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Introduction to Three Dimensional Geometry Class 11 Notes CBSE Maths Chapter 11 [Free PDF Download]

Last updated date: 20th Sep 2024
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Revision Notes for CBSE Class 11 Maths Chapter 11 (Introduction to Three Dimensional Geometry) - Free PDF Download

Introduction to Three Dimensional Geometry is an important chapter that discusses various topics like direction cosine and direction ratios of a line that joins two points. Students will learn about all of these topics and more in this Class 11 Maths Chapter 11 Revision Notes by Vedantu. In these Class 11 Maths Introduction To Three Dimensional Geometry Notes PDF, which is also available for download, we will also talk about the equation of lines and planes in spaces. We will end these revision notes of Class 11 Maths Chapter 11 by solving some questions.

Also, check CBSE Class 11 Maths revision notes for all chapters:

Class 11 Study Materials Important Links

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Introduction to Three Dimensional Geometry Class 11 Notes Maths - Basic Subjective Questions

Section–A (1 Mark Questions)

1. The locus of a point for which x = 0 is _________.

Ans. We know, any point lying in the YZ-plane is of the type (0, y, z) i.e. locus of x = 0 is yz-plane.

2. L is the foot of the perpendicular drawn from a point P (3, 4, 5) on the xy-plane. What are the coordinates of point L.

Ans. Q L is the foot of perpendicular drawn from P(3, 4, 5) on the xy-plane.

$\therefore$ L lies in the xy-plane.

i.e. z-coordinate of L = 0

i.e. coordinates of L is (3, 4, 0).

3. What is the distance from origin to point (a, b, c).

Ans. \text { Distance } \begin{aligned}= & \sqrt{(a-0)^2+(b-0)^2+(c-0)^2} \\& =\sqrt{a^2+b^2+c^2}\end{aligned}

4. Find Mid-point of the line segment joining the points (1, 3, 5) and (2, 4, 3).

Ans. Mid-point of the line segment joining the points $(1,3,5)$ and $(2,4,3)$ is given by

$\left(\frac{1+2}{2}, \frac{3+4}{2}, \frac{5+3}{2}\right)=\left(\frac{3}{2}, \frac{7}{2},4\right)$

5. If a point P lies in yz-plane, then the coordinates of a point on yz-plane is of the form _____.

Ans. If a point $P$ lies in $y z-$ plane, then the coordinates of a point on yz-plane is of the form $(\mathbf{0 ,}, \mathbf{y}, \mathbf{z})$

Section–B (2 Marks Questions)

6. If the distance between the points (a, 0, 1) and (0, 1, 2) is $\sqrt{27}$, then find the value of a.

Ans. Distance between the points (a, 0,1$)$ and $(0,1,2)=\sqrt{27}$

\begin{aligned}& \Rightarrow \sqrt{(0-a)^2+(1-0)^2+(2-1)^2}=\sqrt{27} \\& \Rightarrow a^2+1+1=27 \Rightarrow a^2=25 \\& \therefore a= \pm 5\end{aligned}

7. If a parallelepiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then what is the length of diagonal of the parallelepiped.

Ans. Length of diagonal

$=\sqrt{(3-5)^2+(6-8)^2+(8-10)^2}$ $=\sqrt{(-2)^2+(-2)^2+(-2)^2}$

$=\sqrt{4+4+4}$

$=\sqrt{12}$

$=2 \sqrt{3}$

8. Locate the following points in respective octants:

(i) (1, –1, 3) (ii) (–1, 2, 4)

(iii) (–2, –4, –7) (iv) (–4, 2, –5)

Ans. (i) $(1,-1,3)$ lies in the IV octant (XOY'Z).

(ii) $(-1,2,4)$ lies in the II octant (X'OYZ). (iii) $(-2,-4,-7)$ lies in the VII octant (X'OY'Z').

(iv) $(-4,2,-5)$ lies in the VI octant (X'OYZ').

9. Find the distance between the points P (1, –3, 4) and Q (-4, 1, 2).

Ans. The distance $\mathrm{PQ}$ between the points $\mathrm{P}$ (1, $-3,4)$ and $Q(-4,1,2)$ is

\begin{aligned}P Q & =\sqrt{(-4-1)^2+(1+3)^2+(2-4)^2} \\& =\sqrt{25+16+4} \\& =\sqrt{45}=3 \sqrt{5} \text { units }\end{aligned}

10. Find the distance of the point $(x,y\sqrt{1-x^{2}-y^{2}})$ from the origin.

Ans. Distance of the point $\left(x, y \sqrt{1-x^2-y^2}\right)$

\begin{aligned}& \text { from }(0,0,0) \text { is } \\& \sqrt{(x-0)^2+(y-0)^2+\left(\sqrt{1-x^2-y^2-0}\right)^2} \\& =\sqrt{x^2+y^2+1-x^2-y^2} \\& =1 \\&\end{aligned}

11. Find the centroid of a triangle, the mid-point of whose sides are D (1, 2, –3), E (3, 0, 1) and F (–1, 1, –4).

Ans. $\because$ centroid of a triangle is also the centroid of that triangle whose vertices are the mid-points of the sides of the given triangle.

$\therefore G=\left(\frac{1+3-1}{3}, \frac{2+0+1}{3}, \frac{-3+1-4}{3}\right)$

$=(1,1,-2)$

$=\sqrt{4+9+16}=\sqrt{29}$

$|B C|=\sqrt{(3+1)^2+(5+1)^2+(7+1)^2}$

$=\sqrt{16+36+64}=2 \sqrt{29}$

$|A C|=\sqrt{(3-1)^2+(5-2)^2+(7-3)^2}$

$=\sqrt{4+9+16}=\sqrt{29}$

$\therefore|B C|=|A B|+|A C|$

12. Show that the points A(1,2,3) , B(-1,-1,-1) and C(3,5,7) are collinear.

Ans. $|A B|=\sqrt{(-1-1)^2+(-1-2)^2+(-1-3)^2}$

$\sqrt{4+9+16} = \sqrt{29}$

$\left |BC \right | = \sqrt{(3+1)+(5+1)+(7+1)} = \sqrt{4+9+16} = \sqrt{29}$

$= \sqrt{16+36+64} = 2\sqrt{29}$

$\left | AC \right | = \sqrt{(3-1)+(5-2)+(7-3)} = \sqrt{8} = 2$

$\therefore \left | BC \right |=\left | AC \right |+\left | AC \right |$

Hence, the points are collinear.

13. Find the coordinates of the point which divides the join of P(2,-1,4) and Q(4,3,2) in the ratio 2 : 5 externally.

Ans. Let point $R(x, y, z)$ be the required point.

For external division,

\begin{aligned}& x=\frac{2 \times 4-5 \times2}{2-5}=\frac{8-10}{-3}=\frac{-2}{-3}=\frac{2}{3} \\& y=\frac{2 \times 3-5\times-1}{2-5}=\frac{6+5}{-3}=\frac{11}{-3} \\& z=\frac{2 \times 2-5 \times 4}{2-5}=\frac{4-20}{-3}=\frac{-16}{-3}=\frac{16}{3} \\& \therefore \text { Required point is } R\left(\frac{2}{3}, \frac{-11}{3}, \frac{16}{3}\right) .\end{aligned}

PDF Summary - Class 11 Maths Introduction to Three Dimensional Geometry (Chapter 11)

1. Coordinate Axes

i) Three mutually perpendicular lines are known as Coordinate Axes, i.e., OX, OY and OZ.

2. Coordinate Planes

i) The planes formed by the Coordinate axes are known as Coordinate planes.

ii) The x and y axes form XY plane.

iii) The y and z axes form YZ plane.

iv) The x and z axes form XZ plane.

3. Coordinates of a Point in Space

i) The coordinates of origin, i.e., point O is $(0,0,0)$.

ii) The coordinates of point A is $(x,0,0)$, i.e., this is the coordinates of any point on x-axis.

iii) The coordinates of point B is $(0,0,z)$, i.e., this is the coordinates of any point on z-axis.

iv) The coordinates of point C is $(0,y,0)$, i.e., this is the coordinates of any point on y-axis.

v) Similarly, the coordinates of any point on XY plane is given by $(x,y,0)$, YZ plane is given by $(0,y,z)$ and the XZ plane is given by $(x,0,z)$.

vi) A 3 dimensional coordinate system is divided into 8 section known as Octant. The sign convention of the octants are as follows:

 I II III IV V VI VII VIII x + – – + + – – + y + + – – + + – – z + + + + – – – –

Ex:  The octant of point $(-2,-4,7)$ will be III. (compare from above table).

4. Distance between Two Points

i) Consider two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, as shown in the figure below:

Now from the figure, $\Delta ACB$ is a right angled triangle, so by Pythagoras theorem, we have

$A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}$ ……(i)

Similarly, $\Delta BCH$ is a right angled triangle, so by Pythagoras theorem, we have

$B{{C}^{2}}=C{{H}^{2}}+B{{H}^{2}}$ ……(ii)

From equation (i) and (ii), we have

$A{{B}^{2}}=A{{C}^{2}}+C{{H}^{2}}+B{{H}^{2}}$

From the figure, the coordinates of $BH=x_2-x_1, CH=y_2-y_1,AC=z_2-z_1$.

So, the above equation can be written as,

$A{{B}^{2}}={{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}$

$\Rightarrow AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$

This is the distance between any two points in the three dimensional space.

ii) The distance of any point $(x_1,y_1,z_1)$ from origin is given by:

$\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}+{{z}_{1}}^{2}}$

iii) Rule: The sum of two collinear points is equal to the third collinear point. This is true only in case of three collinear points.

For Example: If A, B, C are three collinear points, then $AB+BC=AC$.

Note: Using the distance formula we can tell whether three points are collinear or not.

5. Section Formula

i) We can find the ratio in which a point divides a line internally using Section Formula.

ii) If AB is a line and C is dividing the line segment internally, then we can find the ratio $m:n$ in which the line AB is divided using the section formula.

iii) Consider two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, as shown in the figure below. Let point C(x,y,z) divide the line segment AB in the ratio $m:n$ internally.

iv) Now draw perpendiculars from A, B and C on the XY plane, such that $AH||CD||BE$. Also draw a line HG through the point C, which is parallel to FE.

v) Now from the figure, quadrilaterals CDFH and DEGC are parallelograms.

vi) Consider $\Delta ACH$ and $\Delta BCG$

These are right angled triangle and have vertically opposite angle, so these two triangles are similar.

Therefore, we have

$\dfrac{m}{n}=\dfrac{AC}{BC}$

$\Rightarrow \dfrac{m}{n}=\dfrac{AH}{BG}$

$\Rightarrow \dfrac{m}{n}=\dfrac{AF-HF}{GE-BE}$

$\Rightarrow \dfrac{m}{n}=\dfrac{AF-CD}{CD-BE}$

Now writing their corresponding coordinates, we get

$\dfrac{m}{n}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-z}$

$\Rightarrow m{{z}_{2}}-mz=nz-n{{z}_{1}}$

$\Rightarrow z(m+n)=m{{z}_{2}}+n{{z}_{1}}$

$\Rightarrow z=\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n}$

vii) Similarly, we can write for other coordinates as,

$x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$

viii) So, the required coordinates of the point dividing internally will be,

$\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)$

ix) Similarly, the required coordinates of the point dividing externally will be,

$\left( \dfrac{m{{x}_{2}}-n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m+n},\dfrac{m{{z}_{2}}-n{{z}_{1}}}{m+n} \right)$

x) Now if the dividing point is a midpoint then $m=1,n=1$, so the coordinates of the point will be,

$\left( \dfrac{{{x}_{2}}+{{x}_{1}}}{2},\dfrac{{{y}_{2}}+{{y}_{1}}}{2},\dfrac{{{z}_{2}}+{{z}_{1}}}{2} \right)$

Maths Chapter 11 Introduction to Three Dimensional Geometry Notes

Understanding The Basics of Three Dimensional Geometry

According to experts, this topic was introduced in the subject of Mathematics to help students understand the different types of figures and shapes. For example, there are various objects like a bed, chair, table, and kitchen utensils are all 3D geometrical shapes.

The Coordinate System in Three Dimensional Geometry

In this section of Class 11 Revision Notes Introduction To Three Dimensional Geometry, students will be able to learn about the coordinate system in three-dimensional geometry. In three dimensional geometry, a coordinate system can be defined as the process of finding the location or position of a point on a coordinate plane. Students can learn more about coordinate systems by going through Class 11 Revision Notes Ch 11 Maths Class 11 Introduction To Three Dimensional Geometry Notes.

Rectangular Coordinate System

A rectangular coordinate system can be described as three lines that are perpendicular to one another. These lines pass through a common point. One of the lines is known as the x-axis, the second line is the y-axis, and the last line is the z-axis. O is the centre or the observer with respect to the positions of all three lines.

It should be noted in these notes of Class 11 Chapter 11 that it is possible for an individual to measure how much distance has been covered by a 3D object by looking at the rectangular coordinate system.

For example, if there is an object with the position coordinates of (3, -4, 5). This means that the object has moved 3 units along the positive x-axis, four unis along the negative y-axis, and five units along the positive z-axis.

Another concept that is related to the Introduction To Three Dimensional Geometry Class 11 Notes is how to calculate distance from the origin. Students should know how to calculate this distance with the help of a formula. We will mention the formula later.

Keeping this image in mind, we can say that the distance from the origin can be calculated by using the Pythagorean theorem.

By using the theorem, the distance of P (x, y, z) from the origin (0, 0, 0) is $\sqrt{x^{2}+y^{2}+z^{2}}$

The formula for calculating the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

And what if a student has to divide a line that is joined by two points? In that case, one should begin by assuming that the two points are P (x1, y1, z1) and Q (x2, y2, z2). If R divides the line PQ into a ratio, then the coordinates of R are:

R = (mx2 + nx1 / m + n, my2 + ny1 / m + n, mz2 + nz1 / m + n)

Projection in 3D space is also a vital topic that students will learn in Class 11 Maths Revision Notes Chapter 11. Let’s assume that we have a line segment. PQ is the projection of that line. Hence, the project is equal to AB cos θ. In this formula, θ is the angle that is between AB and PQ or CD.

Direction Ratios Of A Line And Direction Cosine

In this section of the Ch 11 Class 11 Maths Revision Notes, students will be able to learn about the formulas for direction cosine and direction ratios of a line.

For this, let’s consider that a line L is passing through the origin. This results in the angle of α, β, and γ. This is done with x, y, and z-axes, respectively. Hence, the cosine of these angles is in the direction cosine of the directed line L.

It should also be noted that any three numbers that are proportional to the direction cosines of a line are known as the direction ratios of the line. Hence, the direction cosine of the line L will be l, m, and n. Further, the direction ration a, b, and c will be a = λl, b = λm, c = λn. This is true for non-zero λ ∈ R.

This information can also be depicted as:

L / a = m / b = n / c = k

Also, the value of direction cosine is:

L = ± a / a2 + b2 + c2, m = ± b / a2 + b2 + c2, and n = c / a2 + b2 + c2

You should also keep in mind that if the line is space and does not pass through any origin, then one should draw a line through the origin. This line should be parallel to the given line. This is done while finding the direction of cosines.

After that, you can take one of the directed lines from the origin. This can be done for finding the direction of cosines as two parallel lines have the same set of direction cosines.

There are also some other important formulas related to three-dimensional geometry. Students have to answer questions related to these formulas. These formulas are discussed below.

• The Relation Between The Direction Cosines Of A Line

If there is a line RS that has direction cosines of l, m, and n, then the relation between the line RS and the direction cosine is given by:

L2 + m2 + n2 = 1

• Direction Cosines of a Line Passing Through Two Points

The direction cosines of the line segment that joins the two points P (x1, y1, z1) and Q (x2, y2, z2) can be given as:

x2 - x1 / PQ, y2 - y1 / PQ, z2 - z1 / PQ

Here, PQ = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

Fun Facts about Three Dimensional Space

Did you know that the informal meaning of the term dimension is a geometric setting that has three values or parameters? In mathematics, three-dimensional spaces are also known as Euclidean spaces. These spaces are commonly represented by the symbol R.

Three-dimensional space also serves as a model of the physical universe that has three parameters. However, in this equation, time is not considered. It can also be labelled by various combinations, including depth, height, width, and length.

FAQs on Introduction to Three Dimensional Geometry Class 11 Notes CBSE Maths Chapter 11 [Free PDF Download]

1. Find the direction cosines of a line that makes angles measuring 90 degrees, 135 degrees, and 45 degrees with the x, y, and z axes, respectively.

Let’s assume that the direction cosines of the line are l, m, and n.

In this question, the value of α = 90 degrees, β = 135 degrees, and γ = 45 degrees

We also know that l = cos α, m = cos β, and n = cos γ

This means that the directions of cosine are:

L = cos 90 = 0

M = cos 135 = cos (180 - 45) = -cos 45 = -1 / √2

N = cos 45 = 1 / √2

Hence, the directions cosines of the line are 0, -1 / √2, and 1 / √2

2. Prove that the points (2, 3, 4), (-1, -2, 1), and (5, 8, 7) are collinear

The lines can also be collinear if the direction ratios of the two line segments are proportional to one another.

This means that if assume that the given points are: A (2, 3, 4), B (-1, -2, 1), and C (5, 8, 7), then the direction ratio of the line that is joining the points A (2, 3, 4) and B (-1, -2, 1) can be represented by: ((-1 -2), (-2 -3), (1 - 4)) = (-3, -5, -3)

Here, a1 = -3, b1 = -5, and c1 = -3

The direction ratio of the line that is joining the points B (-1, -2, 1) and C (5 8, 7) is:

[(5 - (-1), (8 - (-2)), (7 - 1)] = (6, 10, 6)

Here, a2 = 6, b2 = 10, and c2 = 6

From these equations it should be clear that the direction ratios of AB and BC are of the same proportions. Hence,

A2 / a1 = 6 / -3 = -2

B2 / b1 = 10 / -5 = -2

C2 / c1 = 6 / -3 = -2

This proves that the points A, B, and C are collinear.

3. From where can I download revision notes of Chapter 11 of Class 11 Maths?

Students can download the revision notes of Chapter 11 of Class 11 Maths from Vedantu. They can follow the given steps for the same:

• Click on the given link CBSE Class 11 Chapter 11

• As the link will open, you will find that the website of Vedantu will appear on your device screen.

• On the page of Vedantu, you will find the notes of Chapter 11 of Class Maths.

• The same page will provide you with the option of “Download PDF”.

4. What should I do to prepare for Chapter 11 of Class 11 Maths?

If the students will follow the provided tips then they can easily prepare Chapter 11 of Class 11 Maths:

• You should study this chapter from the NCERT book as the content is written in very simple language.

• Solve each example and exercise to comprehend the concepts easily.

• Clarify your doubts as soon as possible. If you'll not clear your doubts then there will be chances that you do not understand the chapter.

• Refer to guidebooks or reference books to solve various types of questions and to know about the exam pattern.

5. What do you understand by the term rectangular coordinate system?

The term rectangular coordinate system is defined as the three lines which are perpendicular to each other. These lines have a common intersecting point. The names of the lines are as follows:

• X-axis

• Y-axis

• Z-axis

The central point is written as “O”.

Using this system one can find the distance of the object kept in the 3D plane or space.

For example, position coordinates (4, 5, 6) means that the object has moved 4 units along the positive x-axis, 5 units along the positive y-axis and 6 units along the positive z-axis.

If you want to understand these concepts more accurately then go through the given link CBSE Class 11 Chapter 11. This link will redirect you to the official website of Vedantu where you can access the content related to Chapter 11 for free. Additionally, you can also download its PDF if you want to study offline.

6. Explain the coordinates of the point in space.

The rectangular coordinate system helps in locating the position of a point. These coordinates are measured at perpendicular distances from the three planes. These planes are:

• XY Plane

• YZ Plane

• ZX Plane

The coordinates measured for the point are:

• (x, 0, 0) is measured from the x-axis.

• (0, y, 0) is measured from the y-axis.

• (0, 0, z) is measured from the z-axis

• Coordinates measured from the XY plane are (x, y, 0)

• Coordinates measured from the YZ plane are (0, y, z)

• Coordinates measured from the ZX plane are (x, 0, z)

7. What study materials should be considered by students to prepare Chapter 11 of Class 11 Maths?

For the preparation of exams, students do not understand how and from where they have to study. For a few students, Chapter 11 of Class 11 Maths is very difficult to prepare. Thus they are advised to study from the Maths NCERT book. This book contains lots of questions which will help you to understand the chapter if you solve them. Also, the language of this book is easy to comprehend. Students can also use other reference books for a better understanding of the concepts.