

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 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,

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

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.

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 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

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, and 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:

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

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

Also explore – NCERT Class 11 Solutions