Question

NCERT Solutions Class 11 Maths Chapter 12 Introduction to Three–dimensional Geometry covers the important topics of three-dimensional geometry. Our NCERT Solutions is made with utmost concern to make it easier for students to grasp all thought concepts.

In NCERT Solutions Class 11 we discussed all the important concepts such as:

• Three–dimensional Geometry – three-dimensional geometry deal with the shapes in 3D space and involves three coordinates namely the x-axis, y-axis, and z-axis. In three-dimensional geometry, shapes are in space and there is a total of three parameters needed for their representation and defining their location in the space.

• Coordinate axes and coordinate planes in three dimensions – in a plane we have two mutually perpendicular lines intersecting each other at an angle of 90 degrees we use these two systems of lines to determine the location of any given point. We call these two lines the coordinate axis of the plane and the plane is commonly known as the Cartesian plane. In real life we do not have a two-dimensional system rather we have three-dimensional space and perpendicular lines are intersecting each other. The location of any point is represented using x, y and z. these are called the coordinates concerning the three-dimensional space.

• Coordinates of a point – the pair of numbers used to define the exact location of a point in a two-dimensional plane. The coordinate plane has two axes that are at right angles to each other and these two axes are called the x and y-axis. The coordinate of any point helps us determine how far the point is located from the origin. X-coordinate is called abscissa and y-coordinate is called ordinate.

• Distance between two points – the distance between two points is defined as the length of the line segment joining them. The distance between two points is always positive. The line segments which have equal length are called congruent line segments.

• Section Formula – section formula helps us find out the coordinates of the point which divides the given line in any particular ratio. A line can be divided in two ways, internally and externally.

Our NCERT Solutions Class 11 Maths has solutions to all kinds of queries systematically. Students must refer to our solution for securing good marks in their exams.

Answer 1

NCERT Solutions Class 11 Maths Chapter 12 Introduction to Three–dimensional Geometry

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.

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.

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.

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.

5. 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 b

\(PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_1-z_2)^2}\)

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

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

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

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

6. 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.

Here, PQ + QR = \(\sqrt{14}+2\sqrt{14}=3\sqrt{14}\) = PR 

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

7. 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.

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.

Now, AB² + BC² = \((3\sqrt2)^2+(3\sqrt2)^2\) = 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.

Here, AB = CD = 6, BC = AD = \(\sqrt{43}\)

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.

8. 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). Accordingly, PA = PB

⇒ 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.

9. 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.

On squaring both sides, we obtain

On squaring both sides again, we obtain 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.

Answer 2

10. 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

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

Thus, the coordinates of the required point are \(\left(-\frac45,\frac15,\frac{27}5\right).\)

(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

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

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

11. 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,

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

12. 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

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

Thus, the YZ plane divides the line segment formed by joining the given points in the ratio 2:3.

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

Answer:

The given points are A (2, –3, 4), B (–1, 2, 1), and (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 

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

By taking \(\frac{-k+2}{k+1}=0\), we obtain 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 P. 

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

14. 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

Point B divides PQ in the ratio 2:1. Therefore, by section formula, the coordinates of point B are given by

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).

15. 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).

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

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

∴ Mid-point of AC = Mid-point of BD

⇒ x = 1, y = –2, and z = 8 

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

16. 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.

Since AD is the median, D is the mid-point of B

∴Coordinates of point D =

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

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

Thus, the lengths of the medians of ∆ABC are 7, \(\sqrt{34}\) and 7. 

17. 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

Therefore, coordinates of the centroid of ∆PQR

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

Thus, the respective values of a, b, and c are -2, \(-\frac{16}3\) and 2.

18. 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). 

Accordingly, AP = 5√2

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

Answer 3

19. A point R with x-coordinate 4 lies on the line segment joining the pointsP (2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R. 

[Suppose R divides PQ in the ratio k: 1. The coordinates of the point R are given by 

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

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

Therefore, the coordinates of point R are 

20. 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). 

On using distance formula, we obtain

Now, if PA² + PB² = k² , then

Thus, the required equation is